Prove that if $\lVert T(v)\rVert <\lVert T^*(v)\rVert$ then there exists $u$ such that $\lVert T(u)\rVert >\lVert T^*(u)\rVert$

Question

I was given this question:
V is an inner product space and $T:V \to V$ is a linear transformation, prove that if there exists a $v\in V$ s.t $\lVert T(v)\rVert <\lVert T^*(v)\rVert$ then there exists $u\in V$ such that $\lVert T(u)\rVert >\lVert T^*(u)\rVert$

I didn't really get anywhere with this question, I mean, I saw an inequality with the norm and I thought about the projection on $\text{Im}T^*$ and maybe on $\text{Im}T$ but didn't really get anywhere.
Any help would be greatly appreciated.

Answer 1

Consider the matrix $T^*T-TT^*.$ This matrix is self-adjoint and its trace is equal $0,$ as the trace of $AB$ is the same as of $BA.$ Your assumption implies that $T^*T\neq TT^*.$ Hence the difference has a positive and negative eigenvalues say $\lambda_1<0$ and $\lambda_2>0.$ Let $u_1,u_2$ denote the corresponding unit eigenvectors. Then \begin{multline*}\|Tu_1\|^2=\langle T^*Tu_1,u_1\rangle \\ =\langle TT^*u_1,u_1\rangle +\langle (T^*T-TT^*)u_1,u_1\rangle \\=\|T^*u_1\|^2+\lambda_1< \|T^*u_1\|^2. \end{multline*} Similarly $$\|Tu_2\|>\|T^*u_2\|.$$

The conclusion holds for infinite dimensional spaces under the assumption that $T$ is a Hilbert-Schmidt operator.

Answer 2

This is false. Let $T(x_n)=(x_2,x_3,..)$ on $\ell^{2}$. Then $T^{*}(x_n)=(0,x_1,x_2,...)$. We have $0= \|Te_1\| <1=\|T^{*}e_1\|$ but $\|Tv \|\leq \|T^{*} v\|$ for all $v$.