${{\inf }_{\left\| x \right\|=1}}{{\left\|Ax \right\|}^{2}}={{\inf }_{\left\| x \right\|=1}}{{\left\| {{A}^*}x \right\|}^{2}}$?

Question

Let $\mathcal{B}(F)$ the algebra of all bounded linear operators on a complex Hilbert space $F$.

I see the following remark in a paper:

For an operator $A\in \mathcal{B}\left(F \right)$, if $F$ is infinite-dimensional, then ${{\inf }_{\left\| x \right\|=1}}{{\left\|Ax \right\|}^{2}}$ and ${{\inf }_{\left\| x \right\|=1}}{{\left\| {{A}^*}x \right\|}^{2}}$ may be different.

Is there an example that shows that two expressions are in general different? What about the case $F$ is finite-dimensional?.

Answer 1

If $A$ is invertible, then $$\|x\| = \|A^{-1}Ax\| \leq \|A^{-1}\|\|Ax\|$$ for all $x$, hence $\inf\limits_{\|x\|=1} \|Ax\| \geq \|A^{-1}\|^{-1}$. As for every $\varepsilon > 0$ we can choose $y = Ax$ such that $$\|x\| = \|A^{-1}y\| \geq \|A^{-1}\|\|y\| -\varepsilon = \|A^{-1}\|\|Ax\| -\varepsilon,$$ we must have $$\|A^{-1}\|^{-1} = \inf\limits_{\|x\|=1} \|Ax\|.$$ By standard properties of the adjoint we conclude that $$\inf\limits_{\|x\|=1} \|Ax\| = \inf\limits_{\|x\|=1} \|A^*x\|$$ in this case.

So assume that $A$ (and hence also $A^*$) is not invertible. In finite dimensions this implies that both $A$ and $A^*$ have a non-trivial kernel and so $$\inf\limits_{\|x\|=1} \|Ax\| = 0 = \inf\limits_{\|x\|=1} \|A^*x\|.$$ In infinite dimensions, this is no longer true, see Kavis answer.

Answer 2

Answer for infinite-dimensional case: On $\ell^{2}$ let $A(x_n)=(0,x_1,x_2,...)$. Then $A^{*}(x_n)=(x_2,x_3,...)$. So the first infimum is $1$, But $A^{*} (1,0,0,...)=0$ so the second infimum is $0$.