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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?

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  • $\begingroup$ 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. $\endgroup$ – Daniel Fischer Nov 17 '17 at 19:26
  • $\begingroup$ Should$ f(y) $be in the integrand where you define $K(f)$ $\endgroup$ – Dionel Jaime Nov 17 '17 at 19:35
  • $\begingroup$ @DionelJaime Yes, I have corrected it. $\endgroup$ – user225477 Nov 17 '17 at 19:37
  • $\begingroup$ @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. $\endgroup$ – user225477 Nov 17 '17 at 19:38
  • $\begingroup$ Also should it be $K(f)(x)$ otherwise your function is constant. $\endgroup$ – Dionel Jaime Nov 17 '17 at 19:39

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