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Here is an exercise I had to deal with recently

Let $\phi : f \in C([-\pi,\pi], \mathbb R) \mapsto\int_{[-\pi,\pi]} sin(x+t)f(t)dt $

I proved $\phi$ is an endomophism, and then I had to discuss its eigenvalues, eigenvectors and eigenspaces. So $\phi(f) = \lambda f$ is equivalent to $\lambda f(x) = sin(x)\int_{[-\pi,\pi]}f(t)cos(t)dt + cos(x)\int_{[-\pi,\pi]}f(t)sin(t)dt$ for all $x$ in $[-\pi,\pi]$

If $\lambda \neq 0$ then $f \in span(sin, cos)$ ($1$)

If $\lambda = 0$ then $=0$ and $=0$ so f is in the intersection of 2 hyperplanes (one is orthogonal to $cos$ and the other is orthogonal to $sin$) ($2$)

Can you expand points ($1$) and ($2$) please. Thank you

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    I assume you mean $\langle f,\sin\rangle=0$ and the scalar product is as usual: $\langle f,g\rangle=\int_{[\-pi,\pi]}fg \ dt$. – 2017-01-14
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    Yes of course it's the usual scalar product. Ty, I have replaced cos by sin. – 2017-01-14
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    The definition of $\phi$ is a disguised convolution operator... – 2017-01-14
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    (I've never seen it before) – 2017-01-14

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Point $(1)$ is just the formula you wrote: let $$a:=\langle f,\cos\rangle=\int_{[-\pi,\pi]}f(t)\cos(t)dt$$ and $$b:=\langle f,\sin\rangle=\int_{[-\pi,\pi]}f(t)\sin(t)dt.$$ Then, $\lambda f=a\cos(x)+b\sin(x)$ implies $f=\frac{a}{\lambda}\cos(x)+\frac{b}{\lambda}\sin(x)$.

Point $(2)$ also follows easily: suppose $a\neq 0$, then for some $x$ $a\cos(x)+b\cos(x)\neq 0$, impossible since you are assuming $\lambda=0$.

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    We have to describe precisely the eigenvalues, eigenvectors and eigenspaces. So ($1$) seems to be calculus and for ($2$), what is this intersection of hyperplanes ? – 2017-01-14