# Finding the nullspace and range of an integral operator

I'm trying to determine the nullspace and range of the following integral operator, but I'm having trouble proceeding. Let $K:C([0,1])\to C([0,1])$ be defined by $$Kf(y)=\int_{0}^1 \sin(\pi(x-y))f(y)\,dy.$$ Playing around with several functions, I see that if $f\equiv 1$, then $$K(y)=\int_{0}^1\sin(\pi(x-y))\,dy=-\frac{2\cos(\pi x)}{\pi},$$ and if $f\equiv 0$, then $Kf(y)=0$.

The form $\sin(\pi(x-y))$ in $K$ makes me think that the range might be periodic function in $C([0,1]), but this is a guess with no intuition. Edit: I didn't include this originally, but as Daniel's mentioned, the addition formula implies that $$(Kf)(y)=\sin (\pi x) \int_{0}^1 \cos (\pi y)f(y)\,dy - \cos (\pi x) \int_{0}^1\sin (\pi y)f(y)\,dy.$$ However, I'm not seeing what this implies either. In this form$K$reminds me of the Riemann-Lesbegue lemma for$L^1([0,1]),$but I'm not sure what what that invokes. So my question is ultimately: what should I look for when determining the range and nullspace of integral operators? • Here: The addition theorem. $$\sin \bigl(\pi(x-y)\bigr) = \sin (\pi x) \cos (\pi y) - \cos (\pi x) \sin (\pi y)$$ Generally, hmmm. Any symmetries of the integral kernel help, but I have no universal recipe. – Daniel Fischer Nov 17 '17 at 19:26 • Should$ f(y) $be in the integrand where you define$K(f)$– Dionel Jaime Nov 17 '17 at 19:35 • @DionelJaime Yes, I have corrected it. – user225477 Nov 17 '17 at 19:37 • @DanielFischer, I noticed that but didn't include it in the original question posted. I added more information in the post, but I'm not sure what I get from applying the identity. – user225477 Nov 17 '17 at 19:38 • Also should it be$K(f)(x)\$ otherwise your function is constant. – Dionel Jaime Nov 17 '17 at 19:39