Estimating curvature of oscillatory curve based on global constraints I have a heuristic question about using global constraints of a problem to make estimates of local features of a curve, such as its curvature. 
Consider a suitably well behaved function on $\mathbb{R}$, denoted $f(x,t)$, where $t\in \mathbb{R}^+$ denotes time. Now, $f(x,t)$ is oscillatory and decays in a suitable fashion as $x\to \pm \infty$. Due to symmetries of the problem, I know that the number of waves (in this context I believe it's the same as saying number of zero crossings) of the function remains constant with respect to time.
Quantitatively this is the statement that 
$\frac{d N(t)}{dt} \equiv \frac{d}{dt} \int_{-\infty}^{\infty} |f(x,t)|^2 \ dx =0$. 
Next, by exploiting other conservation laws, I know that the variance of the signal goes to zero, i.e. 
$V(t)  = \int_{-\infty}^{\infty} (x-\bar{x})^2 |f(x,t)|^2 \ dx \to 0 $
in $t$ finite, where 
$\bar{x} = \int_{-\infty}^{\infty} x |f(x,t)|^2\ dx$. 
Now, I want to make a quantitative statement about the magnitude of the curvature, which for a graph is defined as
$\kappa = \frac{|f''|}{(1+f'^2)^{\frac{3}{2}}}$,
where primes $'$ denote differentiation with respect to x. 
My question is, how can I use the information I have about $V(t)$, as well as $N(t)$, to make a statement about the local curvature of $f(x,t)$? Naively, it seems easy to see that as $V(t)$ goes to 0, the curvature must get large, as the number of oscillations stays fixed, but exactly how to quantify this is beyond me. 
Any insights are appreciated. 
Thanks,
Nick 
 A: I don't understand why you think $\int |f|^2$ is related to the number of zero crossing, but perhaps this does not matter. There is no way to bound the curvature from above based on the information given: integrals involving $f$ do not control the derivatives of $f$. To give a lower bound for the curvature, you can argue as follows. Suppose $\kappa\le 1/R$ around $\bar x$. Pick a point $x_0$ near $\bar x$ where $|f|$ is large and attains its maximum (such a point exists because $|f|$ is tightly localized). Let's assume $f(x_0)>0$ to simplify writing. By the curvature assumption, the graph of  $f $ lies above a circle of radius $R$ that is tangent to the graph from below. In a formula, $$f(x)\ge f(x_0)+\sqrt{R^2-(x-x_0)^2}-R,\quad |x-x_0|\le R$$
Actually, the simpler bound $f(x)\ge f(x_0) -R$ should be enough. The conclusion will be that the variance of $f$ is at least of order $R^2$. Hence, the curvature is bounded from below by  $C/\sqrt{V}$.
None of this uses the wavy information about $f$. It's possible that knowing some PDE that $f$ satisfies, one could say much more about the curvature. 
It's also possible that looking at curvature isn't what you want, and $|f''|$ should be substituted for $\kappa$. 
