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Let $X$ be a closed subspace of $L^1[0,1]$. If $\forall f \in X$ exists $p>1$ such that $f \in L^p[0,1]$ then prove that exists $p>1$ such that $X \subseteq L^p[0,1]$.

I presume that I have to use the uniform boundedness principle in $X$ which is a Banach space because it is a closed subspace of the Banach space $L^1[0,1]$.

I believe that I have to define a collection of linear operators which depend on the set of indices $\{p\mid p>1\}$.

Can someone help me to define the appropriate operators to complete my exercise?

Thank you in advance.

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I don't see a natural way to use the uniform boundedness principle here. One can however directly apply Baire's theorem.

As you say, since $X$ is closed in $L^1[0,1]$, it is a Banach space, so a Baire space. For $1 < p < +\infty$, the space $L^p[0,1]$ is reflexive, hence $A(p,K) = \{ f \in L^p[0,1] : \lVert f\rVert_p \leqslant K\}$ is weakly compact. The inclusion $i_p \colon L^p[0,1] \hookrightarrow L^1[0,1]$ is continuous, hence $i_p(A(p,K))$ is weakly compact and therefore weakly closed in $L^1[0,1]$. Since $A(p,K)$ is convex, it follows that $i_p(A(p,K))$ is closed in the norm topology of $L^1[0,1]$, and consequently $X \cap i_p(A(p,K))$ is a closed subset of $X$. Since $L^{p_2}[0,1] \subset L^{p_1}[0,1]$ for $1 \leqslant p_1 \leqslant p_2$, the assumption that every $f\in X$ lies in some $L^p[0,1]$ with $p > 1$ implies that

$$X = \bigcup_{k,n} i_{1 + 1/n} \bigl(A(1+1/n,k)\bigr).$$

By Baire's theorem, there are $n,k \in \mathbb{N}$ such that $i_{1+1/n}\bigl(A(1+1/n,k)\bigr)$ has nonempty interior in $X$. Then it follows that $X \subset L^{1+\frac{1}{n}}[0,1]$.

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