Let $\{e_n\}_{n \in \mathbb{N}}$ be an orthonormal system within $\ell^2$. Fix a sequence $\lambda = (\lambda_1, \ldots , \lambda_n , \ldots) \in \ell^{\infty}$ and define
$ \displaystyle Tf = \sum_{n=1}^{\infty} \lambda_n \langle f, e_n \rangle e_n \quad \forall f \in \ell^2$.
a). Show that the $Tf$ defined above is in $\ell^2$. This implies $T$ is a well-defined operator from $\ell^2$ to $\ell^2$.
b). Show that $T$ is bounded and $\|T\| = \|\lambda\|_{\ell^{\infty}}$.
For part (a), would choosing the sequence $\lambda_n = (0,0,0, \cdots , 0 , \cdots)$ just make the entire summation equal to 0 thus showing that it is in $\ell^2$?
As for part (b), how does one show this operator is bounded? Based on what I have read, I have to show that the norm of $T$ is less than or equal to some constant $M$. How do I apply this to my given situation?
Thanks in advance for taking the time to read this. Any assistance is greatly appreciated since I would really like to solve this problem.
Best Answer
For part a): you cannot choose $\lambda_n=0$ since the sequence $\lambda$ is fixed (for example, we may have $\lambda_n=1$ for each $n$). The $n$-th term of the sequence $Tf$ is $\lambda_n\langle f,e_n\rangle$. Its absolute value is bounded by $\lVert \lambda\rVert_\infty|\langle f,e_n\rangle|$. From this, the inequality $$\sum_{n\geqslant 1} |\lambda_n\langle f,e_n\rangle|^2\leqslant \lVert \lambda\rVert_\infty\sum_{n\geqslant 1} \lvert\langle f,e_n\rangle\rvert^2 $$ should be clear.
From this inequality, it follows that $Tf$ is an element of $\ell^2$ and that $\lVert Tf\rVert_2\leqslant \lVert \lambda\rVert_\infty\lVert f\rVert_2$.
In order to show the equality of norms, consider a particular choice of $f$.