How do you call operators $T$ that satisfy $\langle Tx, y \rangle = - \langle x, T y \rangle$

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 $$$$\tag{1} \int_{a}^{b} u(x) \Psi(\phi)(x) dx = - \int_{a}^{b} \Psi(u)(x) \phi(x) dx.$$$$ 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.

• 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. – Wojowu Apr 12 at 19:36
• physics.stackexchange.com/questions/156911/… may be of interest. – Sylvain Julien Apr 12 at 20:53
• – Keith McClary Apr 13 at 3:24
• @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. – Wojowu Apr 13 at 4:08