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I am reading Rudin Funtional Analysis. Theorem 13.11 gives statements on densely defined symmetric operator $T$ over Hilbert Space $H$,

(a) if $D(T) = H$, then $T=T^*$ and $T$ bounded

(b) if $T=T^*$ and injective, then image of $T$ dense in $H$ and $T^{-1}$ self adjoint

(c) if image of $T$ dense then $T$ injective

(d) if $T$ surjective, then $T=T^*$ and $T^{-1}$ bounded.

I am wondering if the following is also true:

(e) If $T$ is bounded, then $T=T^*$.

I read here that it is true as the top answer says Hermitian implies Self-Adjoint (although I am not familiar with this terminology. Rudin says Hermitian is Self-Adjoint). But how to show this? Thanks

orlando
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1 Answers1

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A densely defined symmetric operator is by definition an operator $T$ with domain $D(T)\underset{dense}{\subset}\mathcal{H}$ such that $$\langle Tx,y\rangle =\langle x,Ty\rangle ,\quad x,y\in D(T)\quad (*)$$ Its adjoint $T^*$ is defined on the domain $$D(T^*)=\{y\in \mathcal{H}\,:\, (\exists v\in\mathcal{H})\,(\forall x\in D(T)) \ \langle Tx,y\rangle =\langle x,v\rangle \}$$ For $y\in D(T^*)$ the element $v$ is unique because the domain $D(T)$ is dense. Thus we may define $T^*y=v.$ It is straightforward that $T^*$ is linear.

By $(*)$ we get $D(T)\subset D(T^*),$ hence the domain $D(T^*)$ is dense. We say that the operator $T$ is self-adjoint if $D(T)=D(T^*).$

If $T$ is bounded, i.e. $\|Tx\|\le c\|x\|$ for all $x\in D(T),$ then $D(T^*)=\mathcal{H}.$ Indeed, for any $y\in \mathcal{H}$ the functional $$D(T)\ni x\mapsto \langle Tx,y\rangle $$ is bounded, hence by the Riesz theorem $$\langle Tx,y\rangle=\langle x,v\rangle,\quad x\in D(T) $$ for a unique element $v\in \mathcal{H}.$ Therefore the operator $T$ is self-adjoint iff $D(T)=\mathcal{H}.$

When $T$ is bounded then by continuity it can be extended uniquely to a bounded operator symmetric operator $\tilde{T}$ such that $D(\widetilde{T})=\mathcal{H}.$ By the previous reasoning $\widetilde{T}$ is self-adjoint.

Summarizing if $T$ is bounded, but originally defined on $D(T)\subsetneq \mathcal{H},$ then $T$ is not self-adjoint, but admits the unique self-adjoint extension.

Ryszard Szwarc
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