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Let $V$ be a vector space. Assume $H_1$ and $H_2$ are subspaces of $V$, and that both $H_1$ and $H_2$ are Hilbert spaces with inner-products $\langle \cdot, \cdot\rangle_1$ and $\langle \cdot,\cdot\rangle_2$ respectively. Let $x\in H_1$, and let $\left\{h_n,\,n\in\mathbb{N}\right\}$ be an orthonormal set in $H_1$ (not necessarily a basis) such that $$\big\Vert x - \sum_{k=1}^n\langle x, h_k\rangle_1 h_k\big\Vert_1 \underset{n\rightarrow\infty}\longrightarrow 0,$$ that is, $x$ lies in the closed linear span of $\left\{h_n\right\}$.

Assume now that $x\in H_2$ and $\left\{h_n\right\}\subset H_2$. Notice that since $\left\{h_n\right\}$ is orthonormal in $H_1$, it is linearly independent in $H_1$ and hence in $V$ and $H_2$, but not necessarily orthogonal in $H_2$. Let $\left\{e_n,\,n\in\mathbb{N}\right\}$ be an orthonormal set in $H_2$ obtained by a Gram-Schmidt process on $\left\{h_n\right\}$, that is, such that $\mbox{span}\left\{e_1,\dots,e_n\right\} = \mbox{span}\left\{h_1,\dots,h_n\right\}$ for each $n$. Is it true that $$\big\Vert x - \sum_{k=1}^n\langle x, e_k\rangle_2 e_k\big\Vert_2 \underset{n\rightarrow\infty}\longrightarrow 0?$$

Edited

In response to user gerw: let me be more specific and maybe we can relate the two norms. Given two probability measures $\mu$ and $\nu$ on $(\mathbb{R},\mathcal{B})$, both equivalent to Lebesgue measure, let $V$ be the vector space of $\mu$-equivalence classes of measurable functions $f:\mathbb{R}\rightarrow\mathbb{R}$ (which is of course equal to the set of $\nu$-equivalence classes of such functions), let $H_1 = L^2(\mu)$ and $H_2 = L^2(\nu)$.

Let $T_\mu:L^2(\mu)\rightarrow L^2(\mu)$ be the positive Hilbert-Schmidt operator defined by $$T_\mu f(x) = \int c(x,y) f(y) d\mu(y),\qquad f\in L^2(\mu)$$ where $c$ is a bounded, measurable kernel. Define $T_\nu$ similarly.

I am given a certain bounded measurable function $\varphi$ such that both $$\big\Vert \varphi - \sum_{k=1}^n\langle \varphi, h^\mu_k\rangle_1 h^\mu_k\big\Vert_1 \underset{n\rightarrow\infty}\longrightarrow 0\quad \mbox{and} \quad\big\Vert \varphi - \sum_{k=1}^n\langle \varphi, h^\nu_k\rangle_2 h^\nu_k\big\Vert_2 \underset{n\rightarrow\infty}\longrightarrow 0$$ hold, where $\left\{h^\mu_n\right\}$ is the orthonormal set of eigenfunctions of $T_\mu$, and similarly for $\left\{h^\nu_n\right\}$. I'd like to show that $\varphi$ is the $\Vert\cdot\Vert_2$-limit of linear combinations of the $h^\mu_n$.

My question can be rephrased as follows: is the $\Vert\cdot\Vert_2$-closure of $\mbox{span}\left\{h_n^\mu\right\}$ equal to the $\Vert\cdot\Vert_2$-closure of $\mbox{span}\left\{h_n^\nu\right\}$?

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    $\begingroup$ Effectively, you ask whether $x$ lies in the closure (w.r.t. $H_2$) of the span of $\{h_n\}$. Since the norms of $H_1$ and $H_2$ cannot be compared without further assumptions, I suspect that the answer will be "False" in general. However, I do not have a counterexample at hand. $\endgroup$ – gerw Jan 10 '16 at 18:43
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Let $H_1 = L^2[0,\pi]$, and let let $H_2$ be the subspace of absolutely continuous functions on $[0,2\pi]$ with first derivative in $L^2$ and with Sobolev norm $$ \|f\|_2 = \sqrt{\|f\|_{L^2}^{2}+\|f'\|_{L^2}^{2}}. $$ Define $h_n(x) = \sin(nx)$ for $n=1,2,3,\cdots$. Then $\{ h_n \} \subset H_2$ as well. The constant function $1$ is in $H_1, H_2$ also. The set $\{ \sqrt{2}\sin(nx)\}_{n=1}^{\infty}$ is a complete orthonormal basis of $H_1$. Hence, $$ \lim_N\left\|1 - 2\sum_{n=1}^{N}(1,\sin(nx))_1\sin(nx)\right\|_{L^2} = 0. $$ The functions $\{ \sin(nx) \}_{n=1}^{\infty}$ are in $H_2$, and $1\in H_2$. However $1$ is not in the closure of the linear span of $\{ \sin(nx)\}_{n=1}^{\infty}$ in $H_2$ because convergence in $H_2$ implies uniform convergence on $[0,\pi]$; to see why, \begin{align} f(x) & =xf(x)+(1-x)f(x) \\ & =\int_{0}^{x}\frac{d}{dt}(tf(t))dt -\int_{x}^{1}\frac{d}{dt}((1-t)f(t))dt \\ & =\int_{0}^{x}\{f(t)+tf'(t)\}dt-\int_{x}^{1}\{-f(t)+(1-t)f'(t)\}dt,\\ |f(x)| & \le \|f\|_{L^2}\|1\|_{L^2}+\|f'\|_{L^2}(\|t\|_{L^2}+\|1-t\|_{L^2}) \\ & \le C\sqrt{\|f\|_{L^2}^2+\|f_{L^2}'\|_1^2} \\ & = C\|f\|_{2}. \end{align}

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