Inequality between operator norm and Hilbert-Schmidt norm

Question

I have seen the following inequality here but I don't know where I can find a proof for it. Could somebody give me a hint to understand it or guide me to a reference please?

$\|AB\|_{HS} \leq \|A\|_{\mathrm{op}} \|B\|_{HS} \ {\rm for}\ A, B ∈ \mathcal{M}_n$.

Answer 1

The key are these properties about positivity of operators: if $0\leq C\leq D$, then $B^*CB\leq B^*DB$; and $C\leq \|C\|_{op}\,I$. You have then, using the positivity of the trace, $$ \|AB\|_{HS}^2=\text{Tr}(B^*A^*AB)\leq \|A^*A\|_{op}\,\text{Tr}(B^*B)=\|A\|^2_{op}\,\|B\|_{HS}^2. $$

Answer 2

Like the OP (see comment to Martin's answer), I had to convince myself of the correctness of Martin's excellent solution, and I worked it out. I'll just copy it here, in case it helps anybody else.

First, an operator (matrix) $A$ is semipositive (positive semi-definite) if $x^T A X \geq 0 \,\forall x$.

$C \leq D$ just means $D-C$ is positive semi-definite.

The first property to note is that if $C \leq D$, then $B^T C B \leq B^T D B$. This is easy. Consider $x^T (B^T D B - B^T C B) x = (Bx)^T (D-C) (Bx) \geq 0$ because $C \leq D$.

Secondly, if $C \leq D$, $Tr(C) \leq Tr(D)$. This follows because the trace is the sum of eigenvalues, hence $Tr(D-C)\geq 0$, and the linearity of trace.

Finally, the third property to note is that for any matrix $A$, we have $A \leq \|A\|_{op}I$. Why? \begin{align} x^T A x &\leq \|x\|\|Ax\| \quad\text{(Cauchy Schwarz)} \\ &= x^T x \frac{\|Ax\|}{\|x\|} \\ &\leq (x^T x) \|A\|_{op} \end{align}

So for any matrix $A$ we have $$A \leq \|A\|_{op}I \Rightarrow B^T A B \leq B^T \|A\|_{op} A \Rightarrow Tr(B^T A B) \leq Tr(B^T \|A\|_{op} A)$$ Finally, \begin{align*} \|AB\|_{HS}^2 &= Tr((AB)^T (AB)) \\ &= Tr(B^T A^T A B) \\ &\leq Tr(B^T (\|A^TA\|_{op} I) B) \\ &= \|A^T A\|_{op} Tr(B^T B) \\ &= \|A\|_{op}^2 \|B\|_{HS}^2 \end{align*}