Rank of linear operator involving integral and inner product

Question

I am studying for my exam in Functional Analysis but I'm confused about the following example:

Consider $\mathcal{C}([-\pi,\pi],\mathbb{K})$ with the $\infty$-norm and define $T:X \to X$ by $$Tf(x)=\int^{\pi}_{-\pi} sin(x-y)f(y) dy$$

T has rank 2

I am confused about how to find out that T has rank 2. I only have experience with how to find the rank with matrices. Can you also find the rank when you consider an inner product for example? For example:

Has finite rank

Thanks a lot in advance!

Answer 1

Recall that $\sin(x - y) = \sin(x) \cos(y) - \sin(y) \cos (x)$, so that \begin{align} Tf(x) &= \int (\sin(x) \cos(y) - \sin(y) \cos (x)) f(y) dy \\&= \sin(x) \left[\int \cos(y) f(y) dy \right] - \cos(x) \left[ \int \sin(y) f(y) dy \right] \end{align}The things in square brackets are just numbers, so $Tf$ is always a linear combination of $\sin(x)$ and $\cos(x)$. They form a subspace of dimension 2, so that the operator $T$ will have rank 2.