# Compute operator norm by image on orthonormal basis

Let $e_n$ a orthonormal basis for a Hilbert space and $T$ a bounded linear operator. Is the following correct?

$$\lVert T \lVert^2 \leq \sup_{n \in \mathbb{N}} \sum_{k \in \mathbb{N}} |\langle e_n,Te_k\rangle|^2$$

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Your inequality is not homogeneous in $T$ (a square is missing in the LHS). The $\max$ should be a $\sup$ and note that by Bessel's inequality, the RHS is $\sup_n\lVert T^*e_n\rVert^2$. – Davide Giraudo Jun 15 '12 at 10:51

Since by Bessel's inequality $$\sum_{k\in\Bbb N}|\langle e_n,Te_k\rangle|^2=\sum_{k\in\Bbb N}|\langle T^*e_n,e_k\rangle|^2=\lVert T^*e_n\rVert^2$$ and $\lVert T\rVert=\lVert T^*\rVert$, the question is: given a linear bounded operator $T\colon H\to H$, do we have $\lVert T\rVert^2=\sup_{n\in\Bbb N}\lVert Te_n\rVert^2$? (the inequality $\geqslant$ is quite obvious)
The answer is no. Indeed, consider $T\colon \ell^2(\Bbb N)\to \ell^2(\Bbb N)$ given by $T(x)(n)=0$ for $n\geq 1$ and $T(x)(0)=\sum_{j=1}^{+\infty}\frac{x(j)}j$ (it is well-defined by Cauchy-Schwarz inequality). Such an operator is linear an bounded. Its norm is reached only for a multiple of the sequence $\{\frac 1j\}_{j\geq 1}$. Denote $e^{(n)}$ the sequence whose $n$-th term is $1$, the other are $0$ (this gives an orthonormal basis). We have $\lVert T(e^{(n)})\rVert=\frac 1n\leq 1<\lVert T\rVert=\frac{\pi}{\sqrt 6}$.
Thank you for your answer. Do you know if there is a different way to estimate the norm in terms of $<e_n, T e_k>$? – Rejnik Jun 15 '12 at 18:28
@Davide Could you explain the notation $T(x)(n)$ and $T(x)(0)$ please, thank you! – Xiao Aug 7 '14 at 10:41
$T(x)(n)$ is the $n$-th term of the sequence $T(x)$. – Davide Giraudo Aug 7 '14 at 11:32