Eigenvalue of a convolution and a restriction?

Question

Let $\epsilon>0$ be small. Let $\eta(t) = \frac{2\epsilon}{\epsilon^2+(2\pi t)^2}$ (the Fourier transform of $x\mapsto e^{-\epsilon |x|}$). Let $V$ be the space of integrable, bounded functions $f:\mathbb{R}\to \mathbb{R}$ that are supported on $[-1/2,1/2]$ and even. We would like to know for which $f\in V$ with $\|f\|_2=1$ the integral $$I_\epsilon(f) = \int_{-\infty}^\infty \eta(t) (1-f\ast f)(t) dt$$ is minimal, and what that minimum is. Since $\epsilon$ and $\eta$ is fixed, this is the same as the problem of maximizing $\langle \eta,f\ast f\rangle$.

A little calculus of variations shows that, for any extremal $f$ as above, $(\eta \ast f)|_{[-1/2,1/2]}$ has to be a multiple $\lambda f$ of $f$. Moreover, $I_\epsilon(f) = 1 - \langle \eta,f\ast f\rangle = 1 - \langle f\ast \eta,f\rangle = 1-\lambda$. A little work suffices to show that $1-\lambda = O(\epsilon)$.

What is (or are) the extremal $f$, however, and what is $\lambda$? We should at least be able to determine $(1-\lambda)/\epsilon$ as $\epsilon\to 0$.

(Looking at what happens when one takes a high power of the operator $f\mapsto (\eta \ast f)|_{[-1/2,1/2]}$ seems promising, but I suspect there is a concise, standard way of solving this kind of problem.)

Answer 1

Let me simultaneously (a) show why it is clear to me that user548030 is an AI, (b) work through some issues raised in its answer.

1. The first paragraph of user548030's answer restates (correctly) one of my remarks, namely, that minimizing $I_\epsilon(f)$ is the same as maximizing what it calls $J_\epsilon(f)$ (this is trivial). Then it expresses (again correctly) $J_\epsilon(f)$ as an integral on the Fourier side. Note it never uses that integral again. (This is not surprising: that integral looks as if it almost gave us the solution, but the information about the support of $f$ is basically lost on the Fourier side, or rather it becomes more awkward to state.)

By the way, this already shows that, if we are dealing with an AI, we are dealing with an AI that can manipulate mathematical expressions.

2. What it says then about what obtains from calculus of variations would be impressive if it were not a simple restatement of what I said (a careless, incorrect one: it's the restriction of $\eta\ast f$ to $[-1/2,1/2]$ that is a multiple of $f$.

Then it says, completely incorrectly, that a Lagrange multiplier is just an eigenvalue. What happens here is that 'Lagrange multiplier' is a buzzword in the area (in fact the name of a useful concept that can be used to show in the proof of what I alluded to) -- and a common symbol used for the multiplier is $\lambda$, which is also a symbol commonly used for an eigenvalue (and which I used for an eigenvalue).

This is a blooper, but not that a beginning graduate student with a tendency to fake it could never do.

3."The support constraint on $f$ imposes a fundamental limitation on how quickly its Fourier transform can decay. Specifically, for any function supported on $[−1/2,1/2]$, its Fourier transform cannot decay faster than $O(1/k^2)$ as $|k|\to \infty$." The first sentence is correct (in that the Fourier transform of a compactly supported function cannot be compactly supported; there are several useful statements here that go by the name of 'uncertainty principle'). The second one is quite wrong: the Fourier transform of $\widehat{f}(t)$ of $t$ can decay faster than any power of $t$ as $|t|\to \infty$ (incidentally, $k$ is a funny choice of letter here); for that to happen, it is enough that $f$ be in $C^\infty$.

4. Now, the function $\sqrt{2} \cos(\pi x)$ is a completely reasonable thing to try; in fact, I suggested it in a post on Facebook (with a change in constants) before a friend shot it down for being naïve. It's a nice function. On the other hand, part of the reason given by user548030 is nonsense ("Its Fourier transform achieves the slowest possible decay rate $𝑂(1/k^2)$") and part is qualitatively sensible but unsupported ("...while being as concentrated as possible near $k=\pm \pi$. This is optimal because we cannot concentrate the transform closer to 0 due to the support constraint")

This is a little frightening. Either user548030 can do a simple Ansatz, or it is snooping friends-only posts on Facebook (or a Facebook friend is involved, likely feeding an AI).

Some further thought shows that this is not a good Ansatz or such a good choice of function. While $\sqrt{2} \cos(\pi x)$ satisfies a boundary condition (vanishing at the ends of the interval), it is not a condition that we set, or indeed one that should be set: a moment's thought shows that an $f$ maximizing $J_\epsilon(f)$ has to be non-negative everywhere (or non-positive everywhere) as well as (obviously) not identically zero, and then $\eta\ast f$ cannot vanish at $1/2$ or $-1/2$; hence, an $f$ with $\lim_{t\to 1/2} f(t) = 0$ or $\lim_{t\to -1/2} f(t) = 0$ cannot be an eigenfunction of $f\mapsto (\eta\ast f)|_{[-1/2,1/2]}$.

Funnily enough, the fact that an eigenfunction $f$ will have to be discontinuous at $1/2$ (and $-1/2$) should mean $\widehat{f}(t)$ will decay no faster than $O(1/t)$ (and so some care will have to be exercised, as $\widehat{f}$ will not be in $L^1$, though of course it will be in $L^2$).

It might also be that there is an eigenfunction $f$ such that $\lim_{t\to \frac{1}{2}^-} f(t)= O(\epsilon)$ (for all that I know).

5. The calculation of the Rayleigh quotient is of course incorrect, since the kernel on the physical (non-Fourier) side is $\eta(y-x)$, not $e^{-\epsilon |x-y|}$. Again, not a mistake a student (or someone careless) would never make, but I think the evidence up to this point was already enough to make it clear we are dealing with an AI (even ignoring the fact that this is "new contributor user548030"). The rest is then useless (or almost useless - see below). Still, I am starting to be impressed even with some of its human-style mistakes (much less so with others).

Using the right kernel here doesn't seem to give something very nice here (but maybe I'm not using a powerful enough symbolic integrator :P).

Here's another way to do things: first, take the Fourier transform of $f = \sqrt{2} \cos \pi x|_{[-1/2,1/2]}$. This is $$\widehat{f}(t) = \frac{\mathrm{sinc}(t+1/2)+\mathrm{sinc}(t-1/2)}{\sqrt{2}},$$ where $\mathrm{sinc}(t) = \frac{\sin \pi t}{\pi t}$. We could of course write $\mathrm{sinc}(t+1/2) = \frac{\cos(\pi t)}{\pi (t+1/2)}$ and $\mathrm{sinc}(t-1/2) = -\frac{\cos(\pi t)}{\pi (t-1/2)}$, and so $$\widehat{f}(t) = \frac{\cos \pi t}{\sqrt{2} \pi \left(\frac{1}{4}-t^2\right)}.$$

Then $$\widehat{\eta\ast f}(t) = \widehat{\eta}(t) \widehat{f}(t) = e^{-\epsilon |t|} \frac{\cos \pi t}{\sqrt{2} \pi \left(\frac{1}{4}-t^2\right)}$$ and so, by Plancherel, $$\langle \eta\ast f, f\rangle = \langle \widehat{\eta\ast f},\widehat{f}\rangle = \int_{-\infty}^\infty e^{-\epsilon |t|} \left(\frac{\cos \pi t}{\sqrt{2} \pi \left(\frac{1}{4}-t^2\right)}\right)^2 dt = 2 \int_0^\infty e^{-\epsilon t} \left(\frac{\cos \pi t}{\sqrt{2} \pi \left(\frac{1}{4}-t^2\right)}\right)^2 dt.$$

Our known computer friends SAGE/sympy/giac/maxima and WolframAlpha are stumped here, but our new friend user548030 is of course completely right that a Taylor expansion around $t = 0$ can be useful here. (Hardly rocket science, but this would again be consistent with a strong undergrad/beginning graduate student in a field close to math who otherwise also bullshits a lot and makes a lot of mistakes.)

We have $e^{-\epsilon t} = 1 - \epsilon t + O((\epsilon t)^2)$ for $t>0$ (even for $t$ large). Hence $$\langle \eta\ast f, f\rangle = 2(I_0 - \epsilon I_1 + O(\epsilon^2) I_2),$$ where $$I_k = \int_0^\infty t \left(\frac{\cos \pi t}{\sqrt{2} \pi \left(\frac{1}{4}-t^2\right)}\right)^2 dt.$$ An easy calculation shows that $I_0=1/2$, $$I_1 = \frac{\pi \mathrm{Si}(\pi)-2}{2\pi^2} = 0.19342\dotsc$$ where $\mathrm{Si}(\pi) = \int_0^\pi \frac{\sin x}{x} dx$, and $$I_2 = 1/8.$$

Hence, $$I_\epsilon(f) = 1 - \langle \eta\ast f,f\rangle = 2 I_1 \epsilon + O(\epsilon^2)\approx 0.38685\epsilon + O(\epsilon^2),$$ where the implied constant is small and can be made explicit very easily.

This shouldn't be optimal (in fact I think one can easily find better functions), but it is something.

For comparison, for any $f$ with $\|f\|_2 = 1$, we trivially have $$I_\epsilon(f) \geq \int_{|t|>1} \eta(t) dt = (1+O(\epsilon^2)) \epsilon \int_{|t|>1} \frac{2}{(2\pi t)^2} dt = \frac{\epsilon}{\pi^2} + O(\epsilon^3) \approx 0.10132 \epsilon + O(\epsilon^3).$$

Also consider $f= 1_{[-1/2,1/2]}$. Then $\widehat{f}(t) = \mathrm{sinc}(t)$, and so $$\langle \eta\ast f, f\rangle = \langle \widehat{\eta\ast f},\widehat{f}\rangle = \int_{-\infty}^\infty e^{-\epsilon |t|} \mathrm{sinc}(t)^2 dt.$$ Here the approximation is trickier because of convergence issues; I'll finish this later. I think there's a closed expression that the symbolic integrators I have barely fail to find. EDIT: silly me, of course one wants to work in physical space for that $f$: $$\begin{aligned}\langle \eta, f\ast f\rangle &= 2 \int_0^\infty \eta(t) dt - 2 \int_0^1 \eta(t) t dt = 1 - 4\epsilon\cdot \int_0^1 \frac{t dt}{\epsilon^2 + (2\pi t)^2}\\ &= 1 - \frac{\epsilon}{2\pi^2}\cdot \log \left(1 + \left(\frac{2\pi}{\epsilon}\right)^2\right),\end{aligned}$$ and so $$I_\epsilon(f) = \frac{\epsilon}{2\pi^2}\cdot \log \left(1 + \left(\frac{2\pi}{\epsilon}\right)^2\right),$$ which, for very small $\epsilon>0$, is much worse than the $\cos$ example above; it is not even $O(\epsilon)$.

In general, if $f$ is absolutely continuous with $f'$ of bounded total variation (as $f=1_{[-1/2,1/2]}$ is not), then $\widehat{f}(t)$ decays at least as fast as $1/t^2$ as $t\to \pm \infty$. Then we can proceed as before, and obtain $$\langle \eta\ast f, f\rangle = 1 - 2\epsilon I_{1}(f) + O_f(\epsilon^2),$$ where $$I_{1}(f) = \int_0^\infty t (\widehat{\eta}(t))^2 dt.$$ So, we want to minimize $I_1(f)$.