# [SOLVED] Orthonormal basis in Hilbert space

• Apr 29th 2010, 04:56 AM
james123
Orthonormal basis in Hilbert space
Hi,
The first part of a problem was to show that if Y is a dense subspace of a separable hilbert space H, then H has an orthonormal basis consisting of elements in Y.
I was able to do this.
The second part of the problem is use this result to prove that if $f\in L^1(\mathbb{R})\setminus L^2(\mathbb{R})$, there is an orthonormal basis $\{g_n\}$ of continuous bounded functions such that $\int_{\mathbb{R}} f(x)g_n(x)\, dx = 0$.

I can see that if we define the linear functional $T$, with domain the set of continuous functions of compact support $C_0(\mathbb{R})$ intersected with $L^2(\mathbb{R})$, by $T(g) = \int_{\mathbb{R}} f(x)g(x) \, dx$, then we want to take $Y$ to be $\ker T$ and $H = L^2(\mathbb{R})$ in the above result. It suffices to show $\ker T$ is dense in $C_0(\mathbb{R})\cap L^2(\mathbb{R})$ since $C_0(\mathbb{R})$ is dense in $L^2(\mathbb{R})$. I know that this is equivalent to $T$ being discontinuous (ie. unbounded), but I'm getting stuck on using that $f$ isn't in $L^2(\mathbb{R})$ to show $T$ is unbounded.

Is this the right idea? If so, how can we use that $f$ is not in $L^2(\mathbb{R})$ to show that $T$ is unbounded?

Many thanks.

Edit: I hadn't solved the problem, but thought the above was too wordy so wished to replace it.