Show that a space $H$ is Hilbert.

Question

Let \begin{equation} H = \{f \in L^2(0,\infty):\int_0^\infty f^2(x)e^{-x}dx \lt\infty\} \end{equation}

Show that $H$ is a Hilbert space with scalar product \begin{equation} \langle f,g \rangle = \int_0^\infty f(x)g(x)e^{-x}dx \end{equation}

I was trying to show that it is complete about the norm coming from scalar product, but I think it's way too difficult. Instead, I was thinking to define a Unitary isomorphism between $H$ and $L^2(0,\infty)$ like this: \begin{equation} U:H\rightarrow L^2(0,\infty) \end{equation}

such that $\langle U(f),U(g)\rangle_{L^2}$ = $\langle f,g\rangle_{H}$.

The idea is that the scala product on $H$ is really similar to $L^2$ 's one. Any suggestions on how to define it properly?

Simply because I'm dealing with these things for the first time. Thank you — James Arten, Oct 23 '18 at 12:16
@KaviRamaMurthy not quite. He is defining it as the subspace of $L^2(\lambda)$ ($\lambda$ is the Lebesgue measure on $(0,\infty)$) which is also in $L^2(\mu)$, and using the $L^2(\mu)$-norm, so it is "missing" things which can blow up subexponentially at infinity, therefore cannot be complete in $L^2(\mu)$. — user10354138, Oct 23 '18 at 12:18
@user10354138 Thanks for your comment. I was wrong and I have deleted my comment. — Kavi Rama Murthy, Oct 23 '18 at 12:50
@JamesArten No, I mean what I wrote, at least in the context of how $U$ is defined in your question. With my definition, you have $$\langle U(f), U(g) \rangle_{L^2} = \int_0^\infty e^{-x/2}f(x)e^{-x/2}g(x) , \mathrm{d}x = \int_0^\infty e^{-x}f(x)g(x) , \mathrm{d}x = \langle f, g\rangle_H.$$ — Theo Bendit, Oct 23 '18 at 13:36
@JamesArten P.S. \langle and \rangle produce $\langle$ and $\rangle$ respectively. — Theo Bendit, Oct 23 '18 at 13:41
@JamesArten As Kavi Rama Murthy's example shows, the map $U$ is not an isomorphism. In particular, this map is not surjective, as the function $e^{-x/2} \in L^2$ has no preimage. — Theo Bendit, Oct 23 '18 at 13:48
This is false. My conjecture is you gave the definition of $H$ wrong; is $H = {f \in L^2(0,\infty):\int_0^\infty f^2(x)e^{-x}dx \lt\infty}$ exactly how $H$ was defined in the statement of the problem? — David C. Ullrich, Oct 23 '18 at 14:39
See, that seems like a silly definition, because it just says $H=L^2((0,\infty))$. — David C. Ullrich, Oct 23 '18 at 14:40
@DavidC.Ullrich yes the exercise states exactly this.. I found actually another version of it where we have $H = {f:\mathbb{R}\rightarrow(0,+\infty): \int_0^\infty f^2(x)e^{-x}dx < \infty}$ — James Arten, Oct 23 '18 at 14:54
That second version is even sillier, because the second $H$ is not even a vector space. Unless maybe you meant $f:(0,\infty)\to\Bbb R$. — David C. Ullrich, Oct 23 '18 at 14:59
Right. Given that your statement of the second version was so totally wrong, I find it hard to believe that you've reproduced the first version correctly. — David C. Ullrich, Oct 23 '18 at 15:17
A simpler approach might be checking validity of the Parallelogram law. — arridadiyaat, Aug 14 '24 at 14:31

Kavi Rama Murthy · Answer 1 · 2018-10-23T23:17:31.097

4

I think this is false. $\{I_{(0,n)}\}$ is a Cauchy sequence in $H$ which is not convergent. If this sequence converges in $H$ then some sequence converges almost everywhere w.r.t. $e^{-x}dx$ hence w.r.t. Lebesgue measure and the limit has to be $1$ almost everywhere w.r.t. Lebesgue measure. But then the limit is not in $L^{2}(0,\infty)$, hence not in $H$.

edited Oct 23 '18 at 23:17

answered Oct 23 '18 at 12:54

Kavi Rama Murthy

359,332

That sequence is not cauchy, is it? – alexp9 Oct 23 '18 at 13:00
It is Cauchy ln the norm of H. – Kavi Rama Murthy Oct 23 '18 at 13:17
Is $I_{(0, n)}$ the indicator function for the interval $(0, n)$? If so, doesn't this converge to the constant function $1$? – Theo Bendit Oct 23 '18 at 13:32
No wait, +1, I see my error. The indicator function doesn't belong to $H$ as $H$ consists only of functions already in $L^2(0, \infty)$. I agree: this is not a Hilbert Space. – Theo Bendit Oct 23 '18 at 13:45
Sorry but if the sequence $I_{(0,n)}$ is not in $H$ from the beginning, does it make sense to consider it to say that the space is not Hilbert? Probably a dumb question but I have to say again that it's the first time I'm studying this. – James Arten Oct 23 '18 at 14:26
@JamesArten What makes you think $I_{(0,n)}$ is not in $H$? – David C. Ullrich Oct 23 '18 at 14:37
Sorry of course it is in $H$ I was thinking about letting $n \rightarrow \infty$ but of course $\chi_{0,n}$ belongs to $L^2$ as long as $n$ is finite. – James Arten Oct 23 '18 at 14:50
@KaviRamaMurthy can you explain better why the sequence shouldn't be convergent in $H$? – James Arten Oct 23 '18 at 15:10
1

@JamesArten We know, or should know, that if $f_n\to f$ in $L^2(\mu)$ then some subsequence tends to $f$ almost everywhere. Hence if the given $f_n$ have a limit in $H$ the llimit must be $1$. But $1\notin H$. – David C. Ullrich Oct 23 '18 at 15:20

score -1 · Answer 2 · answered Oct 23 '18 at 17:14

Ok this is what I got in the end:

Again, we have

\begin{equation} H = \{f \in L^2(0,\infty): \int_0^\infty f^2(x)e^{-x}dx < \infty\} \end{equation} and \begin{equation} \langle f,g \rangle_{H} = \int_0^\infty f(x)g(x)e^{-x}dx \end{equation}

So, (I guess), \begin{equation} ||f||_H = \sqrt{\langle f,f \rangle_{H}} = \left(\int_0^\infty f^2(x)e^{-x}dx \right)^{1/2} \end{equation}

Taking $f_n = \chi_{(0,n)}$ we have $f_n \in L^2(0,\infty) \forall n$ and it is Cauchy in $H$ because, assuming $n >m$ we could consider \begin{equation} ||\chi_{(0,n)}-\chi_{(0,m)}||_H = ||\chi_{(m,n)}||_H = \int_m^n e^{-x}dx = -e^{-n}+e^{-m}<\epsilon \end{equation} For $m,n$ sufficiently large. And also \begin{equation} ||\chi_{(0,n)}-\chi_{(0,\infty)}||_H = \int_0^\infty (\chi_{(0,n)}-\chi_{(0,\infty)})^2e^{-x}dx < \infty \end{equation}

Which step could be wrong?

I'm sorry for my mistakes, I'm here to learn, and also to find people to talk over with.

Thank you

What step could be wrong? What's wrong is you seem to be ignoring the other posted answer and comments. There's no suuch thing as $||\chi_{(0,n)}-\chi_{(0,\infty)}||H$ because $\chi{(0,n)}-\chi_{(0,\infty)}$ is not an element of $H$. — David C. Ullrich, Oct 23 '18 at 17:46

Show that a space $H$ is Hilbert.

2 Answers2