Brezis' Exercise 4.8

Question

I'm trying to solve below exercise in Brezis' Functional Analysis, i.e.,

Exercise 4.8 Let $(\Omega, \mathcal F, \mu)$ be a $\sigma$-finite measure space. Let $X$ be a closed vector subspace of $L^1 (\Omega)$ such that $$ X \subset \bigcup_{q \in (1, \infty]} L^q (\Omega). $$

1. Prove that there is $p>1$ such that $X \subset L^p (\Omega)$.
2. Prove that there is $C \in [0, \infty)$ such that $$ \|f\|_p \le C \|f\|_1 \quad \forall f \in X. $$

Could you confirm if my below attempt is correct?

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Proof 1. It follows from $X$ is closed in $L^1 (\Omega)$ that $X$ is complete w.r.t. $\| \cdot \|_1$. For each $n \in \mathbb N^*$, $$ \begin{align} Y_n &:= \{f \in L^1 (\Omega) : \|f\|_{1+(1/n)} \le n\} \\ X_n &:= X \cap Y_n . \end{align} $$

We have $Y_n$ is closed in $L^1 (\Omega)$. Then $X_n$ is closed in $(X, \|\cdot\|_1)$. Let $f \in X \cap L^q (\Omega)$ for some $q>1$. By interpolation inequality, $$ \|f\|_r \le \|f\|_1^\alpha \|f\|_q^{1-\alpha} \quad \forall r \quad \text{s.t.} \quad \frac{1}{r} = \frac{\alpha}{1} + \frac{1-\alpha}{q}. $$

So $X = \bigcup_n X_n$. By Baire category theorem, there is $p$ such that $$ \operatorname{int}_{(X, \|\cdot\|_1)} (X_p) \neq \emptyset, $$ where $\operatorname{int}_{(X, \|\cdot\|_1)} (X_p)$ is the interior of $X_p$ in $(X, \|\cdot\|_1)$. Fix $f \in \operatorname{int}_{(X, \|\cdot\|_1)} (X_p)$. There is $r>0$ such that $$ B_{(X, \|\cdot\|_1)} (f, r) \subset X_p \subset L^p(\Omega). $$

Here $B_{(X, \|\cdot\|_1)} (f, r)$ is the open ball centered at $f$ with radius $r$ in $(X, \|\cdot\|_1)$. Let $g \in X$ be arbitrary. Because $X$ is a vector space, there is $\lambda>0$ such that $$ \lambda f + (1-\lambda)g \in B_{(X, \|\cdot\|_1)} (f, r). $$

So $\lambda f + (1-\lambda)g \in L^p(\Omega)$. We already have $f \in L^p(\Omega)$. Thus $g \in L^p(\Omega)$. Hence $X \subset X_p$.

2. Consider the embedding $$ T:X \to L^p (\Omega), f \mapsto f. $$

Then $T$ is linear. It suffices to prove that $T$ is continuous. Notice that $X$ and $L^p (\Omega)$ are both Banach spaces. By closed graph theorem, it suffices to prove that the graph of $T$ is closed.

Let $u,u_n \in X$ and $f \in L^p (\Omega)$ such that $(u_n, u_n) \to (u, f)$ in $X \times L^p$. There is a sub-sequence $\varphi$ of $\mathbb N$ such that $u_{\varphi (n)} \to u$ $\mu$-a.e. Clearly, $(u_{\varphi(n)})$ is a Cauchy sequence in $L^p (\Omega)$. So $u_{\varphi (n)} \to u$ in $L^p$. Hence $u=f$ $\mu$-a.e. This completes the proof.

Answer 1

The proof that $X=\bigcup_nX_n$ is not very clear to me. Here is a short proof:

If $f\in X$, then $f\in L_1\cap L_p$ for some $1<p\leq\infty$ and so, $f\in L_r$ for all $1<r<p$. Without loss of generality, we may assume that $1<p<\infty$. By choosing $n\in\mathbb{N}$ large enough so that $\|f\|_1\vee\|f\|_p<n$ and $1+\tfrac1n<p$ we obtain $f\in L_{1+1/n}$.

Here we use the fact that if $0<p<q\leq \infty$, then $L_p\cap L_q\subset L_r$ for all $p<r<q$, and $\|f\|_r\leq \|f\|_p\vee\|f\|_r$, which follows from Hölder's inequality.

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Some other observations:

1. In part (1), the $p$ (which in the context of your argument is an integer) fir which $\operatorname{int}(X_p)\neq\emptyset$ is not the $p$ for which $X\subset L_p$. It is $q:=1+\tfrac1p$ the index for which $X\subset L_q$.
2. in part (b), the fat that $u_{\phi(n)}$ is Cauchy in $L_q$ ($q=1+\tfrac1p$) is not relevant. You have that $u_{\phi(n)}\xrightarrow{n\rightarrow\infty} u$ $\mu-a.s.$ and that $u_{\phi(n)}\xrightarrow{n\rightarrow\infty}f$ in $L_p$ (since $u_n$ converges to $f$ in $L_p$). Then, along a subsequence $psi(n)$ of $\phi$ $u_{psi(n)}\xrightarrow{n\rightarrow\infty}f$ $\mu$-a.s. It then follows that $u=f$ $\mu$-a.s.

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