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 $0<c<1$ 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.
 A: 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 $1$.
