The next few pages build the $L^p$ spaces: functions whose $p$-th power has a finite integral. Before any of that can work, we need three inequalities, and all three come out of one shape. A convex function is one whose graph sags below every chord. That one picture gives Jensen's inequality, then Young's, then Hölder's, then Minkowski's, in a chain where each link is short. This page builds the chain.
A few examples use length on an interval, which is Lebesgue measure $m$. For a continuous $f \ge 0$ on an interval, $\int f \, dm$ is the (improper) Riemann integral you already know.
Pick two points on a graph and connect them with a straight segment, a chord. For some functions the chord always stays on or above the graph, no matter which two points you picked. Those are the convex ones. To say it with symbols, walk from $x$ to $y$ with a dial $\lambda$ that runs from $0$ to $1$: the point $(1 - \lambda)x + \lambda y$ on the axis, the height $(1 - \lambda)\varphi(x) + \lambda \varphi(y)$ on the chord.
Let $-\infty \le a < b \le \infty$. A function $\varphi\colon (a, b) \to \mathbb{R}$ is convex if for all $x, y \in (a, b)$ and every $\lambda \in [0, 1]$, $$\varphi((1 - \lambda)x + \lambda y) \le (1 - \lambda)\varphi(x) + \lambda \varphi(y).$$
The left side is the graph and the right side is the chord, both above the same point. Here it is moving: the dial $\lambda$ slides along the chord, and the gap between chord and graph never goes negative.
Convex functions are physically incapable of frowning.
The gap is zero at both ends (that's where chord and graph share a point) and bulges in the middle. Let's check two familiar functions straight from the definition, and watch one fail.
Show that $x^2$ and $|x|$ are convex on $\mathbb{R}$, and that $\sqrt{x}$ is not convex on $(0, \infty)$.
$x^2$. Expand the chord minus the graph: $$(1 - \lambda)x^2 + \lambda y^2 - \big((1 - \lambda)x + \lambda y\big)^2 = \lambda(1 - \lambda)(x - y)^2.$$ (Multiply out the square; the cross terms collect into $\lambda(1 - \lambda)(x^2 - 2xy + y^2)$.) Since $0 \le \lambda \le 1$, this is never negative.
$|x|$. By the triangle inequality, $|(1 - \lambda)x + \lambda y| \le (1 - \lambda)|x| + \lambda |y|$. That's the definition, word for word.
$\sqrt{x}$. Take $x = 1$, $y = 9$, $\lambda = \frac{1}{2}$. The chord's height over $5$ is $\frac{1}{2}(1 + 3) = 2$, but the graph's is $\sqrt{5} \approx 2.24$. The chord dips below the graph, so $\sqrt{x}$ is not convex.
Notice $|x|$ has a corner at $0$ and is convex anyway: convex functions don't need derivatives. Now a sharper way to say "chords lie above". Take three points $s < t < u$ on the graph. Convexity is exactly the statement that the slopes of the two short chords go up from left to right.
$\varphi\colon (a, b) \to \mathbb{R}$ is convex if and only if, whenever $a < s < t < u < b$, $$\frac{\varphi(t) - \varphi(s)}{t - s} \le \frac{\varphi(u) - \varphi(t)}{u - t}.$$ In that case the slope of the long chord sits between them: $$\frac{\varphi(t) - \varphi(s)}{t - s} \le \frac{\varphi(u) - \varphi(s)}{u - s} \le \frac{\varphi(u) - \varphi(t)}{u - t}.$$
Proof. Every $t$ strictly between $s$ and $u$ is $t = (1 - \lambda)s + \lambda u$ for exactly one $\lambda \in (0, 1)$, namely $\lambda = \frac{t - s}{u - s}$, with $1 - \lambda = \frac{u - t}{u - s}$. So the convexity inequality for the pair $s, u$ at this $\lambda$ reads $$\varphi(t) \le \frac{u - t}{u - s}\varphi(s) + \frac{t - s}{u - s}\varphi(u), \quad\text{that is,}\quad (u - s)\varphi(t) \le (u - t)\varphi(s) + (t - s)\varphi(u).$$ Write $u - s = (u - t) + (t - s)$ on the left and move terms across: $$(u - t)\big(\varphi(t) - \varphi(s)\big) \le (t - s)\big(\varphi(u) - \varphi(t)\big).$$ Dividing by the positive number $(u - t)(t - s)$ gives the slope inequality. Every step reverses, so the slope inequality for all $s < t < u$ gives back the convexity inequality whenever $x \ne y$ and $0 < \lambda < 1$ (call the smaller of $x, y$ "$s$"; if $x > y$, swap the roles of $\lambda$ and $1 - \lambda$). The cases $x = y$, $\lambda = 0$, and $\lambda = 1$ are equalities. So the two conditions are equivalent.
For the long chord, write its slope as an average of the short ones: $$\frac{\varphi(u) - \varphi(s)}{u - s} = \frac{t - s}{u - s} \cdot \frac{\varphi(t) - \varphi(s)}{t - s} + \frac{u - t}{u - s} \cdot \frac{\varphi(u) - \varphi(t)}{u - t}.$$ The weights are positive and add to $1$, so this average lies between the two short slopes. $\blacksquare$
So convex means "slopes increase". The three chords below are the whole theorem in one picture: red, then purple, then blue, each at least as steep as the one before.
For a differentiable function, "slopes increase" should mean "the derivative increases", and it does. This is the test you'll actually use to check that a formula is convex.
Let $\varphi$ be differentiable on $(a, b)$. Then $\varphi$ is convex if and only if $\varphi'$ is increasing: $\varphi'(s) \le \varphi'(t)$ whenever $s < t$. In particular, if $\varphi'' \ge 0$ on $(a, b)$, then $\varphi$ is convex.
Proof. ($\implies$) Let $s < t$ and take $0 < h < t - s$. The three-chord inequality for $s < s + h < t$ gives $\frac{\varphi(s + h) - \varphi(s)}{h} \le \frac{\varphi(t) - \varphi(s)}{t - s}$, and for $s < t - h < t$ it gives $\frac{\varphi(t) - \varphi(s)}{t - s} \le \frac{\varphi(t) - \varphi(t - h)}{h}$. Let $h \to 0^+$: the left quotient tends to $\varphi'(s)$ and the right one to $\varphi'(t)$, so $$\varphi'(s) \le \frac{\varphi(t) - \varphi(s)}{t - s} \le \varphi'(t).$$
($\impliedby$) Let $s < t < u$. By the Mean Value Theorem there are $\xi \in (s, t)$ and $\eta \in (t, u)$ with $$\frac{\varphi(t) - \varphi(s)}{t - s} = \varphi'(\xi) \quad\text{and}\quad \frac{\varphi(u) - \varphi(t)}{u - t} = \varphi'(\eta).$$ Since $\xi < \eta$ and $\varphi'$ is increasing, $\varphi'(\xi) \le \varphi'(\eta)$, which is the slope inequality. So $\varphi$ is convex.
Finally, if $\varphi'' \ge 0$, then $\varphi'$ is increasing (by the Mean Value Theorem again, applied to $\varphi'$). $\blacksquare$
With the second-derivative test, a whole list of functions becomes convex in one line each. These are the ones the rest of the page leans on.
Show that $e^x$ is convex on $\mathbb{R}$, that $t^p$ is convex on $(0, \infty)$ for every $p \ge 1$, and that $-\log x$ is convex on $(0, \infty)$. Is $e^{-x}$ convex?
$(e^x)'' = e^x > 0$. For $t^p$, the second derivative is $p(p - 1)t^{p - 2} \ge 0$ when $p \ge 1$ and $t > 0$. For $-\log x$, the second derivative is $\frac{1}{x^2} > 0$. All three are convex.
$(e^{-x})'' = e^{-x} > 0$, so yes: $e^{-x}$ is convex, even though it's decreasing. Convex is about the slope increasing, not about the function increasing. Its slope $-e^{-x}$ is negative but climbs toward $0$.
That last point is the classic trap, so here it is on its own.
Convex does not mean increasing, and increasing does not mean convex. $e^{-x}$ is convex and decreasing; $\sqrt{x}$ and $\log x$ are increasing and not convex. Also check the interval: $x^3$ is convex on $(0, \infty)$ but not on $\mathbb{R}$, and $\frac{1}{x}$ is convex on $(0, \infty)$ but not on $(-\infty, 0)$. Some books say "convex up" or "concave up" for convex; $-\varphi$ convex is called concave.
Time to try it. Pick a function, drag the two ends of a chord, and hunt for a spot where the chord dips under the graph.
For $x^3$ you only have to straddle $0$; for $\sqrt{x}$ any chord at all will do; and for $x^2$ you can drag all day. Now decide for yourself. The interval always comes with the function, and it matters.
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Here's a bonus you get for free. A convex function on an open interval can't jump. The reason is that increasing slopes trap the graph near any point between two lines that pinch together.
If $\varphi$ is convex on $(a, b)$, then $\varphi$ is continuous on $(a, b)$.
Proof. Fix $x \in (a, b)$ and choose $s, t$ with $a < s < x < t < b$. Write $\sigma(v, w) = \frac{\varphi(w) - \varphi(v)}{w - v}$ for the slope of the chord between $v$ and $w$ (it doesn't matter which is smaller). We claim that for every $y \in (s, t)$ with $y \ne x$, $$\sigma(s, x) \le \sigma(x, y) \le \sigma(x, t).$$
If $x < y < t$: the slope inequality for $s < x < y$ gives $\sigma(s, x) \le \sigma(x, y)$, and the three-chord inequality for $x < y < t$ gives $\sigma(x, y) \le \sigma(x, t)$.
If $s < y < x$: the three-chord inequality for $s < y < x$ gives $\sigma(s, x) \le \sigma(y, x)$, and the slope inequality for $y < x < t$ gives $\sigma(y, x) \le \sigma(x, t)$.
So with $M = \max\{|\sigma(s, x)|, |\sigma(x, t)|\}$, a number that doesn't depend on $y$, $$|\varphi(y) - \varphi(x)| = |\sigma(x, y)| \, |y - x| \le M|y - x| \quad\text{for all } y \in (s, t).$$ The right side goes to $0$ as $y \to x$, so $\varphi$ is continuous at $x$. $\blacksquare$
The proof is a squeeze. Near $x$, the graph is stuck in a bow-tie between the line through $s$ and $x$ and the chord from $x$ to $t$. As $y$ closes in on $x$, the bow-tie pinches shut on the point.
The proof needed room on both sides of $x$: a point $s$ to the left and a point $t$ to the right. At the end of a closed interval there's no room on one side, and continuity really can fail there.
Find a convex function on $[0, 1]$ (same definition, with $x, y \in [0, 1]$) that is not continuous.
Let $\varphi = 0$ on $[0, 1)$ and $\varphi(1) = 1$. A chord between two points of $[0, 1)$ lies on the graph. A chord from $(x, 0)$ to $(1, 1)$ sits at height $\lambda$ over $(1 - \lambda)x + \lambda$, which is in $[0, 1)$ when $\lambda < 1$, where $\varphi = 0 \le \lambda$. So every chord is on or above the graph and $\varphi$ is convex, but it jumps at $1$.
So "open interval" in the theorem isn't decoration. Next: new convex functions from old ones. Lines are convex, and the top edge of a pile of lines is too. Click to drop lines and watch their maximum.
Every top edge you built was convex, however messy the pile. That's a general fact, and limits behave almost as well. Prove both.
(a) Let $\{\varphi_\alpha\}$ be any collection of convex functions on $(a, b)$, and suppose $\varphi(x) = \sup_\alpha \varphi_\alpha(x)$ is finite for every $x$. Prove that $\varphi$ is convex.
(b) Prove that a pointwise limit of convex functions on $(a, b)$ is convex.
(c) If $\varphi_n$ are convex on $(a, b)$, must $\limsup \varphi_n$ be convex (when it's finite everywhere)? Must $\liminf \varphi_n$?
(a) Let $x, y \in (a, b)$ and $\lambda \in [0, 1]$. For each $\alpha$, since $\varphi_\alpha \le \varphi$ and the weights are nonnegative, $$\varphi_\alpha((1 - \lambda)x + \lambda y) \le (1 - \lambda)\varphi_\alpha(x) + \lambda \varphi_\alpha(y) \le (1 - \lambda)\varphi(x) + \lambda \varphi(y).$$ The right side doesn't depend on $\alpha$, so taking the sup over $\alpha$ on the left gives $\varphi((1 - \lambda)x + \lambda y) \le (1 - \lambda)\varphi(x) + \lambda \varphi(y)$. The finiteness matters: $\sup_n nx$ is $\infty$ for every $x > 0$, so it isn't a function to $\mathbb{R}$ at all.
(b) If $\varphi_n \to \varphi$ pointwise, let $n \to \infty$ in $\varphi_n((1 - \lambda)x + \lambda y) \le (1 - \lambda)\varphi_n(x) + \lambda \varphi_n(y)$. Limits keep $\le$.
(c) $\limsup$: yes. Let $g = \limsup \varphi_n$, finite everywhere, and $g_n = \sup_{k \ge n} \varphi_k$, which takes values in $(-\infty, \infty]$. Fix $x, y$ and $\lambda \in (0, 1)$, and put $z = (1 - \lambda)x + \lambda y$. For every $k \ge n$, $\varphi_k(z) \le (1 - \lambda)\varphi_k(x) + \lambda \varphi_k(y) \le (1 - \lambda)g_n(x) + \lambda g_n(y)$, so $g_n(z) \le (1 - \lambda)g_n(x) + \lambda g_n(y)$. Since $g_n(x) \downarrow g(x)$ and $g_n(y) \downarrow g(y)$, both finite, the right side is finite for large $n$ and tends to $(1 - \lambda)g(x) + \lambda g(y)$, while the left side tends to $g(z)$. (For $\lambda = 0$ or $1$ there's nothing to prove.)
$\liminf$: no. $\varphi_n(x) = (-1)^n x$ is linear, so convex, for every $n$. At each $x$ the sequence alternates between $x$ and $-x$, so $\liminf \varphi_n(x) = -|x|$, which is not convex (the chord from $-1$ to $1$ sits at $-1$, below the value $0$ at the midpoint). $\blacksquare$
Now the payoff. Chords lie above a convex graph. Turn that around and you get a line that lies below it: at any point $t$, there's a straight line through $(t, \varphi(t))$ that the graph never dips under. It's the tangent line when $\varphi$ is differentiable, but a corner has one too.
Let $\varphi$ be convex on $(a, b)$ and $t \in (a, b)$. There is a real number $\beta$ such that $$\varphi(s) \ge \varphi(t) + \beta(s - t) \quad\text{for every } s \in (a, b).$$ The line $y = \varphi(t) + \beta(s - t)$ is called a supporting line of $\varphi$ at $t$.
Proof. Let $$\beta = \sup\left\{\frac{\varphi(t) - \varphi(s)}{t - s} : a < s < t\right\}.$$ By the slope inequality, every slope in this set is at most $\frac{\varphi(u) - \varphi(t)}{u - t}$ for any $u \in (t, b)$. So $\beta$ is finite, and $$\frac{\varphi(t) - \varphi(s)}{t - s} \le \beta \le \frac{\varphi(u) - \varphi(t)}{u - t} \quad\text{whenever } s < t < u.$$ For $s < t$, multiply the left inequality by $t - s > 0$: $\varphi(t) - \varphi(s) \le \beta(t - s)$, which rearranges to $\varphi(s) \ge \varphi(t) + \beta(s - t)$. For $u > t$, multiply the right one by $u - t > 0$: $\varphi(u) \ge \varphi(t) + \beta(u - t)$. At $s = t$ it is an equality. $\blacksquare$
Now feed this line a random input. Average the graph over a probability measure and you land above the graph at the average. That's Jensen's inequality, and it needs exactly the total mass $1$.
Let $\mu$ be a positive measure on $(X, \mathcal{M})$ with $\mu(X) = 1$. Let $f \in L^1(\mu)$ be real-valued with $a < f(x) < b$ for every $x \in X$, and let $\varphi$ be convex on $(a, b)$. Then $$\varphi\left(\int_X f \, d\mu\right) \le \int_X (\varphi \circ f) \, d\mu.$$
Proof. Let $t = \int_X f \, d\mu$. First, $a < t < b$. If $a > -\infty$, then $f - a > 0$ at every point. Its integral is $\int_X (f - a) \, d\mu = t - a\,\mu(X) = t - a$, and it can't be $0$: a nonnegative function with integral $0$ is $0$ almost everywhere, and $\mu(X) = 1 > 0$. So $t > a$. The same argument with $b - f$ gives $t < b$.
The supporting line. Take $\beta$ from the lemma at this $t$. Putting $s = f(x)$, $$\varphi(f(x)) \ge \varphi(t) + \beta\big(f(x) - t\big) \quad\text{for every } x \in X.$$ Call the right side $h(x)$. It is integrable (a constant plus a multiple of $f$, on a space of measure $1$).
The integral of $\varphi \circ f$ makes sense. $\varphi$ is continuous (it's convex on an open interval), so $\varphi \circ f$ is measurable. Since $\varphi \circ f \ge h$, its negative part satisfies $(\varphi \circ f)^- \le h^- \le |\varphi(t)| + |\beta| \, |f - t|$, which is integrable. So $\int_X \varphi \circ f \, d\mu = \int (\varphi \circ f)^+ - \int (\varphi \circ f)^-$ is defined, in $(-\infty, \infty]$. If it is $\infty$, there's nothing to prove.
Integrate. Otherwise $\varphi \circ f \in L^1(\mu)$, and integrating $\varphi \circ f \ge h$ gives $$\int_X \varphi \circ f \, d\mu \ge \varphi(t)\,\mu(X) + \beta\left(\int_X f \, d\mu - t\,\mu(X)\right) = \varphi(t) + \beta(t - t) = \varphi(t). \ \blacksquare$$
The proof never needed a picture, but it has one. Put a few weighted points on the graph of $\varphi$. Their weighted average, the centroid, is the point $\left(\int f, \int \varphi \circ f\right)$, and it always sits above the graph at $\int f$. Watch the weights drift.
The orange gap is never negative, because the centroid of points on a convex graph lives inside the region the chords fence off. Two choices of $\varphi$ give inequalities you'll see over and over.
With $\mu(X) = 1$ and real $f \in L^1(\mu)$, what do $\varphi(x) = x^2$ and $\varphi(x) = e^x$ give? Check both with length $m$ on $[0, 1]$ and $f(x) = x$.
$\varphi(x) = x^2$ is convex on $\mathbb{R}$, so $\left(\int f \, d\mu\right)^2 \le \int f^2 \, d\mu$: the square of the average is at most the average of the square. For $f(x) = x$ on $[0, 1]$: $\left(\frac{1}{2}\right)^2 = \frac{1}{4} \le \int_0^1 x^2 \, dx = \frac{1}{3}$.
$\varphi(x) = e^x$ is convex on $\mathbb{R}$, so $e^{\int f \, d\mu} \le \int e^f \, d\mu$. For $f(x) = x$ on $[0, 1]$: $e^{1/2} \approx 1.65 \le \int_0^1 e^x \, dx = e - 1 \approx 1.72$.
In probability, the first one says the variance $\int f^2 - \left(\int f\right)^2$ is never negative. The hypothesis $\mu(X) = 1$ is easy to forget, so look at what happens without it.
Jensen needs $\mu(X) = 1$. With $X = \{1\}$, $\mu = 2\delta_1$, $f = 1$, and $\varphi(x) = x^2$: $\varphi\left(\int f \, d\mu\right) = \varphi(2) = 4$, but $\int \varphi \circ f \, d\mu = 2$. The proof used $\mu(X) = 1$ twice, to turn $\int \varphi(t) \, d\mu$ into $\varphi(t)$ and $\int t \, d\mu$ into $t$. Also watch the direction: the convex function goes on the outside on the small side.
Now set the weights yourself. Drag the values, slide the weights, and switch $\varphi$. The readouts are both sides of Jensen.
The gap closes in exactly two ways: all the values sit at one point, or they all sit on a piece of the graph that's a straight line, like one side of $|x|$. Anywhere the graph genuinely curves, spreading the values costs you.
Point Jensen at a finite set and pick $\varphi = \exp$, and out falls one of the most useful inequalities in all of math. Given positive numbers $\gamma_1, \ldots, \gamma_n$ and weights $\alpha_j > 0$ with $\sum \alpha_j = 1$, the weighted arithmetic mean is $\sum \alpha_j \gamma_j$ and the weighted geometric mean is $\prod \gamma_j^{\alpha_j}$. With equal weights $\frac{1}{n}$ they're the ordinary averages $\frac{\gamma_1 + \cdots + \gamma_n}{n}$ and $\sqrt[n]{\gamma_1 \cdots \gamma_n}$.
If $\gamma_1, \ldots, \gamma_n > 0$ and $\alpha_1, \ldots, \alpha_n > 0$ with $\sum_{j=1}^n \alpha_j = 1$, then $$\prod_{j=1}^n \gamma_j^{\alpha_j} \le \sum_{j=1}^n \alpha_j \gamma_j.$$
Proof. Let $\mu$ be the probability measure on $S = \{1, \ldots, n\}$ with $\mu(\{j\}) = \alpha_j$. Let $f(j) = \log \gamma_j$, which is real, and let $\varphi(s) = e^s$, which is convex on $(-\infty, \infty)$. Every function on a finite set is integrable, so Jensen's inequality gives $$\exp\left(\sum_{j=1}^n \alpha_j \log \gamma_j\right) = \varphi\left(\int_S f \, d\mu\right) \le \int_S \varphi \circ f \, d\mu = \sum_{j=1}^n \alpha_j e^{\log \gamma_j} = \sum_{j=1}^n \alpha_j \gamma_j.$$ The left side is $\prod_j e^{\alpha_j \log \gamma_j} = \prod_j \gamma_j^{\alpha_j}$. $\blacksquare$
Geometric mean on the small side, arithmetic mean on the big side. It's worth seeing that this is real, with numbers.
(a) Compare the two means of $2$ and $8$ with equal weights. (b) Compare them for $8$ and $1$ with weights $\frac{1}{3}$ and $\frac{2}{3}$. (c) Find the smallest value of $x + \frac{4}{x}$ for $x > 0$.
(a) $\sqrt{2 \cdot 8} = 4 \le \frac{2 + 8}{2} = 5$.
(b) $8^{1/3} \cdot 1^{2/3} = 2 \le \frac{1}{3} \cdot 8 + \frac{2}{3} \cdot 1 = \frac{10}{3}$.
(c) Use equal weights on $\gamma_1 = x$ and $\gamma_2 = \frac{4}{x}$: $\sqrt{x \cdot \frac{4}{x}} = 2 \le \frac{1}{2}\left(x + \frac{4}{x}\right)$, so $x + \frac{4}{x} \ge 4$. At $x = 2$ it equals $4$, so the minimum is $4$. No calculus needed: the product $x \cdot \frac{4}{x}$ is fixed, so the sum is smallest when the two terms are equal.
Part (c) is the geometric heart of it: a rectangle and a square with the same area, and the square has the smaller perimeter. Slide the sides and see.
The two squares only match when the rectangle already is a square. Your turn to compute a Jensen gap. The measure is a few weighted points, so every integral is a weighted sum.
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Hölder's inequality pairs two exponents, one for each function, and they have to fit together in a specific way. Here's the fit.
If $p, q > 0$ and $$\frac{1}{p} + \frac{1}{q} = 1, \quad\text{equivalently}\quad p + q = pq,$$ then $p$ and $q$ are called a pair of conjugate exponents.
Both reciprocals are positive and add to $1$, so each is less than $1$: conjugate exponents are automatically bigger than $1$. Solving for $q$ gives $q = \frac{p}{p - 1}$, which is the formula to remember.
Find the exponents conjugate to $2$, $3$, $4$, and $\frac{3}{2}$. What happens as $p$ gets close to $1$?
$q = \frac{p}{p - 1}$: $p = 2$ gives $q = 2$ (the only exponent that's its own partner), $p = 3$ gives $\frac{3}{2}$, $p = 4$ gives $\frac{4}{3}$, and $p = \frac{3}{2}$ gives $\frac{3/2}{1/2} = 3$. The relation is symmetric, so $3$ and $\frac{3}{2}$ are partners both ways.
As $p \to 1^+$, $p - 1 \to 0^+$ and $q \to \infty$. That's why $1$ and $\infty$ are treated as a conjugate pair; the pairing gets its meaning once $\|f\|_\infty$ exists, on The $L^p$ Spaces.
$\frac{1}{p} + \frac{1}{q} = 1$: $p$ and $q$ split the check, and between them they always pay exactly $1$.
Quick practice before we use them. Watch which exponent you're given and which you're asked for.
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Now the two-number inequality that powers Hölder. It's AM–GM in disguise: two numbers, weights $\frac{1}{p}$ and $\frac{1}{q}$, and convexity of $\exp$.
Let $p$ and $q$ be conjugate exponents. For all $a, b \ge 0$, $$ab \le \frac{a^p}{p} + \frac{b^q}{q},$$ with equality exactly when $a^p = b^q$.
Proof. If $a = 0$ or $b = 0$ the left side is $0$ and the right side is $\ge 0$, with equality exactly when both are $0$, that is, when $a^p = b^q$. So let $a, b > 0$ and write $a = e^{s/p}$ and $b = e^{t/q}$ (that is, $s = p \log a$ and $t = q \log b$). Since $\frac{1}{p} + \frac{1}{q} = 1$ and $\exp$ is convex, $$ab = e^{\frac{1}{p}s + \frac{1}{q}t} \le \frac{1}{p}e^s + \frac{1}{q}e^t = \frac{a^p}{p} + \frac{b^q}{q}.$$
Equality. If $s = t$ both sides are $e^s$. If $s < t$, put $c = \frac{1}{p}s + \frac{1}{q}t$, which lies strictly between them. By the Mean Value Theorem, $e^c - e^s = e^\xi(c - s)$ and $e^t - e^c = e^\eta(t - c)$ with $\xi < c < \eta$, so $e^\xi < e^\eta$. Since $\frac{1}{p}(c - s) = \frac{1}{q}(t - c)$ (both equal $\frac{1}{pq}(t - s)$), this gives $\frac{1}{p}(e^c - e^s) < \frac{1}{q}(e^t - e^c)$, which rearranges to $e^c < \frac{1}{p}e^s + \frac{1}{q}e^t$. The case $s > t$ is the same with the roles swapped. So equality holds exactly when $s = t$, that is, when $p \log a = q \log b$, or $a^p = b^q$. $\blacksquare$
There's a second proof you can see with no algebra at all. On the curve $y = x^{p - 1}$, the area under it from $0$ to $a$ is $\frac{a^p}{p}$. Its inverse is $x = y^{q - 1}$ (because $(p - 1)(q - 1) = 1$), so the area to its left from $0$ to $b$ is $\frac{b^q}{q}$. Together the two areas always contain the $a \times b$ rectangle, with some to spare.
The leftover sliver is the slack in Young's inequality, and it vanishes exactly when the corner $(a, b)$ sits on the curve: $b = a^{p - 1}$, the same as $a^p = b^q$. Try to make it vanish yourself.
Now integrate Young's inequality and you get Hölder's. First, some shorthand. For measurable $f\colon X \to [0, \infty]$ and $0 < p < \infty$, write $$\|f\|_p = \left(\int_X f^p \, d\mu\right)^{1/p},$$ a number in $[0, \infty]$. (The $L^p$ Spaces makes this official for complex $f$, by using $|f|$.)
Let $p$ and $q$ be conjugate exponents, $1 < p < \infty$, and let $f, g\colon X \to [0, \infty]$ be measurable. Then $$\int_X fg \, d\mu \le \left(\int_X f^p \, d\mu\right)^{1/p} \left(\int_X g^q \, d\mu\right)^{1/q} = \|f\|_p \, \|g\|_q,$$ with the convention $0 \cdot \infty = 0$.
Proof. Let $A = \|f\|_p$ and $B = \|g\|_q$.
A norm is $0$. If $A = 0$, then $\int f^p \, d\mu = 0$, so $f^p = 0$ almost everywhere, and so $f = 0$ a.e. Then $fg = 0$ a.e. and $\int fg \, d\mu = 0 \le AB$ (which is $0$, even if $B = \infty$). The case $B = 0$ is the same.
A norm is $\infty$. If $A, B > 0$ and one of them is $\infty$, the right side is $\infty$ and there's nothing to prove.
The main case, $0 < A, B < \infty$. Put $F = f/A$ and $G = g/B$, so that $\int F^p \, d\mu = 1$ and $\int G^q \, d\mu = 1$. At every point, Young's inequality gives $$F(x)G(x) \le \frac{F(x)^p}{p} + \frac{G(x)^q}{q}$$ (if $F(x)$ or $G(x)$ is $\infty$, the right side is $\infty$). Integrate: $$\frac{1}{AB}\int_X fg \, d\mu = \int_X FG \, d\mu \le \frac{1}{p}\int_X F^p \, d\mu + \frac{1}{q}\int_X G^q \, d\mu = \frac{1}{p} + \frac{1}{q} = 1.$$ Multiply by $AB$. $\blacksquare$
Notice the trick: scale both functions to have norm $1$, apply Young's inequality at each point, and the right side integrates to exactly $\frac{1}{p} + \frac{1}{q} = 1$. Now set $p = q = 2$.
With $p = q = 2$, Hölder says $\int fg \le \left(\int f^2\right)^{1/2}\left(\int g^2\right)^{1/2}$. Under counting measure on $\{1, \ldots, n\}$ that's $\sum x_j y_j \le \sqrt{\sum x_j^2}\sqrt{\sum y_j^2}$, the Cauchy–Schwarz inequality from Inner Products & Norms.
Cauchy–Schwarz is just Hölder at $p = 2$ that never changed out of its pajamas.
Hölder isn't one inequality, it's a family, one for each $p$. Different choices give different bounds, and some are much better than others.
With length $m$ on $\mathbb{R}$, let $f = 2\chi_{[0, 8]}$ and $g = \chi_{[0, 1]}$. Compute $\int fg \, dm$, and Hölder's bound $\|f\|_p \|g\|_q$ for $p = 3$ and for $p = \frac{3}{2}$.
$fg = 2\chi_{[0, 1]}$, so $\int fg \, dm = 2$.
$p = 3$, $q = \frac{3}{2}$: $\|f\|_3 = (2^3 \cdot 8)^{1/3} = 2 \cdot 8^{1/3} = 4$ and $\|g\|_{3/2} = 1^{2/3} = 1$, so the bound is $4$.
$p = \frac{3}{2}$, $q = 3$: $\|f\|_{3/2} = 2 \cdot 8^{2/3} = 8$ and $\|g\|_3 = 1$, so the bound is $8$.
Both are true, and the first is twice as good. In general $\|f\|_p = 2 \cdot 8^{1/p}$ shrinks toward $2$ as $p$ grows, while $\|g\|_q = 1$ for every $q$, so the bounds close in on the true value $2$.
Here's Hölder doing real work. Averages of an $L^p$ function can't blow up too fast near $0$.
Let $1 < p < \infty$, let $f \ge 0$ be measurable on $(0, \infty)$ with $\|f\|_p < \infty$ (length $m$), and let $F(x) = \frac{1}{x}\int_0^x f \, dm$ be its average over $(0, x)$. Show that $F(x) \le x^{-1/p}\|f\|_p$.
Write $\int_0^x f \, dm = \int f \cdot \chi_{(0, x)} \, dm$ and apply Hölder with the second function $\chi_{(0, x)}$: $$\int_0^x f \, dm \le \|f\|_p \, \|\chi_{(0, x)}\|_q = \|f\|_p \, x^{1/q}.$$ Divide by $x$: $F(x) \le x^{1/q - 1}\|f\|_p = x^{-1/p}\|f\|_p$, since $\frac{1}{q} - 1 = -\frac{1}{p}$. In particular $F(x)$ is finite for every $x > 0$.
So for $1 < p < \infty$, every average $F(x)$ is finite. Is $F$ itself in $L^p$? The Spicy Bonus at the bottom says yes. At $p = 1$ the story flips, and you can prove it now.
Let $f > 0$ on $(0, \infty)$ with $f \in L^1(m)$, and let $F(x) = \frac{1}{x}\int_0^x f \, dm$. Prove that $F \notin L^1(m)$.
Proof. $F$ is measurable: $x \mapsto \int_0^x f \, dm = \int f \chi_{(0, x)} \, dm$ is continuous by dominated convergence (dominated by $f$), so $F$ is continuous on $(0, \infty)$. Let $c = \int_0^1 f \, dm$. Since $f > 0$ everywhere and $m((0, 1)) = 1 > 0$, $c > 0$ (a nonnegative function with integral $0$ vanishes almost everywhere). For $x \ge 1$, since $f \ge 0$, $$F(x) = \frac{1}{x}\int_0^x f \, dm \ge \frac{1}{x}\int_0^1 f \, dm = \frac{c}{x}.$$ So $\int_{(0, \infty)} F \, dm \ge \int_1^\infty \frac{c}{x} \, dx = \infty$, and $F \notin L^1(m)$. $\blacksquare$
When is Hölder tight? Build two step functions and look for the shape where the bound is an equality.
Equality comes when $f^p$ and $g^q$ are proportional: then Young's inequality is an equality at every point, after scaling. Now compute a few bounds yourself.
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Last link in the chain: the triangle inequality for $\|\cdot\|_p$. For $p = 1$ it's free, since $\int (f + g) = \int f + \int g$. For bigger $p$ it takes Hölder.
Let $1 < p < \infty$ and let $f, g\colon X \to [0, \infty]$ be measurable. Then $$\left(\int_X (f + g)^p \, d\mu\right)^{1/p} \le \left(\int_X f^p \, d\mu\right)^{1/p} + \left(\int_X g^p \, d\mu\right)^{1/p},$$ that is, $\|f + g\|_p \le \|f\|_p + \|g\|_p$.
Proof. If $\|f\|_p + \|g\|_p = \infty$ or $\|f + g\|_p = 0$, there's nothing to prove. So assume $\|f\|_p, \|g\|_p < \infty$ and $\|f + g\|_p > 0$. Let $q$ be the conjugate exponent.
Step 1: $\|f + g\|_p < \infty$. For $u, v \ge 0$, $$\left(\frac{u + v}{2}\right)^p \le \frac{u^p + v^p}{2}.$$ For $u, v > 0$ this is convexity of $t^p$ on $(0, \infty)$ (an example above) with $\lambda = \frac{1}{2}$; letting $u$ or $v$ decrease to $0$ keeps it, by continuity; and if $u$ or $v$ is $\infty$, both sides are $\infty$. So $(f + g)^p \le 2^{p - 1}(f^p + g^p)$ at every point, and integrating, $$\|f + g\|_p^p \le 2^{p - 1}\left(\|f\|_p^p + \|g\|_p^p\right) < \infty.$$
Step 2: Hölder twice. Write $$(f + g)^p = f(f + g)^{p - 1} + g(f + g)^{p - 1}.$$ Apply Hölder to each piece, with exponent $p$ on $f$ (or $g$) and $q$ on $(f + g)^{p - 1}$: $$\|f + g\|_p^p \le \|f\|_p \, \|(f + g)^{p - 1}\|_q + \|g\|_p \, \|(f + g)^{p - 1}\|_q = \left(\|f\|_p + \|g\|_p\right)\|(f + g)^{p - 1}\|_q.$$ Since $(p - 1)q = p$, $$\|(f + g)^{p - 1}\|_q = \left(\int_X (f + g)^p \, d\mu\right)^{1/q} = \|f + g\|_p^{p/q} = \|f + g\|_p^{p - 1}.$$ So $\|f + g\|_p^p \le \left(\|f\|_p + \|g\|_p\right)\|f + g\|_p^{p - 1}$. By Step 1 and the assumption, $\|f + g\|_p^{p - 1}$ is a positive finite number, so divide by it. $\blacksquare$
For complex $f$, the case $p = 1$ is the pointwise $|f + g| \le |f| + |g|$ integrated. Its cousin $\left|\int f \, d\mu\right| \le \int |f| \, d\mu$ is the triangle inequality for integrals.
Why does the theorem stop at $p \ge 1$? The proof used Hölder, which needs $p > 1$, and convexity of $t^p$, which fails for $p < 1$. The simplest test is two vectors at right angles, $f = (1, 0)$ and $g = (0, 1)$ under counting measure, where $\|f\|_p = \|g\|_p = 1$ and $\|f + g\|_p = 2^{1/p}$. Watch $p$ slide down past $1$.
Below $p = 1$, the sum is "longer" than its two pieces put together. Here's the same failure for functions on an interval.
With length $m$ on $[0, 1]$, let $f = \chi_{[0, 1/2]}$ and $g = \chi_{[1/2, 1]}$. Compare $\|f + g\|_{1/2}$ with $\|f\|_{1/2} + \|g\|_{1/2}$.
$\|f\|_{1/2} = \left(\int f^{1/2} \, dm\right)^2 = \left(\frac{1}{2}\right)^2 = \frac{1}{4}$, and the same for $g$, so the right side is $\frac{1}{2}$. But $f + g = 1$ almost everywhere on $[0, 1]$, so $\|f + g\|_{1/2} = 1^2 = 1 > \frac{1}{2}$. The triangle inequality fails, so $\|\cdot\|_{1/2}$ is not a norm.
Drag two vectors and the exponent yourself, and try to break the triangle inequality. You'll find it only breaks below $p = 1$.
The $L^p$ Spaces defines $L^p(\mu)$ as the measurable $f$ with $\|f\|_p = \left(\int |f|^p \, d\mu\right)^{1/p} < \infty$. Minkowski, applied to $|f|$ and $|g|$ with $|f + g| \le |f| + |g|$, is exactly what makes $\|\cdot\|_p$ satisfy the triangle inequality there, and Hölder is what pairs $L^p$ with $L^q$.
That's the chain: chords above the graph, a supporting line below it, Jensen, AM–GM, Young, Hölder, Minkowski. Each one came out of the one before it in a few lines.
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, a proof from this page with one step missing.
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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; a few types need their answer choices, and they'll say so.
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Last, the Spicy Bonus. The averages $F(x)$ from the Hölder section are finite for $f \in L^p$, and for $p = 1$ the function $F$ falls out of $L^1$. For $1 < p < \infty$ the averaging map sends $L^p$ back into $L^p$, with a sharp constant. It takes Hölder with a clever extra factor.
(Hardy's inequality.) Let $1 < p < \infty$, let $f \ge 0$ be measurable on $(0, \infty)$ with $\|f\|_p < \infty$ (length $m$), and let $F(x) = \frac{1}{x}\int_0^x f \, dm$. Prove that $$\int_0^\infty F^p \, dm \le \left(\frac{p}{p - 1}\right)^p \int_0^\infty f^p \, dm, \quad\text{that is,}\quad \|F\|_p \le \frac{p}{p - 1}\|f\|_p.$$ Hint: write $xF(x) = \int_0^x f(t) t^{\alpha} t^{-\alpha} \, dt$ with $0 < \alpha < \frac{1}{q}$, use Hölder, integrate in $x$, then choose $\alpha$ well.
Proof. Let $q$ be the conjugate exponent and fix $0 < \alpha < \frac{1}{q}$, so $\alpha q < 1$.
Step 1: Hölder. For $x > 0$, apply Hölder on $(0, x)$ to $f(t)t^\alpha$ and $t^{-\alpha}$: $$xF(x) = \int_0^x f(t) t^\alpha \cdot t^{-\alpha} \, dt \le \left(\int_0^x f(t)^p t^{\alpha p} \, dt\right)^{1/p}\left(\int_0^x t^{-\alpha q} \, dt\right)^{1/q}.$$ Since $\alpha q < 1$, $\int_0^x t^{-\alpha q} \, dt = \frac{x^{1 - \alpha q}}{1 - \alpha q}$. Raise both sides to the power $p$, using $\frac{p}{q} = p - 1$: $$x^p F(x)^p \le (1 - \alpha q)^{1 - p} \, x^{(1 - \alpha q)(p - 1)} \int_0^x f(t)^p t^{\alpha p} \, dt.$$ Since $q(p - 1) = p$, the power of $x$ on the right is $(p - 1) - \alpha p$. Dividing by $x^p$: $$F(x)^p \le (1 - \alpha q)^{1 - p} \, x^{-1 - \alpha p} \int_0^x f(t)^p t^{\alpha p} \, dt.$$
Step 2: integrate in $x$. Everything is nonnegative, so we may swap the order of integration. (That's Tonelli's theorem, coming on the Product Measures & Fubini page: for a nonnegative measurable function of two variables, the order of integration doesn't matter.) The region is $0 < t < x$: $$\int_0^\infty x^{-1 - \alpha p} \int_0^x f(t)^p t^{\alpha p} \, dt \, dx = \int_0^\infty f(t)^p t^{\alpha p} \left(\int_t^\infty x^{-1 - \alpha p} \, dx\right) dt = \int_0^\infty f(t)^p t^{\alpha p} \cdot \frac{t^{-\alpha p}}{\alpha p} \, dt.$$ So $$\int_0^\infty F^p \, dm \le (1 - \alpha q)^{1 - p}(\alpha p)^{-1} \int_0^\infty f^p \, dm.$$
Step 3: the best $\alpha$. Put $u = \alpha q \in (0, 1)$. The constant is $(1 - u)^{1 - p} \cdot \frac{q}{pu} = \frac{q}{p} \cdot \frac{1}{u(1 - u)^{p - 1}}$, smallest when $h(u) = u(1 - u)^{p - 1}$ is largest. Since $h'(u) = (1 - u)^{p - 2}\big((1 - u) - (p - 1)u\big)$, the maximum is at $u = \frac{1}{p}$, where $h = \frac{1}{p}\left(\frac{p - 1}{p}\right)^{p - 1} = \frac{1}{p}\left(\frac{1}{q}\right)^{p - 1}$. The constant becomes $$\frac{q}{p} \cdot p \, q^{p - 1} = q^p = \left(\frac{p}{p - 1}\right)^p.$$ Taking $p$-th roots gives $\|F\|_p \le \frac{p}{p - 1}\|f\|_p$. $\blacksquare$
For complex $f \in L^p$ the same bound holds, since $|F(x)| \le \frac{1}{x}\int_0^x |f| \, dm$. And the constant $\frac{p}{p - 1}$ can't be improved, though proving that takes a sequence of test functions like $f = t^{-1/p}\chi_{(1, N)}$.