Existence of a function such that given integral vanishes 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
$$
\int_0^T \int_0^\infty \frac{u^2-s}{s^{5/2}} \exp{\left( -\frac{u^2}{2s} \right)} f(u) g(s) du ds = 0?
$$
 A: 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 $s\searrow 0$. 
A: 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.
