= Solution
Take $\widehat V\psi=V(x)\psi$ with real $V$, and $\widehat T=-(\hbar^2/2m)d^2/dx^2$. The <operator domain> matters: use sufficiently smooth <wavefunctions> for which the following boundary terms vanish at both infinities. A real multiplication operator gives
$$
\langle f,\widehat Vg\rangle=\int f^*Vg\,dx=\int(Vf)^*g\,dx=\langle\widehat Vf,g\rangle.
$$
Two applications of <integration by parts> give
$$
\langle f,\widehat Tg\rangle-\langle\widehat Tf,g\rangle
=-\frac{\hbar^2}{2m}[f^*g'-f'^*g]_{-\infty}^{\infty}=0.
$$
Thus the <potential energy> and <kinetic energy> operators obey the stated <Hermitian operator> identity on this common domain. For $D=d/dx$, one application of <integration by parts> instead gives
$$
\langle f,Dg\rangle=[f^*g]_{-\infty}^{\infty}-\langle Df,g\rangle=-\langle Df,g\rangle.
$$
The derivative is <anti-Hermitian>, rather than <Hermitian>; the minus sign cannot generally be removed. Multiplication by $-i\hbar$ makes the corresponding <momentum operator> Hermitian. These integrations establish the requested symmetry identities; full <self-adjointness> of an unbounded operator additionally requires equality of its domain with its adjoint's domain.
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