# 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??

• Well, the only possible function here is $f\mapsto \int |f(x)|^2$, so if you can show that it is continious it is closed. – jeanmfischer May 31 '17 at 20:14
• To show that $A_n$ has empty interior, give yourself an $f$ satisfying the inequality and show that for any $\epsilon > 0$ there is a $g \in L^1$ such that $||f - g||_1 < \epsilon$ but $g \notin A_n$. It may be helpful to work on the case $n = 1$ and ask yourself how two functions could be close to each other in the $L^1$ sense but not in the $L^2$ sense. Hint: squaring a function doesn't change the support of the function. – user217285 May 31 '17 at 20:26
• @jeanmfischer the function that you have written is not continuos..atleast I couldn't show it!! – Riju May 31 '17 at 20:34
• @Nitin Can you elaborate more or write an answer? how can I show that the set is closed? – Riju May 31 '17 at 20:35

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 $tf+(1-t)g \in A_n.$

• the proof of $A_{n}$ is closed is fine. But what I don't understand is the following: by $\delta f_{n/\delta}$, I assume you mean $n1_{[0,\delta/n]}$. But we have its one norm as $\delta$ but its two norm is $n \delta$ which is not helping I suppose. – Riju May 31 '17 at 20:59
• And one more thing is that in the first paragraph I suppose you have argued some how that it is enough to show that 0 is not an interior point? – Riju May 31 '17 at 20:59
• But I didn't get the argument.. Why is the set convex. Can you elaborate on that? – Riju May 31 '17 at 21:01
• It seems like my calculation of $\lVert f_k \rVert$ was wrong, so I've fixed my answer accordingly. I also updated my answer to explain why $A_n$ is convex in more detail. The first paragraph is a reduction step, showing that it's sufficient to show 0 is not an interior point. The second paragraph then proves this assertion. You can probably prove directly that every point is not an interior point also, but I think this is a bit cleaner. – ktoi May 31 '17 at 21:16
• Everything seems great but just one last thing I tried to prove the following, If $A \subset X$ contains 0 and A is convex set such whenever a $\in$ A , -a $\in$ A. Then if 0 is not in the interior then $Int(A)=\phi$. That is what you intend to say right? Now I supposed that let $x \in A$ is an interior point , then there exist $\delta$>0 such that $B_{\delta}(x) \subset A$. Now what I want to show that $B_{\delta}(0) \subset A$, which will lead to a contradiction. But I couldn't do so! Can you write a line or two in this direction. Thanks very much though for the wonderful solution. – Riju May 31 '17 at 21:39