Here, I want to know if convolution will preserve the Neumann condition or not. Suppose $K$ is a continuous function and integrable on some interval $[0,L]$, and $u$ is some 'good enouth' function that satisfies the Neumann boundary condition, i.e., $u'(0) = u'(L) = 0$. Define function $H$ by convolution,$$ H[u](x) = \int_0^L K(x-y) u(y) dy$$Does $H[u]$ satisfy Neumann BC?, i.e., Does $H_x [u](0) = H_x[u](L) = 0$ ?
Here, I suppose $u$ to be good enough since I actually don't know which specific condition should be imposed on $u$.
My attempt:
If we assume the following conditions holds: $H$ is self-adjoint, and the Hilbert-Schmidt condition holds, i.e., $\int_0^L \int_0^L K(x-y)^2 dx dy < \infty$, and then we have a discrete sequence of eigenvalues $\{\lambda_j\}$ with finite multiplicity and the corresponding eigenfunctions $\{\phi_j\}$ are an orthogonal basis for $L^2[0,L]$. So if our $u$ here can be written as $ u = \sum_{n=0}^\infty a_n \phi_n$ such that the Neumann boundary condition holds. Then, by plugging this expansion into the convolution directly, I can get what I want.
My question:
- Is there a relatively simpler way without using series expansion to prove this? Like integration by parts (I tried IBP but failed) or other tricks?
- If we have to use the property of Hilbert-Schmidt theory, can we differentiate the series term by term here? Will this introduce new convergence problems so that $u$ requires stronger conditions?