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Does there exists a function $f \in C^2[0,\infty]$ (that is, $f$ is $C^2$ and has finite limits at $0$ and $\infty$) with $f''(0) = 1$, such that for any $g \in L^p(0,T)$ (where $T > 0$ and $1 \leq p < \infty$ may be chosen freely) we get $$ \int_0^T \int_0^\infty \frac{u^2-s}{s^{5/2}} \exp{\left( -\frac{u^2}{2s} \right)} f(u) g(s) du ds = 0? $$

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When you say that $T$ and $p$ can be chosen freely, do you mean that we can just prove the statement for one set of $(T,p)$, or do you mean that we should prove the statement for all $(T,p)$ pairs? – Willie Wong Mar 14 '11 at 10:32
You can take $g(s)$ outside the inner integral; then the inner integral is a continuous function of $s$ on $]0,T[$, so unless it's identically zero you can choose $g$ as an arbitrarily narrow bump function around a non-zero value to make the outer integral non-zero -- so I think you can reduce this question to the question whether $f$ can be chosen such that $\int_0^\infty (u^2-s)\exp\left(-\frac{u^2}{2s}\right)f(u)\mathrm{d}u$ is identically zero. (That appears unlikely.) – joriki Mar 14 '11 at 10:40
@Willie Wong, I meant to prove it for one set of $(T,p)$. – xen Mar 14 '11 at 20:39
@joriki: thanks, now I've got the idea. – xen Mar 14 '11 at 20:40
up vote 6 down vote accepted

I'm pretty sure the answer is no, there exists no such $f$. Here I first give a physical argument.

First define the function $K(u,s) = \frac{1}{\sqrt{s}} \exp (-u^2 / 2s )$. Up to some constant normalisation factors, this is the Heat Kernel.

Observe that your integral can be written as, by an explicit calculuation,

$$ \int_0^T \int_0^\infty (\partial_{uu}^2K)(u,s)f(u)g(s) ~ du ~ ds $$

Now, notice that the condition is linear. If $f$ and $\tilde{f}$ are solutions, then $af + b\tilde{f}$ is also a solution. Next, observe that using the fundamental theorem of calculus, $f(u) = \textrm{constant}$ is a solution: since $\partial_uK$ vanishes both at 0 and infinity. Therefore we can, without loss of generality, assume that the solution we are looking for has $f(0) = 0$. So we can extend $f$ continuously to the negative real line by setting $f(x) = 0$ whenever $x < 0$. Then using that $K$ is a even function:

$$ \int_0^{\infty} \partial_{uu}^2K(u,s)f(u) du = \int_{-\infty}^\infty \partial_{uu}^2K(0-u,s)f(u) du = \partial_{uu}^2 (K_s*f)(0) $$

where $K_s*f$ is the convolution of the heat kernel $K(u,s)$ against $f(u)$ extended to the whole real line. In other words, that is the evaluation of the second derivative of $K_s*f$ at the origin.

Now, $K_s*f$ is a solution to the heat equation with initial data at $s=0$ being $f$. So in particular, up to a constant factor,

$$ \partial_{uu}^2(K_s*f) = \partial_t(K_s*f)$$

So your desired integral condition, since you allow $g$ to be arbitrary, tells you that $K_s*f(0) = 0$ for all $s$. (Which is, in fact, basically what joriki wrote in his comment.)

The condition that $f''(0) = 1$ tells you that the initial temperature fluctuation is non-zero near the origin. As heat is diffusive, to the left of the origin you have no heat content, to the right you start with some non-zero temperature arbitrarily close to the origin. So in arbitrary short time you should feel some heat at the origin.

The mathematical argument follows:

Now, using joriki's comment, the problem reduces to considering on constant $s$ slices. Integrating by parts twice in $u$, we have that your condition implies

$$ \int_0^\infty K(u,s) f''(u) du + \frac{1}{\sqrt{s}} f'(0) = 0 $$

Taking $s\to 0$, the first term converges to some finite value which is non-zero by the assumption that $f''(0) = 1\neq 0$ and $f''$ is continuous. This gives a contradiction as, if $f'(0) = 0$ then the above equation would require $f''(0) = 0$. And if $f'(0) \neq 0$, the above equation shows that the integral $\int_0^\infty K(u,s) f''(u) du \nearrow \infty$ as $s\searrow 0$.

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Thank you very much for the answer! I now see where the problem is. I will edit my question and add some assumptions on $g$. – xen Mar 14 '11 at 20:49

As Joriki pointed out in his comment, this is equivalent to finding an $f(u)$ such that for all $0 \leq s \leq T$ one has $$\int_0^{\infty}(u^2 - s)\exp{(-{u^2 \over 2s})}f(u) \,du = 0$$ Write $\int_0^{\infty}u^2\exp{-({u^2 \over 2s})}f(u)\,du$ as $-\int_0^{\infty}-{u \over s}\exp{(-{u^2 \over 2s})}suf(u)\,du$ and integrate by parts, integrating $-{u \over s}\exp{(-{u^2 \over 2s})}$ to $\exp{(-{u^2 \over 2s})}$, and differentiating the rest. The result is the expression $$\int_0^{\infty}\bigg(s\exp{(-{u^2 \over 2s})}f(u) + suf'(u)\exp{(-{u^2 \over 2s})}\bigg)\,du$$ The first term of this cancels out the second term of your original expression, so what you need is a function $f(u)$ such that for all $0 \leq s \leq T$ one has $$s\int_0^{\infty}uf'(u)\exp{(-{u^2 \over 2s})}\,du = 0$$ You can cancel out the $s$ factor in front, then change variables from $u$ to $\sqrt{u}$ to get $$\int_0^{\infty}f'(\sqrt{u})\exp{(-{u \over 2s})}\,du = 0$$ Lastly, one can replace $s$ by ${1 \over 2s}$ to get that for all $s \geq {1 \over 2T}$ you need $$\int_0^{\infty}f'(\sqrt{u})\exp{(-{su})}\,du = 0$$ This can't happen; the above defines analytic function of $s$ which can't be zero on a segment without being identically zero. So the Laplace transform of $f'(\sqrt{u})$ is identically zero, which for reasonable $f'$ will not happen unless $f'$ is identically zero.

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Thanks for the another answer! – xen Mar 14 '11 at 20:50

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