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Let $X$ be subset of of $L^2(\mathbb R)\cap L^1(\mathbb R)$, dense in $L^2(\mathbb R)$. Let $Tf(x)=\int_{\mathbb R}f(x)dx$ with $f\in X$. Can we say that $\ker T=\{f\in X:Tf(x)=0\}$ is dense in $L^2(\mathbb R)$?

One way of proving or disproving this is to see whether $T$ is continuous or not on $X$ (or $L^2(\mathbb R)$ I'm just lost).

Thanks in advance for your answer.

Math

uniquesolution
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Math
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  • $T$ is not well defined in general. Are you assuming that $X$ is contained in $L^{1}$? – Kavi Rama Murthy Jul 13 '18 at 08:09
  • Yes, also $X\subset L^1(\mathbb R)$ in my case. – Math Jul 13 '18 at 08:27
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    For $X=L^{1}$ the kernel is dense, but I don't know how to prove it in general. What I need is a sequence ${f_n}$ in $X$ such that $\int_{\mathbb R} f_n=1$ and $\sqrt n ||f_n||2 \to 0$. If $X=L^{1}$ we can take $f_n=\frac 1 {n^{2}} I{(0,n^{2})}$. C an you give me more information about $X$? – Kavi Rama Murthy Jul 13 '18 at 08:30
  • Well, I have that both $f$ and $xf$ are in $L^2(\mathbb R)$ and so by the C-S inequality, $f$ is in $L^1(\mathbb R)$. Also, you mean a sequence $f_n$ for which $\int_{\mathbb R}f_n(x)dx=0$, don't you? – Math Jul 13 '18 at 08:38
  • I did mean $\int f_n =1$ What I want to do is take $g_n$ in $X$ converging to $f$ in $L^{2}$ and consider $fI_{(-n,n)}-(\int_n^{n}f) f_n$. This would give a sequence converging to $f$; in cae $X=L^{1}$ this sequence lies in $X$ but I don't know if this works for your $X$. – Kavi Rama Murthy Jul 13 '18 at 08:55
  • Yes, we can if $X$ is a subspace. – uniquesolution Jul 13 '18 at 09:36
  • Sorry to say that I still don't get it... – Math Jul 13 '18 at 10:13
  • @uniquesolution, could you elaborate more and yes $X$ is indeed a subspace – Math Jul 13 '18 at 10:15
  • Any suggestions please – Math Jul 13 '18 at 13:22

1 Answers1

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The answer is yes if $X$ is a subspace.

If $X$ is a subspace then, since $X$ is dense in $L^2(\mathbb{R})$, we have that the $\|\cdot\|_2$-completion of $X$ is precisely $L^2(\mathbb{R})$. If $T$ were bounded then by the Riesz representation theorem, there would exist a unique $g \in L^2(\mathbb{R})$ such that $Tf = \langle f, g\rangle, \forall f \in X$. But the only candidate for such a function is $g \equiv 1$ which is not in $L^2(\mathbb{R})$ so $T$ is not bounded on $X$.

Now we can conclude that $\ker T$ is dense in $X$. Namely, let $(f_n)_n$ be a sequence such that $\|f_n\| = 1$ and $|T(f_n)| \ge n, \forall n\in\mathbb{N}$. Pick arbitrary $f \in X$. Then $\left(f - \frac{Tf}{Tf_n}f_n\right)_n$ is a sequence in $\ker T$ which converges to $f$.

Finally, since $\ker T$ is dense in $X$ and $X$ is dense in $L^2(\mathbb{R})$, if follows that $\ker T$ is dense in $L^2(\mathbb{R})$.

mechanodroid
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  • Great! Before, you wrote your clear answer I also did it my way and I was checking things! So thank you as your solution comes to confirm my slightly different proof. – Math Jul 13 '18 at 17:33