There's a whole family of spaces $L^p(\mu)$, one for each $p$. Two questions come up right away. First: does being in one of them tell you anything about the others? Second: these spaces are full of horrible functions, so is there a small, friendly set of functions that gets arbitrarily close to all of them? This page answers both. The first answer is "it depends on the measure", and we'll see exactly what it depends on. The second answer is "yes": simple functions, and then continuous ones.
On The Lᵖ Spaces, $x^{-a}$ was in $L^p((0, 1))$ exactly when $ap < 1$, and the sequence $\frac{1}{n}$ was in $\ell^p$ exactly when $p > 1$. Blowing up at a point and dragging on forever behave in opposite ways.
Start with the most natural guess: bigger $p$ means a stricter condition, so $L^q \subseteq L^p$ when $p < q$. Or maybe the reverse. On Lebesgue measure on $(0, \infty)$, it's neither, and one function shows why. Watch both ends of $x^{-1/2}$ at once.
Near $0$ the function is a spike: it's huge on a short interval, and raising it to a big power makes things worse. Near $\infty$ it's a tail: it's small on a long interval, and raising it to a big power helps. Here's the computation for any exponent.
On $(0, \infty)$ with Lebesgue measure $m$, let $a > 0$ and $f(x) = x^{-a}$. For which $p > 0$ is $f\chi_{(0, 1)} \in L^p$? For which is $f\chi_{(1, \infty)} \in L^p$? For which is $f \in L^p$?
$|f|^p = x^{-ap}$. By monotone convergence on the intervals $[\frac{1}{n}, 1]$ and $[1, n]$, $$\int_0^1 x^{-ap} \, dx < \infty \iff ap < 1,$$ $$\int_1^\infty x^{-ap} \, dx < \infty \iff ap > 1.$$ So $f\chi_{(0, 1)} \in L^p$ exactly when $p < \frac{1}{a}$, and $f\chi_{(1, \infty)} \in L^p$ exactly when $p > \frac{1}{a}$.
Since $\int_0^\infty = \int_0^1 + \int_1^\infty$, $f \in L^p$ needs both, and no $p$ is both less than and greater than $\frac{1}{a}$ ($p = \frac{1}{a}$ fails both). So $x^{-a}$ is in no $L^p((0, \infty))$ at all.
Now use the two pieces separately. For $p < q$, take $a$ with $p < \frac{1}{a} < q$: then $x^{-a}\chi_{(0, 1)}$ is in $L^p$ but not $L^q$, and $x^{-a}\chi_{(1, \infty)}$ is in $L^q$ but not $L^p$. So on $(0, \infty)$, $L^p \not\subseteq L^q$ and $L^q \not\subseteq L^p$ whenever $p \neq q$. Try it with your own $a$.
Each end has its own set of good exponents, and the sets sit on opposite sides of $\frac{1}{a}$. Since the two ends don't talk to each other, you can mix any spike with any tail. Build a few.
The spike sets a ceiling on $p$ and the tail sets a floor. Keep those two words, because the rest of this page is about measures that forbid one of them.
$L^p[0, 1]$ bans tails, $\ell^p$ bans spikes, and $L^p(\mathbb{R})$ lets in both and then wonders why nobody gets along.
So $L^q$ doesn't sit inside $L^p$ or $L^r$ for $p < q < r$. But a function in $L^q$ can't be too bad: its spikes can't be worse than $L^p$ allows, and its tails can't be worse than $L^r$ allows. To say that, we need to add spaces.
For $0 < p, r \le \infty$, the sum $L^p(\mu) + L^r(\mu)$ is the set of functions $g + h$ with $g \in L^p(\mu)$ and $h \in L^r(\mu)$.
The trick is to cut $f$ at height $1$. Above $1$, raising to a smaller power makes numbers smaller. Below $1$, raising to a bigger power does. So the tall part of $f$ goes down to $L^p$, and the short part goes up to $L^r$.
That picture is the whole proof.
If $0 < p < q < r \le \infty$, then $L^q(\mu) \subseteq L^p(\mu) + L^r(\mu)$. That is, every $f \in L^q(\mu)$ can be written $f = g + h$ with $g \in L^p(\mu)$ and $h \in L^r(\mu)$.
Proof. Let $f \in L^q(\mu)$ and $E = \{x : |f(x)| \ge 1\}$, which is measurable. Let $g = f\chi_E$ and $h = f\chi_{E^c}$, so $f = g + h$.
On $E$ we have $|f| \ge 1$ and $p < q$, so $|f|^p \le |f|^q$. Hence $|g|^p = |f|^p \chi_E \le |f|^q$ everywhere, and $$\int_X |g|^p \, d\mu \le \int_X |f|^q \, d\mu = \|f\|_q^q < \infty.$$ So $g \in L^p(\mu)$.
Off $E$ we have $|h| = |f|\chi_{E^c} < 1$. If $r = \infty$, then $\|h\|_\infty \le 1$, so $h \in L^\infty(\mu)$. If $r < \infty$, then $|h| < 1$ and $q < r$ give $|h|^r \le |h|^q \le |f|^q$, so $\int_X |h|^r \, d\mu \le \|f\|_q^q < \infty$ and $h \in L^r(\mu)$. $\blacksquare$
Notice what the proof bounds: $\|g\|_p^p \le \|f\|_q^q$, the norms raised to different powers. You can't take roots and get $\|g\|_p \le \|f\|_q$, because the roots are different. Here's the split on an actual function.
On $(0, \infty)$ with Lebesgue measure, let $f(x) = x^{-1/3}$ for $0 < x \le 1$ and $f(x) = x^{-2/3}$ for $x > 1$. Show that $f \in L^2$, that $f$ is in neither $L^1$ nor $L^4$, and split $f = g + h$ with $g \in L^1$ and $h \in L^4$.
$|f|^2$ is $x^{-2/3}$ on $(0, 1]$ and $x^{-4/3}$ on $(1, \infty)$. Both integrals are finite ($\frac{2}{3} < 1$ near $0$, $\frac{4}{3} > 1$ near $\infty$), so $f \in L^2$. But $\int_1^\infty x^{-2/3} \, dx = \infty$, so $f \notin L^1$, and $\int_0^1 x^{-4/3} \, dx = \infty$, so $f \notin L^4$.
Cut at height $1$: $x^{-1/3} \ge 1$ on $(0, 1]$, and $x^{-2/3} < 1$ on $(1, \infty)$. So $E = (0, 1]$, $g = x^{-1/3}\chi_{(0, 1]}$ and $h = x^{-2/3}\chi_{(1, \infty)}$. Then $\int g = \int_0^1 x^{-1/3} \, dx = \frac{3}{2}$, so $g \in L^1$, and $\int h^4 = \int_1^\infty x^{-8/3} \, dx = \frac{3}{5}$, so $h \in L^4$.
So $f$ is in neither outer space but in their sum. Is every member of the sum in $L^q$, though? No, and on a nice space it fails badly.
Let $0 < p < q < r \le \infty$ and let $m$ be Lebesgue measure on $[0, 1]$. Show that $L^p(m) + L^r(m) = L^p(m)$, and conclude that $L^p(m) + L^r(m) \neq L^q(m)$.
First, $L^r(m) \subseteq L^p(m)$. If $r = \infty$ and $h \in L^\infty$, then $|h|^p \le \|h\|_\infty^p$ almost everywhere, and $\int_{[0, 1]} \|h\|_\infty^p \, dm = \|h\|_\infty^p < \infty$. If $r < \infty$, then at each point $|h|^p \le 1 + |h|^r$: where $|h| \le 1$, $|h|^p \le 1$, and where $|h| > 1$, $|h|^p \le |h|^r$. So $\int |h|^p \, dm \le 1 + \int |h|^r \, dm < \infty$.
Now if $g \in L^p$ and $h \in L^r \subseteq L^p$, then $g + h \in L^p$, since $|g + h|^p \le 2^p \max\{|g|, |h|\}^p \le 2^p (|g|^p + |h|^p)$. So $L^p + L^r \subseteq L^p$. Conversely, $g = g + 0$ with $0 \in L^r$. So $L^p + L^r = L^p$.
Finally, $f(x) = x^{-1/q}$ has $|f|^p = x^{-p/q}$ with $\frac{p}{q} < 1$, so $f \in L^p = L^p + L^r$. But $|f|^q = x^{-1}$, and $\int_0^1 x^{-1} \, dx = \infty$, so $f \notin L^q$. $\blacksquare$
Now flip it around. If $f$ is in $L^p$ and in $L^r$, it has neither a bad spike (that would break $L^r$) nor a bad tail (that would break $L^p$). So it should be in everything in between. It is, and the norm in between is controlled by the two outer norms. The exponent $\lambda$ in the bound says how far $q$ sits between $p$ and $r$, measured with reciprocals.
On Convexity & Inequalities, Hölder's inequality said $\int fg \le \|f\|_s \|g\|_t$ for conjugate exponents $\frac{1}{s} + \frac{1}{t} = 1$. The proof below is one well-chosen use of it.
If $0 < p < q < r \le \infty$, then $L^p(\mu) \cap L^r(\mu) \subseteq L^q(\mu)$, and for every $f \in L^p(\mu) \cap L^r(\mu)$, $$\|f\|_q \le \|f\|_p^{\lambda} \, \|f\|_r^{1 - \lambda},$$ where $0 < \lambda < 1$ is given by $\frac{1}{q} = \frac{\lambda}{p} + \frac{1 - \lambda}{r}$.
Proof. Since $\frac{1}{r} < \frac{1}{q} < \frac{1}{p}$, the number $\lambda = \left(\frac{1}{q} - \frac{1}{r}\right) \big/ \left(\frac{1}{p} - \frac{1}{r}\right)$ lies strictly between $0$ and $1$ and satisfies the equation.
The case $r = \infty$. Then $\frac{1}{q} = \frac{\lambda}{p}$, so $\lambda = \frac{p}{q}$. For almost every $x$, $|f(x)| \le \|f\|_\infty$, so $$|f(x)|^q = |f(x)|^{q - p} \, |f(x)|^p \le \|f\|_\infty^{q - p} \, |f(x)|^p.$$ Integrating, $\|f\|_q^q \le \|f\|_\infty^{q - p} \|f\|_p^p$. Taking $q$-th roots, $\|f\|_q \le \|f\|_\infty^{1 - p/q} \|f\|_p^{p/q} = \|f\|_p^{\lambda} \|f\|_\infty^{1 - \lambda}$.
The case $r < \infty$. Multiply the defining equation by $q$: $\frac{\lambda q}{p} + \frac{(1 - \lambda) q}{r} = 1$. Both terms are positive, so $s = \frac{p}{\lambda q}$ and $t = \frac{r}{(1 - \lambda) q}$ are conjugate exponents with $1 < s, t < \infty$. Write $|f|^q = |f|^{\lambda q} \, |f|^{(1 - \lambda) q}$ and apply Hölder's inequality: $$\begin{aligned} \int_X |f|^q \, d\mu &\le \left( \int_X |f|^{\lambda q s} \, d\mu \right)^{1/s} \left( \int_X |f|^{(1 - \lambda) q t} \, d\mu \right)^{1/t} \\ &= \left( \int_X |f|^p \, d\mu \right)^{\lambda q / p} \left( \int_X |f|^r \, d\mu \right)^{(1 - \lambda) q / r}. \end{aligned}$$ The right side is $\|f\|_p^{\lambda q} \|f\|_r^{(1 - \lambda) q} < \infty$. Taking $q$-th roots gives the bound, and in particular $f \in L^q(\mu)$. $\blacksquare$
The bound is a weighted geometric mean of the two outer norms. The cleanest way to see where $\lambda$ comes from is to put the exponents on a line by their reciprocals: then $\frac{1}{q}$ is exactly $\lambda$ of the way from $\frac{1}{r}$ to $\frac{1}{p}$, and the bound says $\log \|f\|$ sits under the chord there.
The most common choice is $p = 1$, $r = \infty$, which gives $\lambda = \frac{1}{2}$ at $q = 2$. Here it is with numbers.
On $\mathbb{R}$ with Lebesgue measure, let $f = 2\chi_{[0, 1]} + \chi_{(1, 5]}$. Compute $\|f\|_1$, $\|f\|_\infty$, $\|f\|_2$, and check the interpolation bound with $p = 1$, $q = 2$, $r = \infty$.
$\|f\|_1 = 2 \cdot 1 + 1 \cdot 4 = 6$ and $\|f\|_\infty = 2$. Also $\|f\|_2 = (4 \cdot 1 + 1 \cdot 4)^{1/2} = \sqrt{8} \approx 2.83$.
With $r = \infty$, $\lambda = \frac{p}{q} = \frac{1}{2}$, so the bound is $\|f\|_2 \le \|f\|_1^{1/2} \|f\|_\infty^{1/2} = \sqrt{12} \approx 3.46$. It holds, with room to spare.
When is there no room to spare? Drag the blocks of a step function and find out.
The bound is exact when $|f|$ takes a single value where it isn't zero, and it's loosest when $f$ mixes one tall value with several short ones. Before using it, find $\lambda$ for a few triples.
Interpolate the reciprocals, not the exponents. For $p = 1$, $q = 2$, $r = 4$, the tempting $q = \lambda p + (1 - \lambda) r$ gives $\lambda = \frac{2}{3}$, but the right equation $\frac{1}{2} = \lambda + \frac{1 - \lambda}{4}$ gives $\lambda = \frac{1}{3}$. And when $r = \infty$, the term $\frac{1 - \lambda}{r}$ is $0$, so $\lambda = \frac{p}{q}$.
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Now pick measures that forbid one of the two troubles. Counting measure forbids spikes: every nonempty set has measure at least $1$, so a function can't be big on a set of small measure. Without spikes, only tails matter, and tails get better as $p$ grows. So $\ell^p$ should grow with $p$.
Let $A$ be a nonempty set and $0 < p < q \le \infty$. Then $\ell^p(A) \subseteq \ell^q(A)$, and $\|f\|_q \le \|f\|_p$ for every $f \in \ell^p(A)$.
Proof. The case $q = \infty$. With counting measure the only null set is $\emptyset$, so $\|f\|_\infty = \sup_{\alpha \in A} |f(\alpha)|$. For every $\alpha \in A$, $$|f(\alpha)|^p \le \sum_{\beta \in A} |f(\beta)|^p = \|f\|_p^p.$$ Taking the supremum over $\alpha$ gives $\|f\|_\infty^p \le \|f\|_p^p$, so $\|f\|_\infty \le \|f\|_p$.
The case $q < \infty$. By the first case, $f \in \ell^p(A) \cap \ell^\infty(A)$. The interpolation bound with $r = \infty$ and $\lambda = \frac{p}{q}$, then the first case, give $$\begin{aligned} \|f\|_q &\le \|f\|_p^{p/q} \, \|f\|_\infty^{1 - p/q} \\ &\le \|f\|_p^{p/q} \, \|f\|_p^{1 - p/q} = \|f\|_p. \qquad \blacksquare \end{aligned}$$
In $\mathbb{R}^2$ the norm inequality is a picture: if $\|x\|_q \le \|x\|_p$, then every point of the $\ell^p$ unit ball is in the $\ell^q$ unit ball. So the balls nest, and they grow with $p$.
The inclusion is strict for infinite $A$, and the standard witness is the harmonic sequence.
Compute $\|x\|_1$, $\|x\|_2$, $\|x\|_\infty$ for $x = (3, 4) \in \ell^p(\{1, 2\})$. Then show that $x = (\frac{1}{n})_{n \ge 1}$ is in $\ell^2$ but not in $\ell^1$.
$\|(3, 4)\|_1 = 7$, $\|(3, 4)\|_2 = \sqrt{9 + 16} = 5$, and $\|(3, 4)\|_\infty = 4$. So $4 \le 5 \le 7$, as the proposition says.
For the sequence, $\sum_n \frac{1}{n^2}$ converges (it equals $\frac{\pi^2}{6}$), so $(\frac{1}{n}) \in \ell^2$, while the harmonic series $\sum_n \frac{1}{n}$ diverges, so $(\frac{1}{n}) \notin \ell^1$. Hence $\ell^1 \subsetneq \ell^2$.
Squaring the terms is what saved the sum. Watch the two series side by side.
Now forbid tails instead. If $\mu(X) < \infty$, a function can't be small on a set of huge measure, because there is no set of huge measure. Only spikes are left, and spikes get worse as $p$ grows. So the inclusions should run the other way.
Let $\mu(X) < \infty$ and $0 < p < q \le \infty$. Then $L^q(\mu) \subseteq L^p(\mu)$, and for every $f \in L^q(\mu)$, $$\|f\|_p \le \|f\|_q \, \mu(X)^{\frac{1}{p} - \frac{1}{q}}.$$
Proof. The case $q = \infty$. Here $\frac{1}{q} = 0$. Since $|f|^p \le \|f\|_\infty^p$ almost everywhere, $$\begin{aligned} \|f\|_p &= \left( \int_X |f|^p \, d\mu \right)^{1/p} \\ &\le \left( \|f\|_\infty^p \, \mu(X) \right)^{1/p} = \|f\|_\infty \, \mu(X)^{1/p}. \end{aligned}$$
The case $q < \infty$. The numbers $s = \frac{q}{p}$ and $t = \frac{q}{q - p}$ are conjugate exponents, since $\frac{p}{q} + \frac{q - p}{q} = 1$. Hölder's inequality applied to $|f|^p \cdot 1$ gives $$\begin{aligned} \int_X |f|^p \, d\mu &\le \left( \int_X |f|^{ps} \, d\mu \right)^{1/s} \left( \int_X 1^t \, d\mu \right)^{1/t} \\ &= \|f\|_q^p \, \mu(X)^{1 - p/q}. \end{aligned}$$ Taking $p$-th roots, $\|f\|_p \le \|f\|_q \, \mu(X)^{1/p - 1/q}$. $\blacksquare$
The constant can't be improved: for $f \equiv 1$, $\|f\|_p = \mu(X)^{1/p}$ and $\|f\|_q = \mu(X)^{1/q}$, which is equality. The most used case is $p = 1$, $q = 2$, which is just Cauchy–Schwarz against the function $1$.
On $[0, 4]$ with Lebesgue measure, let $f(x) = x$. Compute $\|f\|_1$ and $\|f\|_2$, and check $\|f\|_1 \le \|f\|_2 \, m([0, 4])^{1/2}$. Then show that $x^{-1/2}$ is in $L^1([0, 4])$ but not in $L^2([0, 4])$.
$\|f\|_1 = \int_0^4 x \, dx = 8$ and $\|f\|_2 = \left( \int_0^4 x^2 \, dx \right)^{1/2} = \sqrt{64/3} \approx 4.62$. The bound is $4.62 \cdot \sqrt{4} \approx 9.24$, and $8 \le 9.24$.
For $g(x) = x^{-1/2}$: $\int_0^4 x^{-1/2} \, dx = 2\sqrt{4} = 4$, so $g \in L^1$. But $|g|^2 = x^{-1}$ and $\int_0^4 x^{-1} \, dx = \infty$, so $g \notin L^2$. The inclusion $L^2 \subseteq L^1$ is strict.
So counting measure and finite measure push the inclusions in opposite directions, and Lebesgue measure on $(0, \infty)$ has both troubles, so it gets no inclusion at all. Here are all three, one after another.
On a finite measure space, bigger $p$ means a smaller club. Exclusive.
The direction also shows up in the norms themselves. Same four values, two measures: watch which way $\|f\|_p$ moves as $p$ grows.
On a probability space, $\mu(X) = 1$ and the constant disappears: the norms climb with $p$. Compute a bound with the constant left in.
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On a probability space, more is true: the norms increase, and you can say exactly when two of them are equal. That's the next exercise.
Let $\mu(X) = 1$ and let $f$ be a complex measurable function on $X$ with $\|f\|_\infty > 0$.
(a) If $\|f\|_s = \infty$ there is nothing to prove. Otherwise $f \in L^s(\mu)$, and the finite-measure bound with $\mu(X) = 1$ gives $\|f\|_r \le \|f\|_s \cdot 1^{1/r - 1/s} = \|f\|_s$.
(b) Exactly when $|f|$ is constant almost everywhere. If $|f| = c$ almost everywhere, then $\|f\|_r = (c^r)^{1/r} = c = \|f\|_s$ for every $r$ and $s$. Conversely, suppose $\|f\|_r = \|f\|_s = M < \infty$; then $M > 0$, because $\|f\|_\infty > 0$ means $f \neq 0$ on a set of positive measure.
If $s = \infty$: $|f| \le M$ almost everywhere, so $M^r - |f|^r \ge 0$ almost everywhere, and $\int_X (M^r - |f|^r) \, d\mu = M^r - \|f\|_r^r = 0$. A nonnegative function with integral $0$ is $0$ almost everywhere, so $|f| = M$ almost everywhere.
If $s < \infty$: let $u = |f|^r$, $k = \frac{s}{r} > 1$, and $c = \int_X u \, d\mu = M^r > 0$. The function $y \mapsto y^k$ lies above its tangent line at $c$, strictly except at $y = c$: $y^k \ge c^k + kc^{k - 1}(y - c)$ for $y \ge 0$, with equality only at $y = c$. Put $y = u(x)$ and integrate (the right side has integral $c^k + kc^{k-1}(c - c) = c^k$): $$\int_X u^k \, d\mu \ge c^k, \quad \text{that is,} \quad \|f\|_s^s \ge \|f\|_r^s.$$ Here both sides equal $M^s$, so the nonnegative function $u^k - c^k - kc^{k - 1}(u - c)$ has integral $0$ and vanishes almost everywhere. So $u = c$ almost everywhere, and $|f| = c^{1/r} = M$ almost everywhere.
(c) If $f \in L^s$, then by (a) $\|f\|_r \le \|f\|_s < \infty$, so $f \in L^r$. The two spaces are equal exactly when there is a $c > 0$ with $\mu(E) \ge c$ for every measurable $E$ with $\mu(E) > 0$ (no spikes are possible).
If such a $c$ exists and $f \in L^r$, then $n^r \mu(\{|f| > n\}) \le \|f\|_r^r$, so $\mu(\{|f| > n\}) < c$ for large $n$, which forces $\mu(\{|f| > n\}) = 0$. So $f$ is essentially bounded by $n$, and $|f|^s \le n^{s - r} |f|^r$ almost everywhere, so $f \in L^s$. Hence $L^r = L^s$.
If no such $c$ exists, there are sets of arbitrarily small positive measure. Fix $N$ with $N \left( \frac{s}{r} - 1 \right) \ge \frac{s}{r}$ (any $N \ge 2$ works if $s = \infty$). Choose sets $A_1, A_2, \ldots$ of positive measure with $\mu(A_1) \le 2^{-N}$ and $\mu(A_{k+1}) \le \min\{\frac{1}{4}\mu(A_k), 2^{-(k+1)N}\}$, and let $F_k = A_k \setminus \bigcup_{j > k} A_j$. The $F_k$ are disjoint, and $$\begin{aligned} \mu(F_k) &\ge \mu(A_k) - \sum_{j > k} \mu(A_j) \\ &\ge \mu(A_k)\left(1 - \tfrac{1}{4} - \tfrac{1}{16} - \cdots\right) = \tfrac{2}{3}\mu(A_k) > 0, \end{aligned}$$ while $a_k = \mu(F_k) \le 2^{-kN}$. Let $f = \sum_k c_k \chi_{F_k}$ with $c_k^r a_k = 2^{-k}$. Then $\int |f|^r = \sum_k 2^{-k} = 1$, so $f \in L^r$. If $s < \infty$, $$\begin{aligned} \int_X |f|^s \, d\mu &= \sum_k c_k^s a_k = \sum_k 2^{-ks/r} a_k^{1 - s/r} \\ &\ge \sum_k 2^{-ks/r} \, 2^{kN(s/r - 1)} \ge \sum_k 1 = \infty, \end{aligned}$$ using $1 - \frac{s}{r} < 0$ and $a_k \le 2^{-kN}$. If $s = \infty$, then $c_k = (2^{-k}/a_k)^{1/r} \ge 2^{k(N - 1)/r} \to \infty$ on sets of positive measure, so $f \notin L^\infty$. Either way $f \in L^r \setminus L^s$. $\blacksquare$
The construction in (c) never used $\mu(X) = 1$: on any measure space, sets of arbitrarily small positive measure let you build a spike that breaks $L^r \subseteq L^s$. The mirror image, with sets of arbitrarily large finite measure, builds a tail. That gives the whole picture.
For some measures, $r < s$ implies $L^r(\mu) \subseteq L^s(\mu)$; for others the inclusion is reversed; and for some, $L^r(\mu)$ and $L^s(\mu)$ don't contain each other when $r \neq s$. Give an example of each, and for $0 < r < s < \infty$ show:
Examples. Counting measure on $\mathbb{N}$: $\ell^r \subseteq \ell^s$ (the proposition on $\ell^p$). Lebesgue measure on $[0, 1]$: $L^s \subseteq L^r$ (finite measure). Lebesgue measure on $(0, \infty)$: neither (the first example on this page).
(i), "if". Let $c = \inf\{\mu(E) : \mu(E) > 0\} > 0$ and $f \in L^r$. For $t > 0$, $t^r \mu(\{|f| > t\}) \le \|f\|_r^r$, so for $t$ large, $\mu(\{|f| > t\}) < c$, hence $= 0$. Then $|f| \le t$ almost everywhere and $|f|^s \le t^{s - r} |f|^r$, so $f \in L^s$.
(i), "only if". If the infimum is $0$, the construction in part (c) of the previous exercise gives $f \in L^r \setminus L^s$.
(ii), "if". Let $C = \sup\{\mu(E) : \mu(E) < \infty\} < \infty$ and $f \in L^s$. Split at height $1$. On $\{|f| \ge 1\}$, $|f|^r \le |f|^s$. The set $G = \{0 < |f| < 1\}$ is the increasing union of $G_n = \{\frac{1}{n} < |f| < 1\}$, and $\mu(G_n) \le n^s \|f\|_s^s < \infty$, so $\mu(G_n) \le C$ and, by continuity from below, $\mu(G) \le C$. On $G$, $|f|^r \le 1$. So $\int |f|^r \le \|f\|_s^s + C < \infty$.
(ii), "only if". If the supremum is $\infty$, fix $N$ with $N\left(1 - \frac{r}{s}\right) \ge \frac{r}{s}$ and choose sets $B_1, B_2, \ldots$ of finite measure with $\mu(B_k) \ge 2\sum_{j < k} \mu(B_j) + 2^{kN + 1}$. Let $G_k = B_k \setminus (B_1 \cup \cdots \cup B_{k-1})$. They are disjoint, with $b_k = \mu(G_k) < \infty$ and $$b_k \ge \mu(B_k) - \sum_{j < k}\mu(B_j) \ge \tfrac{1}{2}\mu(B_k) \ge 2^{kN}.$$ Let $f = \sum_k c_k \chi_{G_k}$ with $c_k^s b_k = 2^{-k}$. Then $\int |f|^s = 1$, but $$\begin{aligned} \int_X |f|^r \, d\mu &= \sum_k c_k^r b_k = \sum_k 2^{-kr/s} \, b_k^{1 - r/s} \\ &\ge \sum_k 2^{-kr/s} \, 2^{kN(1 - r/s)} \ge \sum_k 1 = \infty. \end{aligned}$$ So $f \in L^s \setminus L^r$.
So "neither" happens exactly when both conditions fail: there are sets of arbitrarily small positive measure (spikes) and sets of arbitrarily large finite measure (tails). $\blacksquare$
That rule of thumb, spikes break one inclusion and tails break the other, is all you need to sort a measure space. Sort a few.
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Now the second question. A general function in $L^p$ can be awful, but it can always be approximated by simple functions, as long as $p < \infty$. The approximations have to vanish off a set of finite measure, or they won't even be in $L^p$: a simple function that equals $c \neq 0$ on a set $A$ has $\|s\|_p^p \ge |c|^p \mu(A)$.
On Upper Limits & Simple Functions, every measurable $f \ge 0$ was the pointwise limit of the staircase $s_n = \varphi_n \circ f$, with $0 \le s_1 \le s_2 \le \cdots \le f$.
Let $S$ be the set of complex simple measurable functions $s$ on $X$ with $\mu(\{s \neq 0\}) < \infty$. If $1 \le p < \infty$, then $S \subseteq L^p(\mu)$, and $S$ is dense in $L^p(\mu)$: for every $f \in L^p(\mu)$ and $\varepsilon > 0$ there is $s \in S$ with $\|f - s\|_p < \varepsilon$.
Proof. A simple $s$ takes finitely many values, so $|s| \le M$ for some $M$, and $\|s\|_p^p \le M^p \mu(\{s \neq 0\}) < \infty$. So $S \subseteq L^p(\mu)$.
First, $f \ge 0$. Let $s_n = \varphi_n \circ f$ be the staircase, so the $s_n$ are simple, $0 \le s_n \le f$, and $s_n \to f$ pointwise. Then $s_n \in L^p$. If $s_n \ge c > 0$ on a set $A$, then $c^p \mu(A) \le \|s_n\|_p^p < \infty$; applying this to each of the finitely many nonzero values gives $\mu(\{s_n \neq 0\}) < \infty$, so $s_n \in S$. Also $|f - s_n|^p \le |f|^p$, which is integrable, and $|f - s_n|^p \to 0$ pointwise. By dominated convergence, $$\|f - s_n\|_p^p = \int_X |f - s_n|^p \, d\mu \to 0.$$
General $f \in L^p$. Write $f = (u^+ - u^-) + i(v^+ - v^-)$ with $u = \operatorname{Re} f$ and $v = \operatorname{Im} f$. Each of the four parts is nonnegative and at most $|f|$, so it is in $L^p$, and the first case gives $s^{(1)}, s^{(2)}, s^{(3)}, s^{(4)} \in S$ within $\frac{\varepsilon}{4}$ of $u^+, u^-, v^+, v^-$. Then $s = (s^{(1)} - s^{(2)}) + i(s^{(3)} - s^{(4)}) \in S$, and by Minkowski's inequality (this is where $p \ge 1$ is used), $\|f - s\|_p < 4 \cdot \frac{\varepsilon}{4} = \varepsilon$. $\blacksquare$
The staircase rounds $f$ down to a multiple of $2^{-n}$ and caps it at $n$. Each step closes some of the gap, and dominated convergence says the gap's $L^p$ norm goes to $0$.
On one function you can compute every error exactly.
On $[0, 1]$ with Lebesgue measure, let $f(x) = x$ and $s_n = \varphi_n \circ f$. Compute $\|f - s_n\|_1$ for $n \ge 1$.
For $n \ge 1$ the cap $n$ is never reached (except at $x = 1$ when $n = 1$, a single point), so $s_n(x) = \lfloor 2^n x \rfloor / 2^n$. On each interval $[\frac{k}{2^n}, \frac{k+1}{2^n})$, $s_n = \frac{k}{2^n}$ and $f - s_n$ runs from $0$ up to $2^{-n}$: a triangle of area $\frac{1}{2} \cdot 2^{-n} \cdot 2^{-n}$. There are $2^n$ such intervals, so $$\|f - s_n\|_1 = 2^n \cdot \tfrac{1}{2} \cdot 2^{-2n} = 2^{-n-1} \to 0.$$
The hypothesis $p < \infty$ isn't decoration, and neither is the finite-measure support.
$S$ is not dense in $L^\infty(m)$ on $\mathbb{R}$. The constant $1$ is in $L^\infty$, but every $s \in S$ is $0$ on a set of infinite measure, where $|1 - s| = 1$; so $\|1 - s\|_\infty \ge 1$. (Simple functions without the finite-support condition are dense in $L^\infty$: the staircase converges uniformly on bounded functions. It's the finite support that fails.)
Simple functions are a big improvement, but they jump. On a space with a topology, we'd like continuous approximations. That's exactly what Lusin's theorem is for: it replaces a simple function by a continuous one, wrong only on a set of small measure, and never bigger than the original.
On Lusin's Theorem: if $\mu(\{f \neq 0\}) < \infty$, then for every $\varepsilon > 0$ there is $g \in C_c(X)$ with $\mu(\{f \neq g\}) < \varepsilon$ and $\sup|g| \le \sup|f|$, under the standing hypotheses on $X$ and $\mu$.
Let $X$ be a locally compact Hausdorff space and $\mu$ a measure on it satisfying the standing hypotheses of Lusin's theorem (Lebesgue measure on $\mathbb{R}^n$, for example). For $1 \le p < \infty$, $C_c(X)$ is dense in $L^p(\mu)$.
Proof. First, $C_c(X) \subseteq L^p(\mu)$: a $g \in C_c(X)$ is bounded and vanishes off the compact set $\operatorname{supp} g$, which has finite measure, so $\int |g|^p \le (\sup|g|)^p \mu(\operatorname{supp} g) < \infty$.
By the previous proposition it is enough to approximate each $s \in S$, since if $\|f - s\|_p < \varepsilon$ and $\|s - g\|_p < \varepsilon$ then $\|f - g\|_p < 2\varepsilon$ by Minkowski. Let $s \in S$, $M = \sup_X |s| < \infty$, and $\varepsilon > 0$. The set $\{s \neq 0\}$ has finite measure, so Lusin's theorem gives $g \in C_c(X)$ with $\sup|g| \le M$ and $\mu(\{g \neq s\}) < \varepsilon$. Then $|g - s| \le |g| + |s| \le 2M$, and $g - s = 0$ off $\{g \neq s\}$, so $$\begin{aligned} \|g - s\|_p &= \left( \int_{\{g \neq s\}} |g - s|^p \, d\mu \right)^{1/p} \\ &\le 2M \, \mu(\{g \neq s\})^{1/p} < 2M \varepsilon^{1/p}. \end{aligned}$$ Since $\varepsilon$ is arbitrary, $s$ is a limit of functions in $C_c(X)$, and so is every $f \in L^p(\mu)$. $\blacksquare$
The error bound $2M\varepsilon^{1/p}$ goes to $0$ for every finite $p$, but more slowly as $p$ grows, and at $p = \infty$ the root disappears. Watch one step function and its ramps in three norms at once.
Here's the computation behind the picture, with height $1$.
On $\mathbb{R}$ with Lebesgue measure, let $s = \chi_{[0, 1]}$ and $0 < \delta < \frac{1}{2}$, and let $g_\delta$ be the continuous function that ramps linearly from $0$ to $1$ on $[0, \delta]$, equals $1$ on $[\delta, 1 - \delta]$, ramps back down on $[1 - \delta, 1]$, and is $0$ elsewhere. Compute $\|s - g_\delta\|_p$ for $1 \le p < \infty$, and $\|s - g_\delta\|_\infty$.
On $[0, \delta]$, $|s - g_\delta| = 1 - \frac{x}{\delta}$, so $\int_0^\delta (1 - \frac{x}{\delta})^p \, dx = \frac{\delta}{p + 1}$. The right ramp gives the same, and $s = g_\delta$ everywhere else. So $$\|s - g_\delta\|_p = \left( \frac{2\delta}{p + 1} \right)^{1/p} \to 0 \quad \text{as } \delta \to 0.$$
But $|s - g_\delta| > \frac{1}{2}$ on $(0, \frac{\delta}{2})$, a set of positive measure, and $|s - g_\delta| \le 1$ everywhere, so $\frac{1}{2} \le \|s - g_\delta\|_\infty \le 1$; in fact $|s - g_\delta|$ gets arbitrarily close to $1$ near $0$, so $\|s - g_\delta\|_\infty = 1$ for every $\delta$.
Set your own $p$ and ramp width, and see how slow the big exponents are.
Ramps won't do it in $L^\infty$, and the next exercise shows that nothing continuous will.
$L^\infty$ looked at every continuous function on offer and said: no thanks, I'm keeping my jump.
Show that $\|\chi_{[0, 1]} - g\|_\infty \ge \frac{1}{2}$ for every continuous $g\colon \mathbb{R} \to \mathbb{C}$. Conclude that $C_c(\mathbb{R})$ is not dense in $L^\infty(m)$.
Proof. Suppose $\|\chi_{[0, 1]} - g\|_\infty = c < \frac{1}{2}$. Then $|g(x)| \le c$ for almost every $x < 0$, and $|1 - g(x)| \le c$ for almost every $x \in (0, 1)$. The points where these hold are dense in $(-\infty, 0)$ and in $(0, 1)$ (their complements have measure $0$, so contain no interval), so continuity of $g$ at $0$ gives $|g(0)| \le c$ and $|1 - g(0)| \le c$. Then $$1 \le |g(0)| + |1 - g(0)| \le 2c < 1,$$ a contradiction. So no continuous function, compactly supported or not, comes within $\frac{1}{2}$ of $\chi_{[0, 1]}$ in $L^\infty$. $\blacksquare$
On $\mathbb{R}^n$ there's a bonus. Two continuous functions that agree almost everywhere agree everywhere (if they differed at a point, they'd differ on an open set around it, which has positive measure). So on $C_c(\mathbb{R}^n)$, $\|f - g\|_p$ is a genuine metric, with no equivalence classes needed, and the theorem says $L^p(\mathbb{R}^n)$ is the completion of $C_c(\mathbb{R}^n)$ in that metric, for $1 \le p < \infty$ (completeness is the Riesz–Fischer theorem). For $p = 1$, $\|f\|_1$ on $C_c(\mathbb{R}^n)$ is just the Riemann integral of $|f|$. So $L^1$ is what you get from Riemann-integrable continuous functions by filling in the holes.
Density also makes $L^p$ small in a precise sense: a countable set is enough to get close to everything.
Let $m$ be Lebesgue measure on $[-\pi, \pi]$. Show that $L^p(m)$ is separable (has a countable dense subset) if $1 \le p < \infty$, but $L^\infty(m)$ is not separable.
$1 \le p < \infty$. Let $D$ be the set of finite sums $\sum_j c_j \chi_{I_j}$, where each $c_j$ has rational real and imaginary parts and each $I_j \subseteq [-\pi, \pi]$ is an interval with rational endpoints (intersected with $[-\pi, \pi]$). $D$ is countable. We show it is dense.
By the density of $S$ it is enough to approximate $s = \sum_{j=1}^{n} \alpha_j \chi_{A_j} \in S$. Let $\eta > 0$. For one measurable $A \subseteq [-\pi, \pi]$: outer regularity of Lebesgue measure (from Lebesgue Measure) gives an open $V \supseteq A$ with $m(V \setminus A) < \frac{\eta}{3}$, and we may take $V \subseteq (-\pi - 1, \pi + 1)$. $V$ is a countable disjoint union of open intervals with total length $m(V) < \infty$, so finitely many of them, $J_1, \ldots, J_N$, cover all of $V$ except a set of measure less than $\frac{\eta}{3}$. Shrink each $J_i$ to an interval with rational endpoints, losing less than $\frac{\eta}{3N}$ each, and intersect with $[-\pi, \pi]$; call the union $U$. Then $m(A \,\triangle\, U) < \eta$, and $$\|\chi_A - \chi_U\|_p = m(A \,\triangle\, U)^{1/p} < \eta^{1/p}.$$ Do this for each $A_j$ to get $U_j$, and pick rational $c_j$ with $|c_j - \alpha_j| < \eta$. Then $t = \sum_j c_j \chi_{U_j} \in D$ (a union of finitely many rational intervals is a finite sum of their indicators, after splitting into disjoint pieces), and by Minkowski $$\begin{aligned} \|s - t\|_p &\le \sum_j |\alpha_j| \, \|\chi_{A_j} - \chi_{U_j}\|_p + \sum_j |\alpha_j - c_j| \, \|\chi_{U_j}\|_p \\ &\le \Big(\sum_j |\alpha_j|\Big) \eta^{1/p} + n \eta (2\pi)^{1/p}, \end{aligned}$$ which is small when $\eta$ is.
$L^\infty$. For $-\pi < t < \pi$ let $f_t = \chi_{[-\pi, t]}$. If $t < u$, then $f_u - f_t = \chi_{(t, u]}$ is $1$ on a set of positive measure, so $\|f_t - f_u\|_\infty = 1$. So the open balls $B(f_t, \frac{1}{2})$ are pairwise disjoint, and there are uncountably many of them. A dense set would need a point in each ball, and no countable set has a point in uncountably many disjoint sets. $\blacksquare$
On Orthonormal Bases, a later page, $[-\pi, \pi]$ with its ends glued together becomes the circle $T$, and this exercise says $L^p(T)$ is separable. There, the trigonometric polynomials give a second countable dense set.
So $C_c$ is dense in every $L^p$ with $p < \infty$, and not in $L^\infty$. What is its closure in the sup norm? A uniform limit of continuous functions is continuous, but it can lose compact support: $e^{-x^2}$ is never $0$, yet you can cut it off far out and change it by almost nothing. What survives is the idea of getting small far away.
A complex function $f$ on a locally compact Hausdorff space $X$ vanishes at infinity if for every $\varepsilon > 0$ there is a compact set $K \subseteq X$ with $|f(x)| < \varepsilon$ for all $x \notin K$.
On $\mathbb{R}$, "outside a compact set" means "for $|x|$ large", so this is $f(x) \to 0$ as $x \to \pm\infty$. The functions that do it continuously get a name.
$C_0(X)$ is the set of continuous complex functions on $X$ that vanish at infinity, with the norm $\|f\|_\infty = \sup_{x \in X} |f(x)|$.
Every $f \in C_c(X)$ vanishes at infinity (take $K = \operatorname{supp} f$), so $C_c(X) \subseteq C_0(X)$. If $X$ is compact, every closed set is compact, so $C_c(X) = C_0(X) = C(X)$. On a noncompact space the difference is real.
On $X = \mathbb{R}$, sort each function into $C_c$, $C_0$ but not $C_c$, or neither: $\max\{0, 1 - |x|\}$, $e^{-x^2}$, $\frac{1}{1 + x^2}$, $\arctan x$, $1$. Then do the same for $x(1 - x)$ and $\frac{1}{x}$ on $X = (0, 1)$.
$\max\{0, 1 - |x|\}$ is continuous and $0$ off $[-1, 1]$: it is in $C_c(\mathbb{R})$. $e^{-x^2}$ and $\frac{1}{1 + x^2}$ tend to $0$ at both ends but are never $0$, so their supports are all of $\mathbb{R}$: they are in $C_0$ but not $C_c$. $\arctan x \to \pm\frac{\pi}{2}$ and $1 \to 1$, so neither vanishes at infinity.
On $(0, 1)$, a compact set stays away from both ends, so vanishing at infinity means tending to $0$ at $0$ and at $1$. $x(1 - x)$ does, and it is positive on all of $X$, whose closure in $X$ is $X$ itself, not compact: $C_0$ but not $C_c$. $\frac{1}{x}$ is continuous on $(0, 1)$ but blows up at $0$: not in $C_0$.
Now the theorem: $C_0(X)$ is exactly what $C_c(X)$ fills in to under the sup norm. Multiply $f$ by a Urysohn cutoff that is $1$ on a big compact set, and the product has compact support and is within $\varepsilon$ of $f$.
On Locally Compact Spaces, Urysohn's lemma gave, for compact $K$ inside open $V$, a $g \in C_c(X)$ with $\chi_K \le g \le \chi_V$.
That's the density half. The other half is that nothing outside $C_0(X)$ can be reached, because $C_0(X)$ is already complete. Together:
If $X$ is a locally compact Hausdorff space, then $C_0(X)$ is the completion of $C_c(X)$ in the norm $\|f\|_\infty = \sup_{x \in X} |f(x)|$: $C_c(X)$ is dense in $C_0(X)$, and $C_0(X)$ is complete.
Proof. Every $f \in C_0(X)$ is bounded: with $\varepsilon = 1$, $|f| < 1$ off a compact $K$, and $|f|$ is bounded on $K$ by continuity. So $\|\cdot\|_\infty$ is a norm on $C_0(X)$.
Density. Let $f \in C_0(X)$ and $\varepsilon > 0$. Pick a compact $K$ with $|f| < \varepsilon$ off $K$. Urysohn's lemma (with $V = X$) gives $g \in C_c(X)$ with $\chi_K \le g \le 1$. Then $h = fg$ is continuous and vanishes off $\operatorname{supp} g$, so $h \in C_c(X)$. On $K$, $f - h = f(1 - g) = 0$; off $K$, $|f - h| = |f|(1 - g) \le |f| < \varepsilon$. So $\|f - h\|_\infty \le \varepsilon$.
Completeness. Let $(f_n)$ be Cauchy in $C_0(X)$. Since $|f_n(x) - f_m(x)| \le \|f_n - f_m\|_\infty$, the sequence converges uniformly to some $f$, and $f$ is continuous as a uniform limit of continuous functions. Given $\varepsilon > 0$, pick $n$ with $\|f_n - f\|_\infty < \frac{\varepsilon}{2}$ and a compact $K$ with $|f_n| < \frac{\varepsilon}{2}$ off $K$. For $x \notin K$, $$|f(x)| \le |f(x) - f_n(x)| + |f_n(x)| < \frac{\varepsilon}{2} + \frac{\varepsilon}{2} = \varepsilon.$$ So $f \in C_0(X)$, and $\|f_n - f\|_\infty \to 0$. $\blacksquare$
On $C_c(\mathbb{R}^n)$ the essential sup and the honest sup agree (a continuous function bigger than $c$ at a point is bigger than $c$ on an open set), so this is the $p = \infty$ end of the completion story: the $L^p$-completion of $C_c(\mathbb{R}^n)$ is $L^p(\mathbb{R}^n)$ for $p < \infty$, but its $L^\infty$-completion is only $C_0(\mathbb{R}^n)$. Try the cutoff yourself.
For a function that tends to something other than $0$, no cutoff helps: the error far out is stuck at the limit. Sort some functions.
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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 proofs. Each question shows one line of a proof from this page with a hole in it.
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Then 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 interpolation bound is one instance of a bigger fact: $p \mapsto \log \int |f|^p$ is convex. From that one fact you get which exponents a function can live in, and a surprise about what that set can look like.
Let $f$ be a complex measurable function on $X$, $\mu$ a positive measure, and $\|f\|_\infty > 0$. For $0 < p < \infty$ let $$\varphi(p) = \int_X |f|^p \, d\mu = \|f\|_p^p,$$ and let $E = \{p \in (0, \infty) : \varphi(p) < \infty\}$.
Since $\|f\|_\infty > 0$, $f \neq 0$ on a set of positive measure, so $\varphi(p) > 0$ for every $p$, and $\log \varphi$ makes sense on $E$.
(a) At each point, $|f|^p \le |f|^r + |f|^s$: where $|f| \le 1$, $|f|^p \le |f|^r$, and where $|f| > 1$, $|f|^p \le |f|^s$. So $\varphi(p) \le \varphi(r) + \varphi(s) < \infty$.
(b) Let $r, s \in E$ and $0 < t < 1$, and $p = (1 - t)r + ts$. Hölder's inequality with the conjugate exponents $\frac{1}{1 - t}$ and $\frac{1}{t}$ gives $$\begin{aligned} \varphi(p) &= \int_X |f|^{(1 - t)r} \, |f|^{ts} \, d\mu \\ &\le \left( \int_X |f|^r \, d\mu \right)^{1 - t} \left( \int_X |f|^s \, d\mu \right)^{t} = \varphi(r)^{1 - t} \varphi(s)^t, \end{aligned}$$ so $\log \varphi(p) \le (1 - t) \log \varphi(r) + t \log \varphi(s)$: $\log \varphi$ is convex on $E$.
For continuity, let $p_n \to p$ in $E$. Pick $r, s \in E$ with $r \le p_n \le s$ for all $n$ and $r \le p \le s$: if $p$ is interior to $E$, take $r < p < s$ in $E$ and drop the finitely many $p_n$ outside $[r, s]$; if $p$ is an endpoint of $E$, use $p$ itself for that side. By (a), $|f|^{p_n} \le |f|^r + |f|^s$, which is integrable, and $|f|^{p_n} \to |f|^p$ pointwise (where $f \neq 0$; where $f = 0$ both are $0$). Dominated convergence gives $\varphi(p_n) \to \varphi(p)$.
(c) $E$ can be any interval in $(0, \infty)$: open, closed, half-open, a single point, empty, bounded or not. Work on $(0, \infty)$ with Lebesgue measure and build $f$ from a piece near $0$ (which controls the top of $E$) and a disjoint piece near $\infty$ (which controls the bottom); then $\varphi$ is the sum of the two pieces' integrals, and $E$ is the intersection of their sets of good $p$. For $0 < a, b < \infty$:
Choosing one piece of each kind gives $E = (a, b)$, $[a, b)$, $(a, b]$, $[a, b]$ (a single point when $a = b$), with $a$ replaced by $0$ or $b$ by $\infty$ when the $e^{-x}$ or $\chi_{(0, 1/2)}$ piece is used, and $E = \emptyset$ from $x^{-1/a}$ on both ends (or simply $f = \frac{1}{x}$ on $(0, \infty)$). Every connected subset of $(0, \infty)$ is one of these, so the answer is yes to all three questions: $E$ need be neither open nor closed, it can be a single point, and it can be any connected subset of $(0, \infty)$.
(d) If $\|f\|_r$ or $\|f\|_s$ is $\infty$ there is nothing to prove. Otherwise $r, s \in E$; write $p = (1 - t)r + ts$ with $0 < t < 1$ and $M = \max\{\|f\|_r, \|f\|_s\}$. By (b), $$\begin{aligned} \|f\|_p^p = \varphi(p) &\le \varphi(r)^{1 - t} \varphi(s)^t = \|f\|_r^{r(1 - t)} \, \|f\|_s^{st} \\ &\le M^{(1 - t)r + ts} = M^p, \end{aligned}$$ so $\|f\|_p \le M$. In particular, if $f \in L^r(\mu) \cap L^s(\mu)$ then $\|f\|_p < \infty$, so $f \in L^p(\mu)$. (If instead $f = 0$ almost everywhere, all three norms are $0$.) $\blacksquare$