Operator norm and Hilbert Schmidt norm

Question

I'm looking for a proof of

\begin{equation} ||T||\leq ||T||_{HS}, \end{equation} for which it is sufficient to show \begin{equation} ||Tx|| \leq ||x|| \cdot ||T||_{HS} \forall x\in H, x\not=0 \end{equation} can someone help?

Answer 1

Let $(e_i)_{i\in I}$ be an an orthonormal basis, and $x =\sum_{i\in I} y_i e_i$ with norm $1$, i.e. $\sum_{i\in I} |y_i|^2 = 1$, then

$$\|Tx\| \leq \sum_{i\in I}|y_i|\|Te_i\| \leq \sqrt{\sum_{i\in I} |y_i|^2}\sqrt{\sum_{i\in I}\|Te_i\|^2} = \|T\|_{HS}$$

Answer 2

Or you can use the property of adjoint operators, $$\left \|Tx \right \|^{2}=\sum \left \langle Tx,e_{i} \right \rangle^{2}=\sum \left \langle x,T^{\ast }e_{i} \right \rangle^{2}\leq \sum\left \|T^{\ast }e_{i} \right \|^{2}=\left \|T^{\ast } \right \|_{HS}^{2}=\left \|T\right \|_{HS}^{2}$$