Prove that the closed unit ball of $L^2[a,b]$ is closed in $L^1[a,b]$.
My friends and I have literally been pouring over this problem for days now without success. We've been using Hölder's inequality, and especially its special case the Cauchy-Schwarz Inequality. We've also been using Minkowski's (the triangle) Inequality, and also the inequality:
$||f||_{p_1}\leq(b-a)^{\frac{p_2-p_1}{p_1p_2}}\;||f||_{p_2}$ for $1\leq p_1\leq p_2\leq\infty$
The other inequalities being:
$||fg||_1\leq||f||_p\cdot ||f||_q$ for $\frac{1}{p}+\frac{1}{q}=1$ and if $p=1$ then $q=\infty$ (Hölder's)
$||f+g||_p\leq ||f||_p+||g||_p$ (Triangle)
$||fg||_1\leq ||f||_2\cdot ||g||_2$ (Cauchy-Schwarz)
I'll just say that we've been a lot of places with these inequalities, but nowhere we want to go. We defined $A$ to be the closed unit ball in $L^2[a,b]$:
$$A:=\{\;f\in L^2[a,b] : ||f||_2\leq 1\}.$$
And now we want to consider functions in $L^1[a,b]$ (and thus with the $L^1$ metric), and prove that if every open ball around a function contains a point from $A$, then that function is actually in $A$. Can anyone help us with this? Thanks.
We can generalize to $L^2(X)$ where $X$ is any measure space and other exponents.
As we are the context of metric spaces, we only need to check sequential closeness. Fix $\{f_n\}$ a sequence in the closed unit ball of $L^2(X)$ which converges to $f$ in $L^1$; then $\int_X|f_n|^2\leqslant 1$ for all $n$. Now extract from $\{f_n\}$ a sub-sequence which converges almost everywhere and use Fatou's lemma.