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Let $f\in L^2(\mathbb{R})$. Let $C^{\infty}_{c}(\mathbb{R})$ denote the set of compactly supported smooth functions on $\mathbb{R}$. Suppose that

$$\int_{\mathbb{R}}fg=0 $$ for all $g\in C^{\infty}_{c}(\mathbb{R})$ Show that $f$ is the $0$ element of $L^2 (\mathbb{R})$.

How should I proceed to solve this question.

Tensor_Product
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2 Answers2

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$C_0^{\infty}(\mathbb{R})$ is dense in $L^2(\mathbb{R})$ thus exists a sequence $g_n \in C_0^{\infty}(\mathbb{R})$ such that $g_n \rightarrow f$ in $L^2(\mathbb{R})$.

Also $\int|g_n-f|^2=\int g_n^2+\int f^2 \Rightarrow \int g_n^2 \rightarrow -\int f^2$

Also $\int f^2 \leqslant \int |g_n-f|^2+ \int g_n^2$

Taking limits we have that $$2 \int f^2 \leqslant 0 \Rightarrow f^2=0$$ almost everywhere thus $f=0$ almost everywhere.

Now according to the comment below of @zhw there is an easier solution than the first one:

$$|\int g_nf-\int f^2| \leqslant \int|g_n-f||f| \leqslant \sqrt{\int|g_n-f|^2} \sqrt{\int|f|^2} \rightarrow 0$$

Thus $$0=\lim_{n \rightarrow \infty} \int g_nf=\int f^2$$

Thus $f=0$ almost everywhere.

Marios Gretsas
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  • If $g_n\to f$ in $L^2,$ why not just use $\int f^2 = \lim \int fg_n?$ – zhw. Aug 10 '17 at 22:32
  • @zhw..Is my answer not correct? – Marios Gretsas Aug 10 '17 at 22:36
  • if it is corect then you didn't have to downvoted..if it is not you can point out my mistakes.. – Marios Gretsas Aug 11 '17 at 21:25
  • It may be correct (honestly I didn't read the whole thing), but I still downvoted it because I thought everything after the first line was too long and meandering. Just because a solution is correct doesn't mean it shouldn't be downvoted. Now, why don't you stop after your second line? You've shown $-\int f^2,$ which is $\le 0,$ is the limit of a sequence of numbers that are $\ge 0.$. That by itself implies $\int f^2=0.$ Then you're done and you have a nice short proof. And you don't have to include my comment. – zhw. Aug 11 '17 at 21:42
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    Its clearly your right to upvote or downvote an answer..I just do not downvote a correct answer.I personally do not examine if the answer is long or meandering as you say.If its correct then its fine by me.At the time i was solving it the Cauchy swartz inequality did not pop up to my mind.Also i did not include your comment to protect the validity of my answer,but to help the O.P with more than one solutions – Marios Gretsas Aug 11 '17 at 21:49
  • Correctness is only one component of an answer. A long convoluted weird answer that needlessly goes on forever (I'n not saying your answer did that) might be correct, but it would do more harm than good I think, hence a downvote might be in order. Anyway, we had a good discussion, and I'll be glad to cancel my down vote. – zhw. Aug 11 '17 at 21:59
  • @zhw. thanks for canceling your downvote but my comment was not about a -2 in my rep or something like that because a vote is not the most important.I just wanted to see if i had a mistake..MSE is not a game with votes .It is a contistitution for me and personally has helped me to grow my mathematical maturity.Either way thanks for the conversation. – Marios Gretsas Aug 11 '17 at 22:11
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Suppose $f\neq 0$. Then either $f^{-1}((0, \infty))$ or $f^{-1}((-\infty, 0))$ has positive measure- suppose WTLOG that $\mu(f^{-1}((0, \infty)))>0$. Now, by the continuity of measure $\lim_{n \rightarrow \infty} \mu(f^{-1}([1/n, \infty))) = \mu(f^{-1}((0, \infty)))>0$, hence we can find $N$ such that $\mu(f^{-1}([1/N, \infty)))>0$. Again, by the continuity of measure, $\lim_{m \rightarrow \infty} \mu(f^{-1}([1/N, \infty))\cap [-m,m])=\mu(f^{-1}([1/N, \infty)))>0$ hence we can find $M$ such that $\mu(f^{-1}([1/N, \infty))\cap [-M,M])>0$. Then setting $g=\mathcal{X}_{[-M,M]}$ we immideately see that $\int gf$ $d\mu>0$. This is the contrapositive of what you wanted to show.

Gaute Schwartz
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