Condition for $L^r(\mu)=L^\infty(\mu)$ where $\mu(X)=1$

Question

This is Exercise 3.5 in Rudin's Real and Complex Analysis. Suppose $\mu$ is a positive measure on a set $X$, with $\mu(X)=1$. Rudin askes: Under what conditions do the two spaces $L^r(\mu)$ and $L^s(\mu)$ contain the same functions, where $0<r<s\leq \infty$?

I am trying to show the case where $s=\infty$.

It is easy to show that $L^r(\mu)\supset L^\infty(\mu)$ using the Jensen's inequality. But I have no idea for finding a condtion on $X$ so that $L^r(\mu)=L^\infty(\mu)$. Any hints?

Answer 1

I claim that if $\mu(X) = 1$ and $0 < r < s \leq \infty$, then the condition $L^r(\mu) = L^s(\mu)$ is equivalent to the existence of $c > 0$ such that $\mu(E) = 0$ or $\mu(E) \geq c$ for all measurable sets $E \subset X$.

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"$\Rightarrow$": Assume $L^r(\mu) = L^s(\mu)$. Then the linear map $$ \Phi : L^r(\mu) \to L^s(\mu), f \mapsto f $$ is well-defined. Furthermore, if $f_n \to f$ in $L^r$ and $f_n \to g$ in $L^s$, then also $f_n \to g$ in $L^r$, since (as you noted yourself) $L^s \hookrightarrow L^r$. This implies that $\Phi$ has closed graph. By the closed graph theorem (which also applies to quasi-Banach spaces; in fact, it applies to F-spaces), this means that $\Phi$ is a bounded linear map, say $\| f \|_{L^s} \leq C \cdot \| f \|_{L^r}$ for all $f \in L^r$. For $E \subset X$ measurable with $\mu(E) > 0$, this means $$ [\mu(E)]^{1/s} = \| 1_E \|_{L^s} \leq C \cdot \| 1_E \|_{L^r} = C \cdot [\mu(E)]^{1/r} , $$ and hence $[\mu(E)]^{\frac{1}{r} - \frac{1}{s}} \geq C$, so that finally $\mu(E) \geq C^{1/(\frac{1}{r} - \frac{1}{s})} =: c$, where we used that $\frac{1}{r} - \frac{1}{s} > 0$.

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"$\Leftarrow$": Assume that $\mu(E) = 0$ or $\mu(E) \geq c$ for all measurable $E \subset X$. It suffices to show (why?!) that $L^s (\mu) \hookrightarrow L^\infty(\mu)$. In fact, I claim that if $f \in L^s$, then $\| f \|_{L^\infty} \leq c^{-1/s} \| f \|_{L^s}$. Assume towards a contradiction that this is false; in particular, $\| f \|_{L^s} > 0$. Since $\| f \|_{L^\infty} > c^{-1/s} \| f \|_{L^s}$, there is $\lambda > c^{-1/s} \| f \|_{L^s}$ such that $E := \{ x : |f(x)| \geq \lambda \}$ satisfies $\mu(E) > 0$ and hence $\mu(E) \geq c$. This implies $$ \lambda \cdot c^{1/s} \leq \lambda \cdot [\mu(E)]^{1/s} = \| \lambda \cdot 1_{E} \|_{L^s} \leq \| f \|_{L^s} $$ and hence $\lambda \leq c^{-1/s} \| f \|_{L^s}$, contradicting our choice of $\lambda$.

Answer 2

Let's say $(X,\mu)$ has the "small measure property" (SMP) if there exist pairwise disjoint $E_1,E_2,\dots\subset X$ such that $\mu(E_n)>0$ for all $n$ and $\mu(E_n)\to 0.$

Claim: Let $0<r<\infty.$ Then $L^r=L^\infty$ iff $X$ does not have the SMP.

Proof: $\implies$: Suppose, to reach a contradiction, $X$ has the SMP. Then we can choose pwdj $E_1,E_2,\dots$ such that $0<\mu(E_n)<2^{-n}.$ Define $f = \sum_{n=1}^{\infty} n\chi_{E_n}.$ Then

$$\int_X f^r\,d \mu = \sum_{n=1}^{\infty} n^r\mu(E_n)< \sum_{n=1}^{\infty} n^r2^{-n} < \infty.$$

So $f\in L^r.$ But $f\notin L^\infty$ because $f = n$ on $E_n, n=1,2,\dots,$ contradiction.

$\impliedby$: Let $f\in L^r.$ Set $E_n =\{n\le |f|<n+1\}.$ Note that the $E_n$ are pwdj. We have

$$\infty >\int_X|f|^r\,d\mu = \sum_{n=1}^{\infty}\int_{E_n}|f|^r\,d\mu \ge \sum_{n=1}^{\infty}n^r\mu(E_n).$$

Because the series on the right converges, we have $n^r\mu(E_n)\to 0,$ which certainly implies $m(E_n)\to 0.$ Since $X$ does not have the SMP, only finitely many $E_n$ have positive measure. Thus there is $N$ such that $\mu(E_n)=0$ for $n \ge N.$ This implies $|f|< N$ a.e., and thus $f\in L^\infty$ as desired.