prove $f\in L^2([0,1])$

Question

Assume $f:[0,1]\to R$ is a measurable function such that $fg\in L^1([0,1])$ for all $g\in L^2([0,1])$.prove that $f\in L^2([0,1])$.

My opinion: if I can find a function g such that f+g is in $L^2([0,1])$, then use $$\int(f+g)^2-\int2fg+g^2$$to get $\int f^2$, then we can prove f is in $L^2([0,1])$. Is it right?

I tried to prove f+g is in $L^2([0,1])$, I think if I can prove this, then it should be done — python3, Sep 03 '16 at 16:06
The assumption "For all $g$" here is a little powerful. It should be easier to find some "nice" $g$ that you can use. — iamvegan, Sep 03 '16 at 16:10
@iamvegan Unless $f \geq 0$, using $g = 1$ doesn't show that $f \in L^1$. Rather, you'd want to use $g = \operatorname{sgn}(f)$. — Tom, Sep 03 '16 at 16:25
It is also known as the Banach Steinhaus theorem. Are you self studying? Where did you get this problem? It is a famous problem in big rudin. — abnry, Sep 03 '16 at 16:38
Yeah, I am just beginning to learn real analysis in my school, this problem is from a problem of my homework — python3, Sep 03 '16 at 16:40
What textbook? What are you learning along with it? There are probably other ways to show this result without the big hammer. — abnry, Sep 03 '16 at 16:42

score 5 · Accepted Answer · answered Sep 03 '16 at 16:49

5

We may assume $f\geq 0$. We need to show that $$ \Lambda_f : g\in L^2 \mapsto \int f g \; dx $$ is a bounded linear functional on $L^2$. If this is the case then by Riesz there is $h\in L^2$ so that $\Lambda_f(g)=\int hg \; dx$ and one sees that $f=h$ a.s.

So assume that $\Lambda_f$ is not bounded and let $g_n\geq 0$ be a sequence with $\|g_n\|_{L^2}=1$ so that $\Lambda_f(g_n) \rightarrow \infty$. Extracting a subsequence we may assume that $\Lambda_f(g_n) \geq 4^n$. Then $$ g= \sum_{n\geq 0} \frac{1}{2^n} g_n$$ has $L^2$ norm not greater than $2$ but by monotone convergence $\Lambda_f(g)=+\infty$.

answered Sep 03 '16 at 16:49

H. H. Rugh

35,992

This is along the same lines as this answer with $q=2$. – robjohn Sep 03 '16 at 17:14
@robjohn I wasn't aware. But yes, it seems that there is quite some Poincaré recurrence on this site. Wonder what the invariant measure for that looks like? – H. H. Rugh Sep 03 '16 at 17:40

score -1 · Answer 2 · answered Sep 03 '16 at 16:34

-1

With $g=1\in L^2(0,1)$, the assumption gives $f \in L^1(0,1)$.

With $g= \sqrt{f}\in L^2(0,1)$, it now gives $f^{3/2}\in L^1(0,1)$, ie: $f \in L^{3/2}(0,1)$.

With $g=f^{3/4}\in L^2(0,1)$, it now gives $f^{7/4} \in L^1(0,1)$, ie: $f^{7/8} \in L^2(0,1)$.

etc.. by induction we have that $\forall n \geq 1, f \in L^{2-1/2^n}(0,1)$.

For $n\to \infty$, we get $f\in L^2(0,1)$.

answered Sep 03 '16 at 16:34

anonymus

1,418

5

You can't guarantee that you can pass the limit through. $f=1/x^{1/2}$ is in $L^p(0,1)$ for $1<p<2$ but not $p=2$. – abnry Sep 03 '16 at 16:35

prove $f\in L^2([0,1])$

2 Answers2