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Consider $L_2[0,1]$ with the usual inner product $\langle f, g \rangle = \int_{0}^1 f(t)g(t) \, dt$ and define a new norm

$$ \| f \|^2_{\star} = \sum_{i=1}^\infty \langle f, \phi_i\rangle^2 \lambda_i $$

where $\phi_1, \phi_2, \ldots$ are an orthonormal basis for $L_2[0,1]$ and $\lambda_n$ is a sequence of non increasing postitive numbers such that $\sum_{i=1}^\infty \lambda_i < \infty$.

For some $0<c<1$ the indicator function $I_{[0,c]}(t)$ and define the operator $T : L_2 \to L_2$ such that $T(f) = f I_{[0,c]}$

Is T a bounded operator under $\| \|_{\star}$? (Does there exists $M >0$ such that $ \| T(f) \|_{\star} \leq M \|f\|_{\star}$.

I have tried bounding $\langle f I_{[0,c]} , \phi_i \rangle^2 = (\int_{0}^c f(t) \phi_i(t) \, dt )^2$ for each $i$ but failed.

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The answer depends on the basis $\phi_n$ and its relation to the operator $I_{[0,c]}$. Lets construct an example where it doesn't work.

First note that $I_{[0,c]}$ is a self-adjoint projection with infinite dimensional image and kernel. Let $e_k$ be an ONB of the image and $\widetilde e_k$ an ONB of the kernel.

$\sigma:\Bbb N\to \Bbb N$ will be a bijection that we will describe more later. Let $\phi_{2k}=\frac{e_{\sigma(k)}+\widetilde{e_{\sigma(k)}}}{\sqrt 2}$ and $\phi_{2k+1}=\frac{e_k-\widetilde{e_{k}}}{\sqrt 2}$, $\lambda_{2k}=2^{-k}=\lambda_{2k+1}$.

Now let $f=\frac{e_k+\widetilde{e_k}}{\sqrt 2}$. You have $$\|f\|_{\star}^2=2^{-\sigma^{-1}(k)},\quad I_{[0,c]}f = \frac{e_k}{\sqrt 2}=\frac{\phi_{2\sigma^{-1}(k)}+\phi_{2k+1}}2,\quad \|I_{[0,c]}f\|_\star^2=2^{-\sigma^{-1}(k)-1}(1+2^{\sigma^{-1}(k)-k}).$$

If there exists some permutation of $\Bbb N$ so that the difference $\sigma^{-1}(k)-k$ can get arbitrarily large we are done. It is elementary to define such a permutation.

If on the other hand the $\phi_n$ all either lie in the image or in the kernel of $I_{[0,c]}$, it is simple to check that $I_{[0,c]}$ is a bounded operator of norm $1$.

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  • $\begingroup$ This is great, but I am not sure how to check that $I_{[o,c]}$ is bounded of norm $1$ in the case all $\phi_n$ lie in the image. Could you expand a bit please. $\endgroup$ – Manuel Feb 25 at 20:43
  • $\begingroup$ also I belive itshould be a $\sqrt{2}$ instead of $2$ in the second equality for $I_{[0,c]}f$. $\endgroup$ – Manuel Feb 25 at 20:48
  • $\begingroup$ The factor $\frac12$ is correct, note that $\phi$ already has a $\sqrt2$ factor. For the norm of $I_{[0,c]}$, note that $\langle I_{[0,c]}f, \phi_k\rangle = \langle f, I_{[0,c]}\phi_k\rangle$ by symmetry of $I_{[0,c]}$. The term on the right is however either equal to $\langle f, \phi_k\rangle$ or $0$, depending on whether $\phi_k$ is in the image or in the kernel of $I_{[0,c]}$. What you get is that either terms in the sum $\|I_{[0,c]}f\|_\star^2$ drop out or remain the same, and thus $\| I_{[0,c]}f\|_\star^2≤\|f\|_\star^2$. $\endgroup$ – s.harp Feb 25 at 23:07
  • $\begingroup$ Many thanks. I failed to notice that $e_k$ where 0 after $c$. $\endgroup$ – Manuel Feb 26 at 17:12
  • $\begingroup$ Do you think that it is posible to obtain boundeness imposing other type of conditions over the base $\phi_n$? $\endgroup$ – Manuel Feb 26 at 17:14

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