Let $V$ be a set of $2\pi$-periodic, real, infinitely differentiable functions.
Let $D = \frac{d^2}{dx^2}$. I found that $D$ is a linear transformation on $V$.
Now, with matrices, An $n x n$ matrix $A$ is symmetric if for every $v,w \in \mathbb{R^n}, (Av)\cdot w = v \cdot (Aw).$ So if we extend this to functions, what does it mean for $D$ to be symmetric on $V$.
I thought maybe that it's $\langle Df, g\rangle = \langle f, Dg \rangle$, but I tried proving this with integration by parts and got confused. My instinct tells me this is the correct definition. Can anyone help me out?
I.e.
Prove that for any $f,g \in V$ that
$$ \int_{-\pi}^{\pi} f''(x)g(x)dx = \int_{-\pi}^{\pi} f(x)g''(x)dx $$
Let $u=g(x)$ and $dv=f''(x)dx$. Then by integration by parts,
$$\int_{-\pi}^{\pi}f''(x)g(x)dx=f'(\pi)g(\pi)-f'(-\pi)g(-\pi)-\int_{-\pi}^{\pi}f'(x)g'(x)dx.$$ Here the key assumption you need is that $f$ and $g$ are $2\pi$-periodic, with $[-\pi,\pi]$ being the domain for one $2\pi$-period. With this, the boundary terms vanish, and we get
$$\int_{-\pi}^{\pi}f''(x)g(x)dx=-\int_{-\pi}^{\pi}f'(x)g'(x)dx$$ and we are done. You might object: how are we done? Do we not need to show that $$\int_{-\pi}^{\pi}f''(x)g(x)dx=\int_{-\pi}^{\pi}f(x)g''(x)?$$ However, notice that in the second-to-last equation, we showed that $\int_{-\pi}^{\pi}f''(x)g(x)dx$ is equal to an expression symmetric in terms of $f$ and $g$, so we are indeed finished.