# Densely defined symmetric and bounded operator

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

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.