On Inner Products & Projections, every bounded functional on a Hilbert space turned out to be $x \mapsto \langle x, y \rangle$ for one vector $y$. Extending a functional from a subspace was easy there: you just kept the same $y$.
In a Hilbert space, functionals are cheap. There's one for every vector, and the inner product hands them to you. A general normed space has no inner product, and it isn't even obvious that it has any bounded linear functionals besides $0$. This page shows that it always has plenty. The tool is the Hahn–Banach theorem: a bounded functional on a subspace can always be extended to the whole space without making its norm any bigger.
Then we cash it in. Hahn–Banach separates points from closed subspaces, it builds the second dual, and, in a surprise ending, it produces the Poisson integral: a formula that recovers a function inside the disk from its values on the boundary circle. Along the way we'll see that the extension usually isn't unique, and then find one place where it is.
The proof of Hahn–Banach works one real dimension at a time, using inequalities like $\lambda(x) \le \alpha$. Those only make sense for real numbers. So before anything else, we need to know that a complex functional is secretly a real one. Here's the trick: a complex linear functional is completely determined by its real part.
Let $V$ be a complex vector space.
Proof. (a) Real parts add, and real scalars pass through them, so $u$ is real linear. For any complex number $w$, $\operatorname{Re}(iw) = -\operatorname{Im} w$. With $w = f(x)$ and complex linearity, $\operatorname{Im} f(x) = -\operatorname{Re}(i f(x)) = -\operatorname{Re} f(ix) = -u(ix)$. So $f(x) = u(x) + i \operatorname{Im} f(x) = u(x) - i\,u(ix)$.
(b) Since $u$ is real linear, so is $f$. For the factor $i$, $$f(ix) = u(ix) - i\,u(-x) = u(ix) + i\,u(x) = i\big(u(x) - i\,u(ix)\big) = i f(x).$$ A complex scalar is $a + bi$ with $a, b$ real, so $f((a + bi)x) = a f(x) + b f(ix) = (a + bi) f(x)$.
(c) $|u(x)| = |\operatorname{Re} f(x)| \le |f(x)|$, so $\|u\| \le \|f\|$. Conversely, let $\|x\| \le 1$ with $f(x) \ne 0$, and let $\alpha = |f(x)| / f(x)$, so $|\alpha| = 1$. Then $f(\alpha x) = \alpha f(x) = |f(x)|$ is real, so it equals its real part: $$|f(x)| = u(\alpha x) \le \|u\| \, \|\alpha x\| \le \|u\|.$$ Taking the sup over such $x$ gives $\|f\| \le \|u\|$. $\blacksquare$
Part (c) is the one Hahn–Banach really needs. Throwing away the imaginary part costs nothing in size, because you can always rotate $x$ until $f(x)$ lands on the positive real axis. Here's the formula at work on one functional, $f(z) = (2 + i)z$ on $\mathbb{C}$.
The imaginary part never had an original thought. It's just the real part evaluated at $ix$, with a minus sign.
So $u$ knows everything. Run the recipe once by hand, in two variables.
Let $f\colon \mathbb{C}^2 \to \mathbb{C}$ be complex linear with $\operatorname{Re} f(z, w) = 3\operatorname{Re} z - 2\operatorname{Im} z + \operatorname{Im} w$. Find $f$.
Call the real part $u$. By linearity, $f(z, w) = f(1, 0) z + f(0, 1) w$, so we need two numbers, and the proposition gives each one from two values of $u$.
$f(1, 0) = u(1, 0) - i\,u(i, 0) = 3 - i(-2) = 3 + 2i$, since $(i, 0)$ has $\operatorname{Re} z = 0$ and $\operatorname{Im} z = 1$.
$f(0, 1) = u(0, 1) - i\,u(0, i) = 0 - i(1) = -i$.
So $f(z, w) = (3 + 2i)z - iw$. Check: $\operatorname{Re}\big((3 + 2i)(x + iy)\big) = 3x - 2y$ and $\operatorname{Re}\big(-i(s + it)\big) = t$, which is $3\operatorname{Re} z - 2\operatorname{Im} z + \operatorname{Im} w$.
Notice where the sign flipped: the $\operatorname{Im} z$ coefficient of $u$ was $-2$, and the imaginary part of $f(1, 0)$ came out $+2$. That flip is the whole trap.
The formula is $f(x) = u(x) - i\,u(ix)$, with a minus. If you remember it with a plus, you get the complex conjugate coefficients, and $\overline{3 + 2i} = 3 - 2i$ looks just as believable. Check one value: $\operatorname{Re} f$ of your answer has to give back $u$.
Your turn: recover the functional from its real part.
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Here is the problem we want to solve. A functional lives on a subspace, and we want it on the whole space, without paying for it in norm.
Let $M$ be a subspace of a normed space $X$ and $\lambda\colon M \to \mathbb{C}$ a bounded linear functional. A linear functional $\Lambda$ on $X$ is a norm-preserving extension of $\lambda$ if $\Lambda|_M = \lambda$ and $\|\Lambda\| = \|\lambda\|$.
Extending at all is easy: pick a basis and send the new vectors anywhere. The hard part is the norm. Every extension has $\|\Lambda\| \ge \|\lambda\|$, since it sees all the vectors $\lambda$ sees and more, so the real question is whether we can avoid going up. Start small: add one dimension, in a real space.
Let $X$ be a real normed space, $M$ a subspace, and $\lambda\colon M \to \mathbb{R}$ linear with $|\lambda(m)| \le \|m\|$ for all $m \in M$. Let $x_0 \notin M$ and $N = \{m + c x_0 : m \in M, \ c \in \mathbb{R}\}$. For $\alpha \in \mathbb{R}$, the linear map $f(m + c x_0) = \lambda(m) + c\alpha$ on $N$ satisfies $|f(y)| \le \|y\|$ for all $y \in N$ if and only if $$\sup_{m \in M} \big(\lambda(m) - \|m - x_0\|\big) \;\le\; \alpha \;\le\; \inf_{m \in M} \big(\lambda(m) + \|m - x_0\|\big),$$ and such an $\alpha$ exists.
Proof. Since $x_0 \notin M$, each $y \in N$ is $m + cx_0$ for exactly one pair $(m, c)$, so $f$ is well defined and linear.
The interval is not empty. For $x, y \in M$, $$\lambda(x) - \lambda(y) = \lambda(x - y) \le \|x - y\| \le \|x - x_0\| + \|x_0 - y\|,$$ so $\lambda(x) - \|x - x_0\| \le \lambda(y) + \|y - x_0\|$. Every number on the left is at most every number on the right, so the sup over $x$ is at most the inf over $y$, and both are finite. Any $\alpha$ between them works below.
The bound holds. The two inequalities on $\alpha$ say $|\lambda(m) - \alpha| \le \|m - x_0\|$ for every $m \in M$. Let $y = m + cx_0$. If $c = 0$, $|f(y)| = |\lambda(m)| \le \|m\|$. If $c \ne 0$, then $-m/c \in M$, and $$|f(y)| = |\lambda(m) + c\alpha| = |c| \, \big|\lambda(-m/c) - \alpha\big| \le |c| \, \big\|{-m/c} - x_0\big\| = \|m + cx_0\|.$$
Only that interval works. If $|f(y)| \le \|y\|$ on $N$, then for $m \in M$, $\alpha - \lambda(m) = f(x_0 - m) \le \|x_0 - m\|$ and $\lambda(m) - \alpha = f(m - x_0) \le \|m - x_0\|$, which are the two inequalities. $\blacksquare$
The lemma doesn't just say "some $\alpha$ works". It names all of them: every number between a sup and an inf. Watch those two numbers form in the plane with the norm $\|(x, y)\|_1 = |x| + |y|$, with $M$ the $x$-axis, $\lambda(t, 0) = t$, and $x_0 = (0, 1)$.
With the $\ell^1$ norm the gap between sup and inf is a whole interval. With the Euclidean norm it pinches down to a single number. Let's do both by hand.
On $\mathbb{R}^2$, let $M$ be the $x$-axis and $\lambda(t, 0) = t$. Find every norm-preserving extension of $\lambda$ (a) for $\|(x, y)\|_1 = |x| + |y|$ and (b) for $\|(x, y)\|_2 = \sqrt{x^2 + y^2}$.
In both norms $\|(t, 0)\| = |t|$, so $\|\lambda\| = 1$. An extension is $\Lambda(x, y) = x + \alpha y$ with $\alpha = \Lambda(0, 1)$, so we need the allowed $\alpha$ for $x_0 = (0, 1)$.
(a) $\|(t, 0) - (0, 1)\|_1 = |t| + 1$. The lower end is $\sup_t (t - |t| - 1) = -1$ (any $t \ge 0$), and the upper end is $\inf_t (t + |t| + 1) = 1$ (any $t \le 0$). So $\Lambda(x, y) = x + \alpha y$ works for every $\alpha \in [-1, 1]$: infinitely many extensions.
(b) $\|(t, 0) - (0, 1)\|_2 = \sqrt{t^2 + 1}$. Now $t - \sqrt{t^2 + 1} < 0$ and it tends to $0$ as $t \to \infty$, so the sup is $0$. Likewise $t + \sqrt{t^2 + 1} > 0$ tends to $0$ as $t \to -\infty$, so the inf is $0$. Only $\alpha = 0$ works: the extension $\Lambda(x, y) = x$ is unique.
There's a picture behind this. $\|\Lambda\| = 1$ means the line $\{\Lambda = 1\}$ never enters the open unit ball, and $\Lambda(1, 0) = 1$ pins that line to the point $(1, 0)$. So the extensions are the lines through $(1, 0)$ that touch the ball without cutting in. The $\ell^1$ ball has a corner there, which leaves room to rock the line. The round ball doesn't. Try it.
Corners give a fan of extensions, and smooth or flat spots give exactly one. A norm doesn't have to be one of the famous three, so build your own ball and look for both.
One dimension at a time is fine when there are finitely many to add. In general there are uncountably many, and there's no "next" one. That's a job for Zorn's lemma.
On Orthonormal Bases, Zorn's lemma built a maximal orthonormal set: if every chain in a nonempty partially ordered set has an upper bound, then there is a maximal element. Same move here, with partial extensions in place of orthonormal sets.
Now the theorem. The real case is the lemma plus Zorn, and the complex case is the real case plus the proposition from the first section.
(Hahn–Banach) Let $X$ be a normed space, $M$ a subspace, and $\lambda\colon M \to \mathbb{C}$ a bounded linear functional. Then $\lambda$ has a norm-preserving extension: there is a bounded linear $\Lambda\colon X \to \mathbb{C}$ with $\Lambda|_M = \lambda$ and $\|\Lambda\| = \|\lambda\|$.
Proof. If $\lambda = 0$, take $\Lambda = 0$. Otherwise replace $\lambda$ by $\lambda / \|\lambda\|$; so assume $\|\lambda\| = 1$, that is, $|\lambda(m)| \le \|m\|$ on $M$. Since every extension has $\|\Lambda\| \ge \|\lambda\| = 1$, it is enough to find an extension with $|\Lambda(x)| \le \|x\|$ for all $x$.
Step 1: the real case. Suppose $X$ is a real normed space and $\lambda$ is real. Let $\mathcal{S}$ be the set of pairs $(N, f)$ with $N$ a subspace, $M \subseteq N \subseteq X$, and $f\colon N \to \mathbb{R}$ linear with $f|_M = \lambda$ and $|f(y)| \le \|y\|$ on $N$. Order it by $(N_1, f_1) \le (N_2, f_2)$ when $N_1 \subseteq N_2$ and $f_2|_{N_1} = f_1$. It is nonempty, since $(M, \lambda) \in \mathcal{S}$.
Let $\{(N_i, f_i)\}$ be a chain, and $N = \bigcup_i N_i$. It is a subspace: any two vectors of $N$ lie in some $N_i$ and $N_j$, and one of these contains the other because the chain is totally ordered, so their sum and scalar multiples lie in $N$. Define $f(y) = f_i(y)$ for $y \in N_i$. This is well defined because the $f_i$ agree where their domains overlap, and it is linear, extends $\lambda$, and has $|f(y)| \le \|y\|$. So $(N, f)$ is an upper bound for the chain.
By Zorn's lemma there is a maximal $(Z, F) \in \mathcal{S}$. If $Z \ne X$, pick $x_0 \notin Z$; the lemma extends $F$ to $Z + \mathbb{R}x_0$ with the same bound, a strictly bigger element of $\mathcal{S}$, contradicting maximality. So $Z = X$, and $\Lambda = F$ works.
Step 2: the complex case. Regard $X$ as a real normed space (same vectors and norm, real scalars only), so $M$ is a real subspace. Let $u = \operatorname{Re} \lambda$. It is real linear on $M$, and by part (c) of the proposition, $\|u\| = \|\lambda\| = 1$. Step 1 gives a real linear $U\colon X \to \mathbb{R}$ with $U|_M = u$ and $|U(x)| \le \|x\|$. Define $$\Lambda(x) = U(x) - i\,U(ix).$$ By part (b), $\Lambda$ is complex linear, and its real part is $U$, so by part (c), $\|\Lambda\| = \|U\| \le 1$. For $m \in M$, $im \in M$ too, so $\Lambda(m) = u(m) - i\,u(im) = \lambda(m)$ by part (a). $\blacksquare$
The proof never builds $\Lambda$ by hand; Zorn just promises it's there. Here is the Zorn step as a picture: partial extensions nested inside each other, a chain of them, and its union sitting on top.
Hahn–Banach: Zorn's Lemma's greatest hit.
The theorem is an existence statement, and that's all it is. The example above already showed it has nothing to say about uniqueness. We'll come back to that. First, what it buys.
Here's the first payoff. If a point sits a positive distance away from a subspace, some bounded functional is zero on the subspace and not zero at the point. The functional "sees" the gap. The distance from $x_0$ to $M$ is $\operatorname{dist}(x_0, M) = \inf_{m \in M} \|x_0 - m\|$, and it is positive exactly when $x_0 \notin \overline{M}$.
Let $M$ be a subspace of a normed space $X$ and $x_0 \in X$. Then $x_0 \in \overline{M}$ if and only if every $\Lambda \in X^*$ with $\Lambda|_M = 0$ also has $\Lambda(x_0) = 0$. In fact, if $x_0 \notin \overline{M}$ and $\delta = \operatorname{dist}(x_0, M)$, there is $\Lambda \in X^*$ with $\Lambda|_M = 0$, $\Lambda(x_0) = 1$, and $\|\Lambda\| \le 1/\delta$.
Proof. ($\implies$) Let $\Lambda \in X^*$ with $\Lambda|_M = 0$. If $x_0 \in \overline{M}$, there are $m_n \in M$ with $m_n \to x_0$, and since $\Lambda$ is continuous, $\Lambda(x_0) = \lim \Lambda(m_n) = 0$.
($\impliedby$) Suppose $x_0 \notin \overline{M}$. The complement of $\overline{M}$ is open, so some ball around $x_0$ misses $M$, and $\delta = \operatorname{dist}(x_0, M) > 0$. Let $N = \{m + cx_0 : m \in M, \ c \in \mathbb{C}\}$. Each vector of $N$ has exactly one such form (because $x_0 \notin M$), so $\lambda(m + cx_0) = c$ is a well-defined linear functional on $N$, with $\lambda|_M = 0$ and $\lambda(x_0) = 1$. For $c \ne 0$, $-m/c \in M$, so $$\|m + cx_0\| = |c| \, \big\|x_0 - (-m/c)\big\| \ge |c| \, \delta,$$ that is, $|\lambda(m + cx_0)| = |c| \le \|m + cx_0\| / \delta$. For $c = 0$ both sides of $|\lambda(m)| \le \|m\|/\delta$ are fine since $\lambda(m) = 0$. So $\|\lambda\| \le 1/\delta$, and Hahn–Banach extends $\lambda$ to $\Lambda \in X^*$ with the same norm. $\blacksquare$
Look at the inequality in the middle: the distance is a lower bound for $\|x_0 - m\|$, so it ends up in the denominator of the norm. Closer points need steeper functionals. In the plane you can watch the level lines of $\Lambda$ crowd together as $x_0$ moves in.
In $\mathbb{R}^2$, a functional that vanishes on a line is fixed up to scale, so there's no choice to make: $\Lambda$ is forced, and $\|\Lambda\| = 1/\delta$ exactly. Drag $x_0$ and change the norm to see how the distance and the norm trade off.
Why the closure and not just $M$? Because a bounded functional can't tell $M$ from $\overline{M}$. Here's a subspace where that matters.
In $c_0$, let $M = c_{00}$ be the sequences with only finitely many nonzero terms, and $x_0 = (1, \frac{1}{2}, \frac{1}{3}, \ldots)$. Show that $x_0 \notin M$, but every $\Lambda \in c_0^*$ that vanishes on $M$ also vanishes at $x_0$.
$x_0$ has infinitely many nonzero terms, so $x_0 \notin M$. But the truncations $(1, \frac{1}{2}, \ldots, \frac{1}{N}, 0, 0, \ldots)$ lie in $M$ and are within $\frac{1}{N+1}$ of $x_0$ in the sup norm, so $x_0 \in \overline{M}$. By the corollary, every bounded $\Lambda$ vanishing on $M$ has $\Lambda(x_0) = 0$.
You can also see it directly: each unit vector $e_n$ is in $M$, so $\Lambda(e_n) = 0$ for all $n$, and on Banach Spaces & Bounded Operators we saw that a bounded functional on $c_0$ is determined by the numbers $\Lambda(e_n)$. So $\Lambda = 0$.
In fact $c_{00}$ is dense in $c_0$, and the example is a special case of a test for density. It's a short consequence of the corollary.
Let $M$ be a subspace of a normed space $X$. Prove that $M$ is dense in $X$ if and only if the only $\Lambda \in X^*$ with $\Lambda|_M = 0$ is $\Lambda = 0$.
Proof. ($\implies$) If $\overline{M} = X$ and $\Lambda|_M = 0$, the corollary gives $\Lambda(x) = 0$ for every $x \in \overline{M} = X$, so $\Lambda = 0$.
($\impliedby$) If $M$ is not dense, pick $x_0 \notin \overline{M}$. The corollary gives $\Lambda \in X^*$ with $\Lambda|_M = 0$ and $\Lambda(x_0) = 1$, so a nonzero $\Lambda$ vanishes on $M$. $\blacksquare$
We've been writing $X^*$ for "the bounded functionals on $X$". Time to make it official. It's a normed space in its own right, with the norm from Banach Spaces & Bounded Operators.
The dual space of a normed space $X$ is $X^* = \{\Lambda\colon X \to \mathbb{C} : \Lambda \text{ is bounded and linear}\}$, with the norm $\|\Lambda\| = \sup\{|\Lambda(x)| : \|x\| \le 1\}$.
The dual is always a Banach space, even when $X$ isn't complete: the values $\Lambda(x)$ live in $\mathbb{C}$, which has no holes. (The proof is short, and it's planned as the Spicy Bonus on Baire Category & Its Consequences.) The first thing Hahn–Banach tells us about $X^*$ is that it is big: every nonzero vector has a functional that measures it exactly.
Let $X$ be a normed space and $x_0 \in X$, $x_0 \ne 0$. There is $\Lambda \in X^*$ with $\|\Lambda\| = 1$ and $\Lambda(x_0) = \|x_0\|$.
Proof. Let $M = \{cx_0 : c \in \mathbb{C}\}$ and $\lambda(cx_0) = c\|x_0\|$. Then $|\lambda(cx_0)| = |c| \, \|x_0\| = \|cx_0\|$, so $\|\lambda\| = 1$, and $\lambda(x_0) = \|x_0\|$. Hahn–Banach extends $\lambda$ to $\Lambda \in X^*$ with $\|\Lambda\| = 1$. $\blacksquare$
That functional is called a norming functional for $x_0$. In $\mathbb{C}^2$ you can find them by hand, and how many there are depends on the norm.
Find every norming functional $\Lambda(x) = a x_1 + b x_2$ for $x_0 = (3, -4)$ in $\mathbb{C}^2$ with (a) $\|x\|_1 = |x_1| + |x_2|$ and (b) $\|x\|_\infty = \max\{|x_1|, |x_2|\}$. Then (c) do the same for $x_0 = (1, 1)$ with $\|\cdot\|_\infty$.
The norm of $\Lambda$ is the dual norm of $(a, b)$: $\max\{|a|, |b|\}$ for $\|\cdot\|_1$, and $|a| + |b|$ for $\|\cdot\|_\infty$ (Banach Spaces & Bounded Operators).
(a) $\|x_0\|_1 = 7$, so we need $\max\{|a|, |b|\} = 1$ and $3a - 4b = 7$. But $|3a - 4b| \le 3|a| + 4|b| \le 7$, with equality only when $a = 1$ and $b = -1$. So $\Lambda(x) = x_1 - x_2$, and it's the only one.
(b) $\|x_0\|_\infty = 4$, so we need $|a| + |b| = 1$ and $3a - 4b = 4$. Now $|3a - 4b| \le 3|a| + 4|b| \le 4(|a| + |b|) = 4$, and equality needs $a = 0$ and $b = -1$. So $\Lambda(x) = -x_2$, again the only one.
(c) $\|(1, 1)\|_\infty = 1$, and $\Lambda(x) = t x_1 + (1 - t) x_2$ has $|t| + |1 - t| = 1$ and $\Lambda(1, 1) = 1$ for every $t \in [0, 1]$. Infinitely many: $(1, 1)$ is a corner of the unit square.
A norming functional is a norm-preserving extension of a functional on the line through $x_0$, so in the plane you can find all of them the same way as before: a line through $x_0/\|x_0\|$ that supports the unit ball. Corners again mean many, and flat or round spots mean one. Find the possible values.
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Two consequences fall right out of the norming functional. Bounded functionals separate points: if $x \ne y$, a norming functional for $x - y$ gives them different values. And $X$ sits inside the dual of its dual. Each $x \in X$ gives a functional on $X^*$, "evaluate at $x$": $\lambda_x(\Lambda) = \Lambda(x)$.
The second dual (or bidual) of a normed space $X$ is $X^{**} = (X^*)^*$, the dual space of $X^*$.
So $\lambda_x \in X^{**}$, and $x \mapsto \lambda_x$ is the natural way to put $X$ inside $X^{**}$. Prove that it keeps every length.
Let $X$ be a normed space.
Proof. (a) $x - y \ne 0$, so the corollary gives $\Lambda \in X^*$ with $\Lambda(x - y) = \|x - y\| \ne 0$, that is, $\Lambda(x) \ne \Lambda(y)$.
(b) $\lambda_x$ is linear: $\lambda_x(\Lambda_1 + c\Lambda_2) = \Lambda_1(x) + c\Lambda_2(x)$, because addition and scalar multiplication in $X^*$ are pointwise. It is bounded: $|\lambda_x(\Lambda)| = |\Lambda(x)| \le \|\Lambda\| \, \|x\|$, so $\|\lambda_x\| \le \|x\|$. The map $x \mapsto \lambda_x$ is linear because each $\Lambda$ is: $\lambda_{x + cy}(\Lambda) = \Lambda(x) + c\Lambda(y)$.
For the reverse inequality, if $x = 0$ both sides are $0$. Otherwise the corollary gives $\Lambda \in X^*$ with $\|\Lambda\| = 1$ and $\Lambda(x) = \|x\|$, so $\|\lambda_x\| \ge |\lambda_x(\Lambda)| = \|x\|$. So $\|\lambda_x\| = \|x\|$: the map is an isometry. $\blacksquare$
An isometry is one-to-one, so $X$ is a copy of part of $X^{**}$. Is it all of it? In finite dimensions, yes, by counting dimensions. For Hilbert spaces and for $L^p$ with $1 < p < \infty$, also yes, even though they're infinite-dimensional. But not always.
On Banach Spaces & Bounded Operators, an Exercise proved $c_0^* = \ell^1$ and the Spicy Bonus proved $(\ell^1)^* = \ell^\infty$, both through $\Lambda x = \sum_n y_n x_n$.
Show that the image of $c_0$ in $c_0^{**}$ is not all of $c_0^{**}$.
Identify $c_0^*$ with $\ell^1$ (each $y \in \ell^1$ is the functional $\Lambda_y x = \sum_n y_n x_n$), and then $c_0^{**}$ with $(\ell^1)^* = \ell^\infty$ (each $z \in \ell^\infty$ is the functional $y \mapsto \sum_n z_n y_n$). Both identifications keep norms.
For $x \in c_0$, $\lambda_x(\Lambda_y) = \Lambda_y(x) = \sum_n y_n x_n$. That's the functional on $\ell^1$ given by $z = x$ itself. So the image of $c_0$ in $c_0^{**} = \ell^\infty$ is just $c_0$.
And $c_0 \ne \ell^\infty$: the constant sequence $(1, 1, 1, \ldots)$ is bounded but doesn't tend to $0$. The functional it gives on $c_0^*$, $\Lambda_y \mapsto \sum_n y_n$, is in $c_0^{**}$ but isn't $\lambda_x$ for any $x \in c_0$.
The same two spaces give one more surprise. Every $y \in \ell^1$ gives a functional on $\ell^\infty$ too, so it's tempting to guess $(\ell^\infty)^* = \ell^1$. Hahn–Banach says no.
Proof. (a) For $x \in \ell^\infty$, $\sum_n |y_n x_n| \le \|x\|_\infty \sum_n |y_n| = \|x\|_\infty \|y\|_1$, so the series converges absolutely, $\Lambda_y$ is linear, and $|\Lambda_y x| \le \|y\|_1 \|x\|_\infty$.
(b) $c_0$ is a closed subspace of $\ell^\infty$ (Banach Spaces & Bounded Operators). Let $\mathbf{1} = (1, 1, 1, \ldots)$. For $x \in c_0$, $\|\mathbf{1} - x\|_\infty \ge \lim_n |1 - x_n| = 1$, so $\operatorname{dist}(\mathbf{1}, c_0) = 1$ and $\mathbf{1} \notin \overline{c_0} = c_0$. The corollary on separating a point from a subspace gives $\Lambda \in (\ell^\infty)^*$ with $\Lambda|_{c_0} = 0$ and $\Lambda(\mathbf{1}) = 1$.
(c) Suppose $\Lambda = \Lambda_y$ for some $y \in \ell^1$. Each unit vector $e_n$ is in $c_0$, so $y_n = \Lambda_y(e_n) = 0$ for every $n$. Then $y = 0$ and $\Lambda = 0$, contradicting $\Lambda(\mathbf{1}) = 1$. $\blacksquare$
Notice that nobody can write that $\Lambda$ down. Hahn–Banach, through Zorn, promises it exists and refuses to say more.
The $\ell^1$ example showed that a norm-preserving extension can fail to be unique in two dimensions. Here's the same thing in a function space, with the most natural functional there is: evaluation at a point.
Let $X = L^\infty(T)$ with $\|f\|_\infty$, let $M = C(T)$, and let $\lambda(f) = f(0)$ on $M$. Show that $\lambda$ has infinitely many norm-preserving extensions to $X$.
On $C(T)$, $|f(0)| \le \|f\|_\infty$ and $\lambda(1) = 1$, so $\|\lambda\| = 1$. (A continuous function's essential sup is its sup, since $\{|f| > a\}$ is open, and a nonempty open set has positive measure.)
Let $C'$ be the elements of $L^\infty(T)$ that have a representative continuous on $T \setminus \{0\}$ with one-sided limits $f(0-)$ and $f(0+)$ at $0$. Two such representatives agree almost everywhere, hence everywhere off $0$, so the limits are well defined. $C'$ is a subspace containing $C(T)$. For $0 \le \alpha \le 1$ put $$\lambda_\alpha(f) = \alpha f(0-) + (1 - \alpha) f(0+).$$ Each $\lambda_\alpha$ is linear, equals $f(0)$ when $f$ is continuous, and has $|\lambda_\alpha(f)| \le \alpha|f(0-)| + (1 - \alpha)|f(0+)| \le \|f\|_\infty$. (The representative has $|f| \le \|f\|_\infty$ at every point of $T \setminus \{0\}$: at a point of continuity where $|f| > \|f\|_\infty$, a whole interval around it would be above $\|f\|_\infty$, and an interval has positive measure. So the one-sided limits obey the same bound.) So each $\lambda_\alpha$ is a norm-preserving extension of $\lambda$ to $C'$, and Hahn–Banach extends each one to $X$ with norm $1$.
They're all different. Let $h(t) = 1 - t/\pi$ for $0 < t \le \pi$ and $h(t) = 0$ for $-\pi \le t \le 0$. It is continuous except at $0$ (at $t = \pm\pi$ both pieces are $0$), with $h(0-) = 0$ and $h(0+) = 1$. So $h \in C'$ and $\lambda_\alpha(h) = 1 - \alpha$, a different number for each $\alpha$.
On continuous functions everybody agrees. At a jump, each $\alpha$ picks its own compromise between the two sides.
An $L^\infty$ function has no value at $0$. Hahn–Banach is happy to make one up. Several, actually.
So where does uniqueness come from when it does happen? Always from the shape of the unit ball.
The Spicy Bonus on Inner Products & Projections proved that in a Hilbert space, the norm-preserving extension is unique. Hilbert balls are round, like the $\ell^2$ ball in the plane.
Don't read "the Hahn–Banach extension" as if there's only one. There's a unique one when the dual space's unit ball has no flat pieces, as in a Hilbert space; when the space's own unit ball has corners, as in $\ell^1$, $\ell^\infty$, $C(T)$, and $L^\infty$, expect a whole family. On a prelim, "extend $\lambda$" means "show an extension exists" unless the problem says otherwise.
Now the surprise. For point evaluation, there's one setting where the extension is unique, and finding it means writing it down. That's where the rest of the page is headed. First we need a fact about polynomials.
Let $D = \{z \in \mathbb{C} : |z| < 1\}$ be the open unit disk, $\overline{D}$ the closed disk, and $T$ the unit circle, where we write a function on $T$ as a $2\pi$-periodic function of $t$ through $t \mapsto e^{it}$. Polynomials on the disk have a striking habit: their size is largest on the edge.
For every polynomial $p(z) = \sum_{k=0}^{n} a_k z^k$ with complex coefficients, $$\max_{z \in \overline{D}} |p(z)| = \max_{|z| = 1} |p(z)|.$$
Proof. $\overline{D}$ is compact and $|p|$ is continuous, so $|p|$ attains its maximum at some $z_0 \in \overline{D}$. If $|z_0| = 1$ we're done, so suppose $|z_0| < 1$ and $|p(z_0)| \ge |p(z)|$ for all $z \in \overline{D}$. We show $p$ is constant, so its maximum is attained on the circle too.
Expanding powers of $z = z_0 + (z - z_0)$ rewrites $p$ as $p(z) = \sum_{k=0}^{n} q_k (z - z_0)^k$, with $q_0 = p(z_0)$. Choose $\rho > 0$ with $|z_0| + \rho \le 1$, so the circle $z_0 + \rho e^{i\theta}$ lies in $\overline{D}$. Since $\frac{1}{2\pi}\int_{-\pi}^{\pi} e^{ij\theta} e^{-ik\theta} \, d\theta$ is $1$ when $j = k$ and $0$ otherwise, $$\frac{1}{2\pi}\int_{-\pi}^{\pi} \big|p(z_0 + \rho e^{i\theta})\big|^2 \, d\theta = \frac{1}{2\pi}\int_{-\pi}^{\pi} \Big|\sum_{k=0}^{n} q_k \rho^k e^{ik\theta}\Big|^2 \, d\theta = \sum_{k=0}^{n} |q_k|^2 \rho^{2k}.$$ The integrand is at most $|p(z_0)|^2 = |q_0|^2$, so the left side is at most $|q_0|^2$. Hence $\sum_{k \ge 1} |q_k|^2 \rho^{2k} \le 0$, so $q_k = 0$ for every $k \ge 1$, and $p = q_0$ is constant. $\blacksquare$
The heart of the proof is that average. Around any interior point, the average of $|p|^2$ on a circle is $|p(z_0)|^2$ plus something nonnegative. So somewhere on that circle, $|p|$ is at least as big as at the center, and strictly bigger unless $p$ is constant.
Whatever drama a polynomial has on the disk, it saves for the edge.
Go hunting for the maximum yourself. Then check the average from the proof at any point you like.
Every record you set in the middle got beaten closer to the edge. The proof says exactly why: pick any center and any circle around it, and compare the average of $|p|^2$ on the circle with $|p|^2$ at the center.
When the coefficients cooperate, you can find the maximum exactly: the triangle inequality bounds $|p|$ on the circle, and you look for the point where every term points the same way.
Find $\max\{|z^2 - 3| : |z| \le 1\}$.
By the theorem, the maximum is on $|z| = 1$. There, $|z^2 - 3| \le |z|^2 + 3 = 4$. Equality needs $z^2$ and $-3$ to point the same way, so $z^2 = -1$, which happens at $z = \pm i$: $|i^2 - 3| = |-4| = 4$. The maximum is $4$.
The center is a poor guess: $|p(0)| = 3$. So is $z = 1$: $|1 - 3| = 2$.
The theorem passes to uniform limits of polynomials, and those limits form the space where our functional will live.
The disk algebra $A(D)$ is the closure of the polynomials in $C(\overline{D})$ with the sup norm $\|f\|_\infty = \sup_{z \in \overline{D}} |f(z)|$.
So $f \in A(D)$ when polynomials converge to $f$ uniformly on the closed disk. Every polynomial is in it, and so is anything given by a power series that converges absolutely on $\overline{D}$, like $\sum z^n / n^2$. The maximum principle comes along for free.
For every $f \in A(D)$, $\sup_{z \in \overline{D}} |f(z)| = \sup_{|z| = 1} |f(z)|$.
Proof. Let $p_n$ be polynomials with $p_n \to f$ uniformly on $\overline{D}$. Since $\big|\sup|g| - \sup|h|\big| \le \sup|g - h|$ over any set, $\sup_{\overline{D}} |p_n| \to \sup_{\overline{D}} |f|$ and $\sup_{|z| = 1} |p_n| \to \sup_{|z| = 1} |f|$. By the theorem the two sequences are equal term by term, so their limits are equal. $\blacksquare$
Before moving on, practice the exact computation.
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The polynomials in $z$ are not dense in $C(\overline{D})$, which is why $A(D)$ is a closure and not all of $C(\overline{D})$. The function $\bar{z}$ is continuous on $\overline{D}$, but $|\bar z - p(z)|$ can't be made small for every polynomial $p$: on the circle, $\frac{1}{2\pi}\int_{-\pi}^{\pi} \big(e^{-it} - p(e^{it})\big) e^{it} \, dt = 1$, so $\sup |\bar z - p| \ge 1$. Stone–Weierstrass needs $\bar z$ in the algebra, and polynomials in $z$ alone don't have it.
Here's the setup. The corollary says a function in $A(D)$ is as big on the circle as anywhere, so restricting to the circle loses no size: $f \mapsto f|_T$ is a linear isometry from $A(D)$ onto a subspace $A_T$ of $C(T)$. In particular it's one-to-one, so the values on the circle determine $f$. Then how do you get $f(z)$ for $z$ inside from the values on the circle?
Fix $z \in D$. On $A_T$, define $\lambda_z(f|_T) = f(z)$: point evaluation, read off the boundary. It's linear, and $|\lambda_z(f|_T)| = |f(z)| \le \sup_{\overline{D}} |f| = \sup_T |f|$, so $\|\lambda_z\| \le 1$; and $\lambda_z(1) = 1$, so $\|\lambda_z\| = 1$. Hahn–Banach extends it to some $\Lambda$ on $C(T)$ with $\|\Lambda\| = 1$. We don't know yet which extension we got. The next lemma says it doesn't matter much: any such extension is positive.
Let $\Lambda$ be a linear functional on $C(T)$ with $\|\Lambda\| = 1$ and $\Lambda(1) = 1$. Then $\Lambda$ is positive: $\Lambda(g) \ge 0$ whenever $g \ge 0$. Also $\Lambda(g)$ is real when $g$ is real, and $\Lambda(\bar g) = \overline{\Lambda(g)}$ for every $g \in C(T)$.
Proof. Real functions go to real numbers. Let $g$ be real with $|g| \le 1$, and write $\Lambda(g) = a + ib$. For real $r$, $|g + ir|^2 = g^2 + r^2 \le 1 + r^2$ pointwise, and $\Lambda(g + ir) = a + i(b + r)$ because $\Lambda(1) = 1$. Since $\|\Lambda\| = 1$, $$a^2 + (b + r)^2 = |\Lambda(g + ir)|^2 \le \|g + ir\|_\infty^2 \le 1 + r^2,$$ that is, $a^2 + b^2 + 2br \le 1$ for every real $r$. Letting $r \to \pm\infty$ forces $b = 0$, so $\Lambda(g)$ is real. Every real $g$ is a real multiple of one with $|g| \le 1$, so $\Lambda(g)$ is real for every real $g$.
Conjugates. Write $g = g_1 + ig_2$ with $g_1, g_2$ real. Then $\Lambda(\bar g) = \Lambda(g_1) - i\Lambda(g_2)$, which is the conjugate of $\Lambda(g) = \Lambda(g_1) + i\Lambda(g_2)$ since both values are real.
Positive. Let $g \ge 0$, $g \ne 0$; replacing $g$ by $g/\|g\|_\infty$, assume $0 \le g \le 1$. Then $h = 2g - 1$ is real with $|h| \le 1$, so $\Lambda(h)$ is real and $\Lambda(h) \ge -|\Lambda(h)| \ge -\|h\|_\infty \ge -1$. Hence $\Lambda(g) = \frac{1}{2}\big(\Lambda(1) + \Lambda(h)\big) \ge \frac{1}{2}(1 - 1) = 0$. $\blacksquare$
Both hypotheses matter. Here's what happens when the norm is too big.
Let $\Lambda(g) = 2g(0) - g(\pi)$ on $C(T)$. Show that $\Lambda(1) = 1$ but $\Lambda$ is not positive, and find $\|\Lambda\|$.
$\Lambda(1) = 2 - 1 = 1$. The function $g(t) = \frac{1}{2}(1 - \cos t)$ is $\ge 0$, with $g(0) = 0$ and $g(\pi) = 1$, so $\Lambda(g) = -1 < 0$: not positive.
The norm: $|\Lambda(g)| \le 2|g(0)| + |g(\pi)| \le 3\|g\|_\infty$, and $g(t) = \cos t$ has $\|g\|_\infty = 1$ and $\Lambda(g) = 2 + 1 = 3$. So $\|\Lambda\| = 3$, not $1$, and the lemma doesn't apply.
A positive functional on $C(T)$ is exactly what the Riesz representation theorem eats.
On The Riesz Representation Theorem, every positive linear functional on $C_c(X)$, for $X$ locally compact Hausdorff, was integration against a unique regular Borel measure. $T$ is compact, so $C_c(T) = C(T)$.
So $\Lambda(g) = \int_T g \, d\mu_z$ for some positive measure $\mu_z$ on the circle. That's already a formula for $f(z)$ in terms of boundary values, but $\mu_z$ is a mystery. The rest of the page identifies it. The answer has a name.
For $0 \le r < 1$ and real $\varphi$, the Poisson kernel is $$P_r(\varphi) = \frac{1 - r^2}{1 - 2r\cos\varphi + r^2}.$$
It looks pulled out of a hat. It's really a geometric series in disguise, and the series is what makes it compute.
For $0 \le r < 1$ and real $\varphi$, $\displaystyle P_r(\varphi) = \sum_{n \in \mathbb{Z}} r^{|n|} e^{in\varphi}$, and the series converges absolutely and uniformly in $\varphi$. In particular $P_r(\varphi) > 0$, and for every integer $n$, $$\frac{1}{2\pi}\int_{-\pi}^{\pi} P_r(\theta - t) \, e^{int} \, dt = r^{|n|} e^{in\theta}.$$
Proof. The terms have size $r^{|n|}$, and $\sum_n r^{|n|} < \infty$, so the series converges absolutely and uniformly. Let $w = re^{i\varphi}$, so $|w| = r < 1$. The terms for $n$ and $-n$ are conjugates, so $$\sum_{n \in \mathbb{Z}} r^{|n|} e^{in\varphi} = 1 + 2\operatorname{Re} \sum_{n=1}^{\infty} w^n = \operatorname{Re}\Big(1 + \frac{2w}{1 - w}\Big) = \operatorname{Re} \frac{1 + w}{1 - w} = \frac{\operatorname{Re}\big((1 + w)(1 - \bar w)\big)}{|1 - w|^2}.$$ The numerator is $\operatorname{Re}(1 - \bar w + w - |w|^2) = 1 - r^2$, since $w - \bar w$ is imaginary, and $|1 - w|^2 = 1 - 2r\cos\varphi + r^2$. That's $P_r(\varphi)$, and it is positive because $1 - r^2 > 0$ and $|1 - w|^2 > 0$.
For the integral, put $\varphi = \theta - t$ and integrate term by term (allowed by uniform convergence): $\frac{1}{2\pi}\int_{-\pi}^{\pi} e^{ik(\theta - t)} e^{int} \, dt$ is $e^{in\theta}$ when $k = n$ and $0$ otherwise, so only the term $k = n$ survives, leaving $r^{|n|} e^{in\theta}$. $\blacksquare$
The closed form is the one to compute with, and the series is the one to prove things with. Check that they agree at a few points.
Let $r = \frac{1}{2}$. Compute $P_r(0)$, $P_r(\frac{\pi}{2})$, and $P_r(\pi)$ from the formula, and check $P_r(0)$ and $P_r(\pi)$ against the series.
The numerator is $1 - r^2 = \frac{3}{4}$. The denominators are $1 - 1 + \frac{1}{4} = \frac{1}{4}$ at $\varphi = 0$, $1 - 0 + \frac{1}{4} = \frac{5}{4}$ at $\varphi = \frac{\pi}{2}$, and $1 + 1 + \frac{1}{4} = \frac{9}{4}$ at $\varphi = \pi$. So $P_r(0) = 3$, $P_r(\frac{\pi}{2}) = \frac{3}{5}$, and $P_r(\pi) = \frac{1}{3}$.
Series at $\varphi = 0$: $1 + 2\sum_{n \ge 1} (\frac{1}{2})^n = 1 + 2 = 3$. At $\varphi = \pi$, $e^{in\pi} = (-1)^n$, so $1 + 2\sum_{n \ge 1} (-\frac{1}{2})^n = 1 + 2 \cdot (-\frac{1}{3}) = \frac{1}{3}$. Both match.
Watch the kernel as $r$ climbs toward $1$: it piles up around $\varphi = 0$ while its average stays exactly $1$ (that's the case $n = 0$ above).
That pile-up is the whole intuition: averaging against $P_r(\theta - t)$ weighs the boundary values near angle $\theta$ most heavily, more and more so as the point moves out to the edge. Give it a name.
The Poisson integral of $g \in C(T)$ at $z = re^{i\theta} \in D$ is $$\frac{1}{2\pi}\int_{-\pi}^{\pi} P_r(\theta - t) \, g(t) \, dt.$$
Now the main theorem. It says the Poisson integral gives back every $f \in A(D)$, and it settles the uniqueness question we left open.
Let $z = re^{i\theta} \in D$. The point evaluation $\lambda_z(f|_T) = f(z)$ on $A_T$ has exactly one norm-preserving extension to $C(T)$, namely the Poisson integral $g \mapsto \frac{1}{2\pi}\int_{-\pi}^{\pi} P_r(\theta - t) \, g(t) \, dt$. In particular, for every $f \in A(D)$, $$f(re^{i\theta}) = \frac{1}{2\pi}\int_{-\pi}^{\pi} \frac{1 - r^2}{1 - 2r\cos(\theta - t) + r^2} \, f(e^{it}) \, dt.$$
Proof. Hahn–Banach gives at least one norm-preserving extension. Let $\Lambda$ be any of them, and write $e_n(t) = e^{int}$.
The moments of $\Lambda$. For $n \ge 0$, $e_n$ is the restriction to $T$ of the polynomial $z^n$, which is in $A(D)$, so $\Lambda(e_n) = \lambda_z(e_n) = z^n = r^n e^{in\theta}$. For $n < 0$, $e_n = \overline{e_{-n}}$, and $\Lambda$ satisfies the lemma's hypotheses ($\|\Lambda\| = 1$ and $\Lambda(1) = \lambda_z(1) = 1$), so $\Lambda(e_n) = \overline{\Lambda(e_{-n})} = \overline{z^{-n}} = r^{|n|} e^{in\theta}$. So $\Lambda(e_n) = r^{|n|} e^{in\theta}$ for every $n \in \mathbb{Z}$.
The Poisson functional has the same moments. Let $\Phi(g) = \frac{1}{2\pi}\int_{-\pi}^{\pi} P_r(\theta - t) \, g(t) \, dt$. By the lemma on the kernel, $\Phi(e_n) = r^{|n|} e^{in\theta} = \Lambda(e_n)$. By linearity, $\Lambda$ and $\Phi$ agree on every trigonometric polynomial $\sum_{|n| \le N} c_n e_n$.
They agree everywhere. Both are bounded on $C(T)$: $\|\Lambda\| = 1$, and since $P_r > 0$ with average $1$ (the case $n = 0$), $|\Phi(g)| \le \|g\|_\infty \frac{1}{2\pi}\int_{-\pi}^{\pi} P_r(\theta - t) \, dt = \|g\|_\infty$. The trigonometric polynomials are dense in $C(T)$ (Fourier Series, from Fejér's theorem). Two bounded functionals that agree on a dense set agree everywhere, so $\Lambda = \Phi$.
So every norm-preserving extension equals $\Phi$: there is exactly one. For $f \in A(D)$, $f(z) = \lambda_z(f|_T) = \Phi(f|_T)$, which is the formula. $\blacksquare$
On Fourier Series, Fejér's theorem proved that every continuous function on $T$ is a uniform limit of trigonometric polynomials. It does the same job here as everywhere else: pass from the $e_n$ to everything.
So the mystery measure is $d\mu_z = P_r(\theta - t) \, \frac{dt}{2\pi}$, and the uniqueness of $\mu_z$ in Riesz matches the uniqueness of the extension. Point evaluation on $L^\infty(T)$ had a whole family of extensions; point evaluation inside the disk, read off the boundary, has exactly one. The difference is that the values inside the disk are already pinned down by the values on the circle.
The proof is also a recipe. On $e^{int}$ the Poisson integral just multiplies by $r^{|n|}$ and replaces $t$ by $\theta$, so it handles $\cos nt$ and $\sin nt$ the same way. Move the point around and watch the weights.
The Poisson integral makes sense for any continuous $g$ on the circle, not just restrictions of functions in $A(D)$, and it always fills in the disk. Paint some boundary values and see what it does.
Two things to notice. The value at the center is the plain average of the boundary values (at $r = 0$ the kernel is $1$). And nothing inside is ever hotter than the hottest point on the edge, because the Poisson integral is an average with positive weights. Now compute one by hand.
Let $g(t) = 1 + 4\sin t + 8\cos 2t$. Find its Poisson integral at $z = \frac{1}{2}i$, and check the theorem on $f(z) = z^2$ at the same point.
Here $r = \frac{1}{2}$ and $\theta = \frac{\pi}{2}$. Since $\cos nt = \frac{1}{2}(e_n + e_{-n})$ and $\sin nt = \frac{1}{2i}(e_n - e_{-n})$, the moments give $\cos nt \mapsto r^n \cos n\theta$ and $\sin nt \mapsto r^n \sin n\theta$, and the constant stays put. So the value is $$1 + 4 \cdot \tfrac{1}{2} \sin\tfrac{\pi}{2} + 8 \cdot \tfrac{1}{4} \cos \pi = 1 + 2 - 2 = 1.$$ Compare the boundary value at $t = \frac{\pi}{2}$: $g = 1 + 4 - 8 = -3$. The integral is not the boundary value.
For $f(z) = z^2$, the boundary function is $e^{2it} = e_2$, and its Poisson integral is $r^2 e^{2i\theta} = \frac{1}{4} e^{i\pi} = -\frac{1}{4}$. Directly, $f(\frac{1}{2}i) = -\frac{1}{4}$. They match, as the theorem promises.
Damp each frequency by $r^n$, then evaluate at $\theta$. Your turn.
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The animation of the kernel showed a peak that grows and tails that die. Prove both, with numbers.
Let $0 < r < 1$. Prove that
Proof. (a) Since $-1 \le \cos\varphi \le 1$, the denominator $1 - 2r\cos\varphi + r^2$ lies between $(1 - r)^2$ (at $\cos\varphi = 1$) and $(1 + r)^2$ (at $\cos\varphi = -1$). The numerator $1 - r^2 = (1 - r)(1 + r)$ is positive, so $$\frac{(1 - r)(1 + r)}{(1 + r)^2} \le P_r(\varphi) \le \frac{(1 - r)(1 + r)}{(1 - r)^2},$$ which simplifies to the claim, with equality at $\varphi = \pi$ and $\varphi = 0$.
(b) For $\delta \le |\varphi| \le \pi$, $\cos\varphi \le \cos\delta$, so the denominator is at least $1 - 2r\cos\delta + r^2 = (1 - r)^2 + 2r(1 - \cos\delta) \ge 2r(1 - \cos\delta)$. Hence $$P_r(\varphi) \le \frac{1 - r^2}{2r(1 - \cos\delta)},$$ and $1 - \cos\delta > 0$ is fixed while $1 - r^2 \to 0$ and $2r \to 2$. So the sup tends to $0$. $\blacksquare$
On Fourier Series, the Fejér kernel was positive, averaged to $1$, and piled up at $0$, and that's why Fejér means of a continuous function converge uniformly. The Dirichlet kernel isn't positive, and some Fourier series diverge. The Poisson kernel has all three good properties, with $r \to 1$ in place of $N \to \infty$, and the same argument shows that the Poisson integral of a continuous $g$ tends to $g$ at the boundary.
That's every proof on the page. Check that you can put them back together.
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Every question so far came right after its idea, which is a hint. Down here the hint is gone. First, true or false: a claim about anything on this page, some of them the kind that sounds right and isn't. Decide, then read why.
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Next, the mixed set. Every question type on this page except true or false goes into one hat, and each click draws a new one at random. Flip the switch to type your answer; the types that need their answer choices will say so.
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Last, the Spicy Bonus. The separation corollary gave a functional with norm at most $1/\delta$. In fact you can always hit the distance exactly, and the distance from a point to a subspace is a maximum over functionals. Here's the precise statement.
Let $M$ be a subspace of a normed space $X$, $x_0 \in X$, and $\delta = \operatorname{dist}(x_0, M)$. Prove that $$\delta = \max\{|\Lambda(x_0)| : \Lambda \in X^*, \ \|\Lambda\| \le 1, \ \Lambda|_M = 0\},$$ and in particular that the maximum is attained.
Proof. Every such $\Lambda$ gives at most $\delta$. Let $\Lambda \in X^*$ with $\|\Lambda\| \le 1$ and $\Lambda|_M = 0$. For every $m \in M$, $$|\Lambda(x_0)| = |\Lambda(x_0 - m)| \le \|\Lambda\| \, \|x_0 - m\| \le \|x_0 - m\|.$$ Taking the inf over $m$, $|\Lambda(x_0)| \le \delta$.
Some $\Lambda$ gives exactly $\delta$. If $\delta = 0$, $\Lambda = 0$ does. Otherwise $x_0 \notin \overline{M}$, so in particular $x_0 \notin M$, and every vector of $N = \{m + cx_0 : m \in M, \ c \in \mathbb{C}\}$ has exactly one such form. Define $\lambda(m + cx_0) = c\delta$, a linear functional on $N$ with $\lambda|_M = 0$ and $\lambda(x_0) = \delta$.
Its norm is at most $1$: for $c \ne 0$, $-m/c \in M$, so $\|m + cx_0\| = |c| \, \|x_0 - (-m/c)\| \ge |c|\delta = |\lambda(m + cx_0)|$, and for $c = 0$, $\lambda(m) = 0$.
Its norm is at least $1$: choose $m_k \in M$ with $\|x_0 - m_k\| \to \delta$. Then $x_0 - m_k \ne 0$ and $\lambda(x_0 - m_k) = \delta$, so $$\|\lambda\| \ge \frac{|\lambda(x_0 - m_k)|}{\|x_0 - m_k\|} = \frac{\delta}{\|x_0 - m_k\|} \to 1.$$
So $\|\lambda\| = 1$. Hahn–Banach extends $\lambda$ to $\Lambda \in X^*$ with $\|\Lambda\| = 1$, $\Lambda|_M = 0$, and $\Lambda(x_0) = \delta$. With the first part, the sup is $\delta$ and $\Lambda$ attains it. $\blacksquare$