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I have a positive continuous function $f(x)$ that is strictly monotonically decreasing on the interval $(0,1)$. That is, $f(x)>0$ and $f'(x)<0$ for $x\in(0,1)$. I want to show that the integral $$I = \int_{0}^{1}\! f(x)\cos(k\pi x)dx$$ is always positive, for all $k = 0,1,2,...$

Edit: As pointed out in the comments, it's not possible to show this for general $f(x)$, but I have numerical eveidence to suggest that this is true for $$f(x) = {\frac {\left( 3\,{\epsilon}^{2}-\cos \left( \pi\,x \right) +1 \right) }{ \left( 2\,{\epsilon}^{2} -\cos \left( \pi\,(x) \right) +1 \right) ^{3/2}}}$$ where $0<\epsilon<1$.

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  • $\begingroup$ where did you get this problem? $\endgroup$ – Levon Minasian Oct 30 at 11:39
  • $\begingroup$ I'm trying to show that the eigenvalues of a particular Fredholm integral operator are positive. The eigenfunctions are the Fourier modes, but as $f(x)$ is even, it suffices to show the above integral is positive. $\endgroup$ – Ben94 Oct 30 at 11:42
  • $\begingroup$ I can't see any useful way of restricting $f$ so that your conjecture is true. As it stands, it is false for all $k\ge 2$. $\endgroup$ – TonyK Oct 30 at 13:20
  • $\begingroup$ @TonyK, hmm yep perhaps I need to redefine the problem. I have a particular f(x) that I am looking at where this integral remains positive for all $k$ (by numerically evaluating the integral). I thought it was due to the above properties, but as you pointed out, there must be something special about the $f(x)$ I'm looking at. I'll write out the function explicitly in the question $\endgroup$ – Ben94 Oct 30 at 13:28
  • $\begingroup$ A counterexample for $k=2$: I find that $\int_{0}^{1} (1-x^2) \cos 2\pi x~dx=-1/(2\pi^2)$ $\endgroup$ – Z Ahmed Oct 30 at 14:05
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For $k=1$: $$I=\int_{0}^{1} f(x) \cos \pi x ~ dx ~~~(1)$$ Use: $\int_{0}^{a} f(x) dx=\int_{0}^{a} f(a-x) dx$, then $$I=\int_{0}^{1} -f(1-x) \cos \pi x ~dx~~~~~(2)$$ Add (1) and (2), to gt $$2I=\int_{0}^{1}[f(x)-f(1-x)] \cos \pi x ~dx~~~~(3)$$ $x <1-x, x \in (0,1/2)$, as $f(x)$ is decreasing function, we get $$f(x)>f(1-x), x \in (0,1/2), ~f(x)< f)1-x), x \in (1/2,1)~~~~~~(4)$$ Then $$2I=\int_{0}^{1/2} [f(x)-f(1-x)] \cos \pi x~dx+\int_{1/2}^{1} [f(x)-f(1-x)] \cos \pi x ~dx~~~~(5)$$ $\cos \pi x <0, ~in~ (0,1/2)$ and $\cos \pi x >0 ~ in~ (1/2,1)$. So it follows that $$2I>0$$

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  • $\begingroup$ That's a really interesting way to show this for the $k=1$ case. Would it be possible to extend this to general $k$ by dividing up the integral into more pieces, as you do in the last part of your proof? $\endgroup$ – Ben94 Oct 30 at 13:52
  • $\begingroup$ @Ben94 Thanks, I have just mentioned a counterexample for $k=2$ in the question. $\endgroup$ – Z Ahmed Oct 30 at 14:07
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$I$ can be negative. For instance, with $k=2$, take the function $g(x)=1$ if $x\le\frac34$, and $g(x)=0$ if $x>\frac34$. Then $$\int_0^1 g(x)\cos(2\pi x)dx=\int_0^\frac34\cos(2\pi x)dx=-\frac{1}{2\pi}$$ Of course, $g$ doesn't satisfy your conditions; but you can easily construct a continuous, differentiable, strictly decreasing $f$ that does satisfy them, and is sufficiently close to $g$ that the integral is still negative.

You can use this idea to construct a counter-example for any integer $k\ge 2$ $-$ just replace $\frac34$ with $\frac{3}{2k}$.

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  • $\begingroup$ Ok yep, I should have said that $f(x)$ is a continuous function. Then would we be able to at least show that the integral is at least non-negative? I have edited the question to include that $\endgroup$ – Ben94 Oct 30 at 11:26
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    $\begingroup$ @Ben94 You can make $f$ continuous and as close to $g$ as you want (in whatever norm, which makes the integral as close to $-1/(2\pi)$ as you want), so continuity makes no difference. $\endgroup$ – runway44 Oct 30 at 11:53

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