I was wondering if someone could give me a few examples of some self adjoint operators, I understand this topic in terms of matrices with inner product spaces. However I have seen integrals that are inner products and I don't know any examples for the inner product space on an integral?

Could someone give me an example of a self adjoint operator and non self adjoint operator on the integral inner product space please! Thank you :)

Sure. Let $$K\colon [0,1]\times [0,1] \to \Bbb C$$ be a continuous function and define $$T_K\colon \mathcal{C}([0,1], \Bbb C) \to \mathcal{C}([0,1], \Bbb C)$$ by $$T_K(f)(s) = \int_0^1 K(t,s) f(t)\,{\rm d}t.$$Assume that the inner product in $$\mathcal{C}([0,1], \Bbb C)$$ is given by $$\langle f,g\rangle = \int_0^1 f(s) \overline{g(s)}\,{\rm d}s.$$If $$K$$ satisfies the Hermitian symmetry $$K(s,t) = \overline{K(t,s)}$$, then $$T_K$$ is self-adjoint. Because $$\langle T_K(f),g\rangle = \int_0^1 T_K(f)(s)\overline{g(s)}\,{\rm d}s = \int_0^1\int_0^1 K(t,s)f(t)\overline{g(s)}\,{\rm d}t\,{\rm d}s,$$as well as $$\langle f, T_K(g)\rangle = \int_0^1 f(s) \overline{T_K(g)(s)}\,{\rm d}s = \int_0^1\int_0^1 f(s) \overline{K(t,s) g(t)}\,{\rm d}t\,{\rm d}s.$$Using the Hermitian symmetry of $$K$$ and renaming $$t \leftrightarrow s$$ if needed, we have $$\langle T_K(f),g\rangle = \langle f,T_K(g)\rangle$$. For a non self-adjoint operator, take $$T_K$$ where $$K$$ does not satisfy that Hermitian symmetry.

• Could you give an example of such f, g, K etc just for clarity sake? Thank you though!
– user635953
Commented Dec 9, 2020 at 21:52
• Sure, compute both integrals with $K \equiv 1$ and $f(t)=g(t)=t$ to get a feeling for what happens, it's easy enough. Commented Dec 9, 2020 at 22:02
• Sorry another question! When would K not satisfy hermitian symmetry?
– user635953
Commented Dec 9, 2020 at 22:18
• Try to write any example, it will probably fail. Like $K(t,s)=1+{\rm i}s$, etc. Commented Dec 9, 2020 at 22:23
• Thank you very much
– user635953
Commented Dec 9, 2020 at 22:30