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Is there any other solution for $f(x)=\int_0^\pi t f(t)\,\mathrm{d}t+\cos x $? I found one solution as follows but I have no clue how to prove that this solution is unique.

Let $\int_0^\pi t f(t)=\lambda$, we have

\begin{align} \lambda &=\int_0^\pi t f(t) \,\mathrm{d}t\\ &=\int_0^\pi t (\lambda+\cos t)\,\mathrm{d}t\\ &= \frac{4}{\pi^2-2} \end{align}

Thus $f(x)=\frac{4}{\pi^2-2}+\cos x$.

Question

Is there any other solution? How to prove the uniqueness of this solution?

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Assume that you have $f,g$ such that $$\cos(x)=f(x)-\int_0^\pi tf(t)dt=g(x)-\int_0^\pi tg(t)dt.$$ It then follows that $$0=f(x)-g(x)-\int_0^\pi tf(t)dt+\int_0^\pi tg(t)dt=f(x)-g(x)-\int_0^\pi t(f(t)-g(t))dt.$$ So you get $$f(x)-g(x)=\int_0^\pi t(f(t)-g(t))dt=C$$ a constant. So write $f(x)=g(x)+C$ and calculate $$C=\int_0^\pi t(f(t)-g(t))dt=\int_0^\pi t(g(t)+C-g(t))dt\\ =\int_0^\pi t(g(t)-g(t))dt+\int_0^\pi tCdt=C\int_0^\pi tdt =C\frac{\pi^2}{2}.$$

The only value of $C$ such that this equation can hold is $C=0$. Consequently, solutions are unique.

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Suppose $g$ is another solution, so $$ g(x)=\int_0^\pi tg(t)\,dt+\cos x $$ Then $f'(x)=g'(x)$ and therefore $g(x)=k+f(x)$. Hence $$ g(x)=\int_0^\pi t(k+f(t))\,dt+\cos x= \int_0^\pi kt\,dt+f(x)=\frac{k\pi^2}{2}+f(x) $$ Hence $k\pi^2/2=k$, so $k=0$.

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  • $\begingroup$ @ArtificialStupidity You can't see the downvote; I do. Anyway, I apologize for attributing it to you. $\endgroup$ – egreg Feb 13 at 14:47
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The problem can be write in this way: $f'(x)=-sin(x) ;\hspace{0.1cm} f(\pi /2)=\lambda$ because $f$ is continous and lipschitz=> Picard–Lindelöf theorem can by applied=>solution exist and is unique.The proof for Picard–Lindelöf is long you can find a proof on wikipedia.

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