# Is this multiplication operator bounded for this special norm?

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 $$0 the indicator function $$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.

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 $$1$$.

• 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. – Manuel Feb 25 at 20:43
• also I belive itshould be a $\sqrt{2}$ instead of $2$ in the second equality for $I_{[0,c]}f$. – Manuel Feb 25 at 20:48
• 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 $\| I_{[0,c]}f\|_\star^2≤\|f\|_\star^2$. – s.harp Feb 25 at 23:07
• Many thanks. I failed to notice that $e_k$ where 0 after $c$. – Manuel Feb 26 at 17:12
• Do you think that it is posible to obtain boundeness imposing other type of conditions over the base $\phi_n$? – Manuel Feb 26 at 17:14