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Thread: [SOLVED] Orthonormal basis in Hilbert space

  1. #1
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    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 $\displaystyle f\in L^1(\mathbb{R})\setminus L^2(\mathbb{R})$, there is an orthonormal basis $\displaystyle \{g_n\}$ of continuous bounded functions such that $\displaystyle \int_{\mathbb{R}} f(x)g_n(x)\, dx = 0$.

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

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

    Many thanks.

    Edit: I hadn't solved the problem, but thought the above was too wordy so wished to replace it.
    Last edited by james123; May 1st 2010 at 06:00 AM.
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