True or False; Functional Analysis 
Given $T: V \to W$ with $V,W$ being Hilbert Spaces. We always have $\| T^ *\| = \| T \|$.

I think it is true because of Riesz' Theorem, but I am not sure if a proof is necessary. 
EDIT: In case notations have conflicts. $T^*$ is adjoint of $T$
 A: Only Cauchy-Schwarz is needed to prove that
$$
\Vert z\Vert=\sup\limits_{\Vert y\Vert\leq 1}|\langle z,y\rangle|
$$
Then we get
$$
\begin{align}
\Vert T^*\Vert
&=\sup\limits_{\Vert x\Vert\leq 1}\Vert T^*(x)\Vert\\
&=\sup\limits_{\Vert x\Vert\leq 1}\sup\limits_{\Vert y\Vert\leq 1}|\langle T^*(x),y\rangle|\\
&=\sup\limits_{\Vert x\Vert\leq 1}\sup\limits_{\Vert y\Vert\leq 1}|\overline{\langle T(y),x\rangle}|\\
&=\sup\limits_{\Vert y\Vert\leq 1}\sup\limits_{\Vert x\Vert\leq 1}|\langle T(y),x\rangle|\\
&=\sup\limits_{\Vert y\Vert\leq 1}\Vert T(y)\Vert\\
&=\Vert T\Vert
\end{align}
$$
A: Note that
$$\|T^\ast(x)\|^2=\langle T^\ast(x),T^\ast(x)\rangle\leq \|x\|\|T\|\|T^\ast(x)\|\quad  \forall\ x\in V,$$
where we used the inequality of Cauchy-Schwarz and that $T\in L(V,W)$ (this hypothesis should be added). So 
$$\|T^\ast(x)\|\leq \|T\|\|x\|\quad  \forall\ x\neq 0, $$
$T^\ast\in L(V,W)$ and $\|T^\ast\|\leq \|T\|$. Similarly, 
$$\|T(x)\|^2=\langle T(x),T(x)\rangle=\langle x,T^\ast(T(x))\rangle\leq \|x\|\|T^\ast\|\|T(x)\|.$$
Thus, $\|T\|\leq \|T^\ast\|$ and follows the inequality.
