# existence of adjoints of bounded operators

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 $\widetilde{T}$.

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

 $f(x)=\langle x,u\rangle$

i.e.

 $\langle\widetilde{T}x,z\rangle=\langle x,u\rangle.$

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

 $\langle Tx,z\rangle=\langle x,u\rangle\;\;\text{for every}\;x\in\mathscr{D}(T).$

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

 $\sup_{z\neq 0}\frac{\|T^{*}z\|}{\|z\|}=\sup_{\begin{subarray}{c}z\neq 0\\ T^{*}z\neq 0\end{subarray}}\;\frac{|\langle T^{*}z,T^{*}z\rangle|}{\|T^{*}z\|% \|z\|}\leq\sup_{\begin{subarray}{c}z\neq 0\\ x\neq 0\end{subarray}}\;\frac{|\langle x,T^{*}z\rangle|}{\|x\|\|z\|}=\sup_{% \begin{subarray}{c}z\neq 0\\ x\neq 0\end{subarray}}\;\frac{|\langle Tx,z\rangle|}{\|x\|\|z\|}\leq\|T\|$

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

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

Title existence of adjoints of bounded operators ExistenceOfAdjointsOfBoundedOperators 2013-03-22 17:33:44 2013-03-22 17:33:44 asteroid (17536) asteroid (17536) 4 asteroid (17536) Theorem msc 47A05 bounded operators   have (bounded) adjoints