If $T:X \to Y$ is a linear homeomorphism, is its adjoint $T^*$ a linear homeomorphism? $X$ and $Y$ denote Hilbert spaces. If $T:X \to Y$ is a linear homeomorphism, is its adjoint $T^*$ a linear homeomorphism? Homeomorphism means continuous map with continuous inverse.
I think the answer is yes, the only thing I am unable to show is what is the inverse of $T^*$ and if it is continuous?
 A: Let $S$ be inverse to $T$. Then $ST=1_X$ and $TS=1_Y$. Apply to this equalities $^*$ functor to get $T^*S^*=1_{X^*}$ and $S^*T^*=1_{Y^*}$. This means that $T^*$ is invertible and what is more $(T^*)^{-1}=S^*$
In this proof I implicitly assumed that $T^*\in \mathcal{B}(Y^*,X^*)$ whenever $T\in\mathcal{B}(X,Y)$. This is indeed true. For the begining recall one of the corollaries of Hahn-Banach theorem
$$
\Vert g\Vert=\sup\{|g(y)|:y\in\operatorname{Ball}_{Y}\}
$$
where $g\in Y^*$. Then
$$
\begin{align}
\Vert T^*\Vert
&=\sup\{\Vert T^*(g)\Vert: g\in\operatorname{Ball}_{X^*}\}\\
&=\sup\{\Vert T^*(g)(x)\Vert: g\in\operatorname{Ball}_{Y^*}, x\in\operatorname{Ball}_{X}\}=\\
&=\sup\{| g(T(x))|: g\in\operatorname{Ball}_{Y^*}, x\in\operatorname{Ball}_{X}\}=\\
&=\sup\{\Vert T(x)\Vert: x\in\operatorname{Ball}_{X}\}=\\
&=\Vert T\Vert
\end{align}
$$
A: The proof of boundedness of $T^\ast$:
$$\|T^\ast(x)\|^2 = \langle T^\ast x, T^\ast x\rangle= \langle TT^\ast x, x\rangle \leq \|TT^\ast x\|\|x\| \leq \|T\|\|T^\ast x\|\|x\|,$$ where the second to last inequality is Cauchy-Schwarz. If $\|T^\ast x\| = 0$, then there's nothing much to say. Otherwise, we divide and get $$\|T^\ast(x)\| \leq \|T\|\|x\|.$$ So $\|T^\ast\|$ is bounded and $\|T^\ast \| \leq \|T\|.$ 
The reason that $T^\ast$ is a bijection is given in other answsers. The inverse of $T^\ast$ is bounded since $(T^\ast)^{-1}$ is the adjoint of a continuous function, namely $T^{-1}$. (Or if you like shiny tools, you can use the open mapping theorem!)
Edit: Once you know about reflexivity of Banach spaces, the above will give $\|T^\ast\| \leq \|T^{\ast\ast}\| = \|T\|$ and so $\|T^\ast\| = \|T\|$ follows.
