Show that $A_{n}$ are no-where dense sets. For positive integer $n$ Let,
$$A_{n}=\left\{\ f \in L^{1}[0,1] | \int|f(x)|^{2}dx \leq n\right\}.\ $$
Show that $A_{n}$ is a closed set for each $n$ in $L^{1}[0,1]$ with empty interior.
I am not getting a clue!! I suppose it says to prove that $A_{n}$ are no-where dense sets. I was trying to define some continuous functions so that I can these $A_{n}$'s as inverse image of closed sets but couldn't do so. But $L^{2}[0,1] \subset L^{1}[0,1]$. Will that yield something??
Thanks in advance!!
 A: To show $A_n$ is nowhere dense, first note that $A_n$ is convex (as $x \mapsto x^2$ is convex) and $-f \in A_n$ whenever $f \in A_n.$ So then if $A_n$ wasn't nowhere dense, then there is $g \in L^1[0,1]$ and $\delta > 0$ such that $B(g,\delta) \subset A_n$ and so $B(0,\delta) \subset A_n$ using the above properties.
To show this gives a contradiction, we need to find a function $f \in L^1[0,1]$ such that $\lVert f \rVert_1 \leq \delta$ but $\lVert f \rVert_2 \geq \sqrt{n}.$ For this note that for $k>1,$ $f_k = k 1_{[0,1/k]}$ satisfies $\lVert f_k \rVert_1 = 1$ while $\lVert f_k \rVert_2 = \sqrt{k}$ so the function $f = \delta f_{n/\delta^2}$ will do the trick.
To show $A_n$ is closed, suppose $f_m \in A_n$ is a sequence such that $f_m \rightarrow f$ in $L^1[0,1].$ Then a general result in measure theory allows us to pass to an a.e. convergent subsequence $f_{m_k} \rightarrow f.$ Then by Fatou's lemma,
$$ \int_0^1 |f(x)|^2 dx = \int_0^1 \liminf_{k\rightarrow \infty} |f_{m_k}(x)|^2 dx \leq \liminf_{k \rightarrow \infty} \int_0^1 |f_{m_k}(x)|^2 dx \leq n.$$
So $f \in A_n.$
Side note: I think there should be an easier way of proving that $A_n$ is closed, without passing to a subsequence. Some kind of measure theoretic argument is likely unavoidable though, since the $\lVert \cdot \rVert_2$ norm is stronger than the $\lVert \cdot \rVert_1$ norm.
Edit in response to comments: To show $A_n$ is convex, note that for $f,g \in A_n$ and $t \in [0,1],$ note that for all $x \in [0,1],$
$$ \left| tf(x) + (1-t)g(x)\right|^2 \leq \left( t|f(x)| + (1-t)|g(x)|\right)^2 \leq t |f(x)|^2 + (1-t) |g(x)|^2. $$
The first step is the triangle inequality and the second holds as $x \mapsto x^2$ is convex (second derivative is everywhere positive). Now integrating both sides gives,
$$ \int_0^1 |tf(x)+(1-t)g(x)|^2 dx \leq t \int_0^1 |f(x)|^2 dx + (1-t) \int_0^1 |g(x)|^2 dx \leq n. $$
So $tf+(1-t)g \in A_n.$
