existence of adjoints of bounded operators

Let $\mathscr{H}$ be a Hilbert space and let $T:𝒟(T)\subset \mathscr{H}⟶\mathscr{H}$ be a densely defined linear operator.

Theorem - If $T$ is bounded (http://planetmath.org/ContinuousLinearMapping) then its adjoint $T^{*}$ is everywhere defined and is also bounded.

Proof : Since $T$ is densely defined and bounded, it extends uniquely to a bounded (everywhere defined) linear operator on $\mathscr{H}$, which we denote by $\tilde{T}$.

For each $z\in \mathscr{H}$, the function $f:\mathscr{H}⟶ℂ$ defined by$f(x)=⟨\tilde{T}x,z⟩$ defines a bounded linear functional on $\mathscr{H}$. By the Riesz representation theorem there exists $u\in \mathscr{H}$ such that

$$f(x)=⟨x,u⟩$$

i.e.

$$⟨\tilde{T}x,z⟩=⟨x,u⟩.$$

Since $\tilde{T}$ extends $T$, we also have that for every $z\in \mathscr{H}$ there exists $u\in \mathscr{H}$ such that

$$⟨Tx,z⟩=⟨x,u⟩\text{for every}x\in 𝒟(T).$$

We conclude that $T^{*}$ is everywhere defined. To see that it is bounded one just needs to check that

$$\underset{z\ne 0}{sup}\frac{\parallel T^{*}z\parallel }{\parallel z\parallel }=\underset{\begin{array}{c}z\ne 0\\ T^{*}z\ne 0\end{array}}{sup}\frac{|⟨T^{*}z,T^{*}z⟩|}{\parallel T^{*}z\parallel \parallel z\parallel }\le \underset{\begin{array}{c}z\ne 0\\ x\ne 0\end{array}}{sup}\frac{|⟨x,T^{*}z⟩|}{\parallel x\parallel \parallel z\parallel }=\underset{\begin{array}{c}z\ne 0\\ x\ne 0\end{array}}{sup}\frac{|⟨Tx,z⟩|}{\parallel x\parallel \parallel z\parallel }\le \parallel T\parallel$$

where the last inequality comes from the Cauchy-Schwarz inequality and the fact that $T$ is bounded. $\mathrm{\square }$

Remark - This theorem shows in particular that bounded linear operators $T:\mathscr{H}⟶\mathscr{H}$ have bounded adjoints $T^{*}:\mathscr{H}⟶\mathscr{H}$.