Hermiticity of operator on square integrable functions I'm trying to solve this problem for my Quantum Mechanics class but its my first time dealing with infinite dimensional spaces so I'm a little lost here. I need to show that the operator defined by
$$
i\frac{d}{dx}\delta(x-x_0)
$$
is not Hermitian in the $L^2(\mathbb{R})$ space.
I tried calculating the integrals
$$
\int\limits^{\infty}_{-\infty}\phi^*(x)i\frac{d}{dx}\phi(x)dx\\
\int\limits^{\infty}_{-\infty}\left(\phi(x)i\frac{d}{dx}\right)^*\phi(x)dx
$$
but i didnt get anywhere, maybe i made a mistake but I'm not sure.
Any help very welcome!
 A: The Hermitian adjoin $S^\dagger$ of an operator $S$ satisfies $\langle u,Sv\rangle=\langle S^\dagger u,v\rangle$.
With $S\phi:=i\frac{\mathrm{d}}{\mathrm{d}x}\big(\delta(x-x_0)\phi(x)\big)$ we can work out
$$ \begin{array}{ll} 
\langle u,Sv\rangle & \displaystyle = \int_{\mathbb{R}}\overline{u(x)}~i\frac{\mathrm{d}}{\mathrm{d}x}\Big(\delta(x-x_0)v(x)\Big)~\mathrm{d}x \\[5pt]
 & = \displaystyle i\int_{\mathbb{R}} \overline{u(x)} \Big( \delta'(x-x_0)v(x)+\delta(x-x_0)v'(x) \Big)\,\mathrm{d}x \\[5pt]
& = \displaystyle i\int_{\mathbb{R}} \delta'(x-x_0)\overline{u(x)}v(x)\,\mathrm{d}x+i\int_{\mathbb{R}}\delta(x-x_0)\overline{u(x)}v(x)\,\mathrm{d}x \\[5pt]
& = \displaystyle -i\Big(\overline{u'(x_0)}v(x_0)+\overline{u(x_0)}v'(x_0)\Big)+i\overline{u(x_0)}v'(x_0) \\[5pt]
& =  -i\overline{u'(x_0)}v(x_0) \\[5pt]
& = \displaystyle \int_{\mathbb{R}}\overline{i\delta(x-x_0)\frac{\mathrm{d}}{\mathrm{d}x} u(x)}\, v(x)\,\mathrm{d}x  \end{array} $$
Therefore, $\Big(i\frac{\mathrm{d}}{\mathrm{d}x}\delta(x-x_0)\Big)^\dagger = i\delta(x-x_0)\frac{\mathrm{d}}{\mathrm{d}x}$ which is a different operator (as you can show by example); in other words $S^\dagger\ne S$ so $S$ is not Hermitian.
One could also use the facts $(iS)^\dagger=-iS^\dagger$ and $(AB)^\dagger=B^\dagger A^\dagger$ along with the fact $\delta(x-x_0)$ is Hermitian and $\frac{\mathrm{d}}{\mathrm{d}x}$ is anti-Hermitian (via by-parts integration and functions decaying at $\pm\infty$).
