First, some definitions I know an linear bounded operator $T: \mathscr{H}_1 \to \mathscr{H}_2$ between to Hilbert spaces always has an so called adjoint operator $T^*: \mathscr{H}_2 \to \mathscr{H}_1$ that satisfies $$ \langle Tx, y \rangle_{\mathscr{H}_2} = \langle x, T^* y \rangle_{\mathscr{H}_1} $$ We call $T$ self-adjoint if $T = T^*$.

In the context of partial differential equations (in one dimensionional spaces) we define $v$ to be the weak derivative of $u$ if all test functions $\phi \in \mathcal{C}_{\text{c}}^{\infty}((a,b))$ satisfy $$ \int_{a}^{b} u(x) \phi'(x) dx = - \int_{a}^{b} v(x) \phi(x) dx. $$ Here, $u,v \in L^1_{\text{loc}}((a,b))$ and $a,b \in \mathbb{R}$. $\phi'$ is the classical derivative and coincides with the weak derivative since $\phi \in \mathcal{C}^{\infty}$.

I want to know if you can regard the operator $\Psi$ defined by $u \mapsto u'$, where $u'$ is the weak derivative can be understood as an kind of "anti-selfadjoint" operator, since we have \begin{equation} \tag{1} \int_{a}^{b} u(x) \Psi(\phi)(x) dx = - \int_{a}^{b} \Psi(u)(x) \phi(x) dx. \end{equation} and if the is any name for such operators.

I also see that there are some problems: $L^1_{\text{loc}}((a,b))$ probably isn't a Hilbert space and I think $\langle f,g \rangle := \int_{a}^{b} f(x) g(x) dx$ should be a scalar product but I'm not sure. Note that here we assume $\mathscr{H}_1 = \mathscr{H}_2$ with notation from above, which should simplify things a little.

Edit The formula (1) obviously holds if $\Psi$ maps a function to it's classical derivative, because then, this formula is just integration by parts, which is where this formula and intuition for weak derivative originates from.

Edit 2: Perhaps related (not duplicate, I hope...) this question. Also related: this question.

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    $\begingroup$ You can restrict attention to elements of $L^2((a,b))\subseteq L^1_{loc}((a,b))$, which is a Hilbert space. Then $\langle f,g\rangle=\int_a^b f(x)g(x)dx$ is a perfectly legitimate inner product. $\endgroup$ – Wojowu Apr 12 at 19:36
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    $\begingroup$ physics.stackexchange.com/questions/156911/… may be of interest. $\endgroup$ – Sylvain Julien Apr 12 at 20:53
  • $\begingroup$ Skew-Hermitian. $\endgroup$ – Keith McClary Apr 13 at 3:24
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    $\begingroup$ @ViktorGlombik.Actually, for any bounded interval, $L^2((a,b))\subseteq L^1((a,b))$, as follows from Cauchy-Schwarz. Hence $L^2((a,b))\subseteq L^1_{loc}((a,b))$ for any, possible unbounded, interval. $\endgroup$ – Wojowu Apr 13 at 4:08

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