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Suppose $1<p<\infty$ and let $L^1$ and $L^p$ denote the usual Lebesgue spaces on $[0,1]$. Let $$A=\{f\in L^1:\|f\|_p\leq 1\}.$$ Show $A$ is closed in $L^1$.

I took a sequence $\{f_n\}$ in $A$ and assumed it converges to $f$ in $L^1$. I am having trouble showing $\|f\|_p\leq 1$.

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" assumed it converges to $f$ in $L^1$. I was able to prove $f$ itself is in $L^1$" I'm confused. There's nothing to prove there. – Alex Becker Jul 9 '12 at 20:32
up vote 4 down vote accepted

A subsequence $\{f_{n_k}\}$ converges to $f$ almost everywhere. Indeed, pick $\{f_{n_k}\}$ a subsequence such that $\lVert f-f_{n_k}\rVert\leq 2^{-k}$. Then $$\lambda\{x\mid |f(x)-f_{n_k}(x)|\geq j^{-1}\}\leq j2^{-k}$$ so $$\lambda\bigcup_{k\geq k_0}\{x\mid |f(x)-f_{n_k}(x)|\geq j^{-1}\}\leq j2^{-k_0+1}$$ and the set on which $\{f_{n_k}\}$ doesn't converge to $0$ is of null measure. By Fatou's lemma, $$\int |f|^p=\int\liminf_k|f_{n_k}|^p\leq \liminf_k\int |f_{n_k}|^p\leq 1.$$

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To show $A$ is closed, show its complement is open. Suppose $f_0 \in L^1 \backslash A$.

Case 1: suppose $f_0 \in L^p$, with $\|f_0\|_p > 1$. Since $(L^p)^* \sim L^q$ where $1/p + 1/q = 1$, there is $g \in L^q$ such that $\|g\|_q \le 1$ and $\int_{[0,1]} f_0 g > 1$. If $g_M = \max(\min(g, M),-M)$, for some $M$ we have $\int_{[0,1]} f_0 g_M > 1$ (using Dominated Convergence). Of course $\|g_M\|_q \le \|g\|_q \le 1$. Since $g_M$ is bounded, $\{f \in L^1: \int_{[0,1]} f g_M > 1\}$ is a neighbourhood of $f_0$ in $L^1$, and it is disjoint from $A$.

Case 2: suppose $f_0 \notin L^p$, i.e. $\int_[0,1] |f_0|^p = \infty$. If $E(M) = \{x \in [0,1]: |f_0(x)| \le M$, then $\lim_{M \to \infty} \int_{E(M)} |f_0|^p = \infty$, so there is some $M$ such that $1 < \int_{E(M)} |f_0|^p < \infty$. Apply the previous case, with $[0,1]$ replaced by $E(M)$, obtaining some bounded $g_M$ on $E(M)$ with $\int_{E(M)} f_0 g_M > 1$. We take $g_M$ to be $0$ outside $E(M)$, and again $\{f \in L^1: \int_{[0,1]} f g_M > 1$ is a neighbourhood of $f_0$ in $L^1$ disjoint from $A$.

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