Prove $\|A\| \leq \|A\|_{HS}$, where $\|A\|$ is the operator norm of A

Question

I Am trying to solve the following problem Let $H_1 ,H_ 2$ be Hilbert spaces. Let $A \in B(H_1 ,H_2 )$ be a Hilbert-Schmidt operator. For a complete orthonormal sequence $( u_n )$ in $H_1$, define the Hilbert-Schmidt norm $\|.\|_{HS}$ by

$\|A\|_{HS}=\left(\sum^\infty_{n=1}\|A(u_n)\|^2\right)^\frac{1}{2} $

Prove $\|A\| \leq \|A\|_{HS}$, where $\|A\|$ is the operator norm of A.

My current attempt is as follows:

Let $x \in H_1$ such that $ x=\sum^\infty_{n=1} \langle x,u_n\rangle u_N$ and $\|x\|=\left(\sum^\infty_{n=1}|\langle x,u_n\rangle|^2\right)^\frac{1}{2} $. Then we have,

$\|A(x)\|=\|A\left(\sum^\infty_{n=1} \langle x,u_n\rangle u_N\right)\| =\|\sum^\infty_{n=1}A\left( \langle x,u_n\rangle u_n\right)\| = \|\sum^\infty_{n=1} \langle x,u_n\rangle A\left(u_n\right)\| \leq \|\sum^\infty_{n=1}A(u_n)\|\ \|\sum^\infty_{n=1}\langle x,u_n\rangle\|$

I am unsure where to go from here and feel I have made a mistake? Thank you in advance.

Answer 1

In the last inequalities you have used an inequality of the form: $$\left\Vert\sum_n a _n b_n \right\Vert \leq \left\Vert\sum_n a _n \right\Vert \left|\sum_n b_n \right|$$ which is not true. However you can use a Cauchy-Schwartz inequality which gives: $$\sum_n \left\Vert a _n\right\Vert |b_n| \leq \left(\sum_n \left\Vert a _n \right\Vert^2 \right)^\frac{1}{2} \left(\sum_n \left| b _n \right|^2 \right)^\frac{1}{2} $$

Answer 2

Besides the standard argument that Delta-u mentions, you could also note that the HS norm does not depend on the orthonormal basis used (basically because $\|A\|_{HS}=\operatorname{Tr}(A^*A)^{1/2}$). Then, given $x$ with $\|x\|=1$, you could form an orthonormal basis $\{v_n\}$ with $v_1=x$. Then $$ \|A\|_{HS}^2=\sum_n\|Av_n\|^2\geq\|Av_1\|^2=\|Ax\|^2. $$