# Find a linear operator such that $\|T\| > \sup_{\|f\| \leq 1} \left|\langle Tf, f \rangle\right|$

Find a linear operator $T: \mathbb C^2 \to \mathbb C^2$ such that $$\|T\| > \sup_{\|f\| \leq 1} \left|\langle Tf, f \rangle\right|,$$ where $\langle \cdot, \cdot \rangle$ is the standard inner product.

I know that $$\|T\| := \sup_{f \neq 0} \frac{\|Tf\|}{\|f\|} = \sup_{\|f\|=1} \|Tf\| = \sup_{\|f\|\leq 1} \|Tf\|.$$ This does not seem to be helpful. Any help, please? Thank you!

• $\langle\cdot\rangle$ is the standard scalar product or another scalar product? – Martín-Blas Pérez Pinilla Nov 18 '14 at 7:49
• @Martín-BlasPérezPinilla It is standard inner product. – user156460 Nov 18 '14 at 7:50

Hint: what happens if $T$ is a rotation?
• Then $\|T\| = 1$, but I do NOT see why this will give strict $>$ unless one assumes that rotation of a non-zero angle. Right? – user156460 Nov 18 '14 at 21:15