On finite measurable space $X$, the whole of $L^p(X)$ is closed in $L^1(X)$ iff there is a const $C$ st $||f||_p\leq C||f||_1, \forall f \in L^p(X)$

Question

On finite measurable space $(X, \mathcal{M}, \mu)$, the whole of $L^p(X, \mu)(p>1)$ is closed in $L^1(X,\mu)$ iff there is a const $C$ st $||f||_p\leq C||f||_1, \forall f\in L^p(X)$, iff both $L^p(X,\mu)$ and $L^1(X,\mu)$ are finite dimensional.

At first, I tried to prove if there is no such const $C$ st $||\frac{f}{||f||_1}||_p\leq C, \forall f\in L^p(X) \text{ and } ||f||_1 \neq 0$, then for each $n\in \mathbb{N_+}$, there exists $f_n\in L^p(X), ||f_n||_1=1 $ st $||f_n||_p >n.$ We can't claim that $\{f\in L^1: ||f||_1=1\}$ is compact because we don't know whether $L^1(X,\mu)$ is finite dimensional. So I don't know how to find a subsequence converge $f_{n_k}$ convergent to $f_0$ which is not $L^p$ integrable.

Answer 1

• Suppose that $L^p$ is closed in $L^1$: in order to prove the existence of $C$, we only need to show that $L^1\subset L^p$, then we use the closed graph theorem. Let $f\in L^1$; define $f_n:=f\chi_{\{|f|\leqslant n\}}$. Then the sequence $(f_n)_{n\geqslant 1}$ is an element of $L^p$, which converges in the $L^1$ norm to $f$. By closeness, $f$ belongs to $L^p$.

• If $\lVert f\rVert_p\leqslant C\lVert f\rVert_1$, then taking $f=\chi_A$ we can see that $\mu(A)\geqslant \alpha\gt 0$ for some universal constant $\alpha$ (depending on $C$). We thus obtain that $\lVert f\rVert_\infty\leqslant \alpha^{-1}\lVert f\rVert_1$. Consider $f_c(x):=\sum_{j=1}^Nc_jf_j(x)$ for some set $f_1,\dots,f_N$ of orthonormal elements of $L^2(X,\mathcal M,\mu)$. Then we can show that the dimension of $L^\infty(X,\mathcal M,\mu)$ is finite. Indeed, let $(c_k)_{k\geqslant 1}$ be a dense subset of the unit ball of $\mathbf R^n$. Then for each $k$ there is some $N_k$ of measure $0$ for which $|f_{c_k}(x)|\leqslant \alpha^{-1}\sqrt N$ if $x\in X\setminus N_k$ and by density, there is some $N$ of measure $0$ for which the equality $|f_c(x)|\leqslant \alpha^{-1}\sqrt N$ holds for each $x\in X\setminus N$ and each $c$ in the unit ball of $\mathbf R^n$. Considering $f_{c(x)}(x)$ for an appropriate $c(x)$ we obtain that $\sqrt N\leqslant \alpha^{-1}$.