Let $M$ be the set of maxima of the function $k:\mathbb{R}\to \mathbb{R}$. We define the function $$L(t) = \sum_{y\in M} k(y)\delta(t-y)$$ and there by the step function $$\Gamma(t) = \int_0^t L(\tau)d\tau$$.

I'd like to know, if there is any single equation based (kind of closed form in terms of $k,\delta,k',k''$ and possibly signum function) expression without explicitly using the set $M$?

I mean is there any distributional way of writing this function?

PS : Assume $k$ is smooth!

  • $\begingroup$ I suspect that there is no nice formula. You can always change the function $k$ near a maximum to multiply it (add small wiggles). Then essentially the function $L$ gets multiplied at that point. This leads to a drastic change (much larger jump) for the function $\Gamma$ despite the change of $k$ being small. This is not a proof; this is just why a nice equation seems unlikely to me. $\endgroup$ – Joonas Ilmavirta Dec 14 '15 at 21:04

While I still agree with my comment that there might not be a nice formula, there is something. I just don't find composing a delta function with a function very nice.

For a function $f$ with non-degenerate zeros we can naturally interpret $\delta(f(t))$ as $\sum_{a\in f^{-1}(0)}|f'(a)|^{-1}\delta(t-a)$. If $\eta$ is a smooth function, we can multiply by it to obtain $$ \eta(t)\delta(f(t)) = \sum_{a\in f^{-1}(0)}\eta(a)|f'(a)|^{-1}\delta(t-a). \tag{1} $$ In fact, it suffices that $\eta$ is smooth near zeros of $f$ (elsewhere the function vanishes). We can use this idea to write a sum over zeros without explicitly using the zero set.

If we assume that $k''\neq0$ whenever $k'=0$, we can write $$ L(t) = \sum_{y\in M}k(y)\delta(t-y) = \sum_{y,k'(y)=0}H(-k''(y))k(y)\delta(t-y) = -H(-k''(t))k(t)k''(t)\delta(k'(t)), $$ where $H$ is the Heaviside step function. I used $f=k'$ and $\eta(t)=-H(-k''(t))k(t)k''(t)$ in formula (1). I used $H$ to pick only those critical points that are local maxima, and I assumed that every critical point is a local minimum or a local maximum. Notice that the final expression contains no explicit sum.

To get $\Gamma$, you can formally integrate this expression for $L$: $$ \Gamma(t) = \int_0^tL(s)ds = -\int_0^tH(-k''(s))k(s)k''(s)\delta(k'(s))ds. $$ There might be a clever way to integrate by parts (formally) to get a nicer expression. I don't know if these formulas are of any use, but I hope they are in the desired spirit.

| cite | improve this answer | |
  • $\begingroup$ Thanks very much, +1; Couldn't ask for a better expression. It seems very right to me, I never got the idea of composition of functions earlier. $\endgroup$ – Rajesh Dachiraju Dec 15 '15 at 4:16
  • $\begingroup$ But I still need to understand from the middle expression, how you got the right most expression where summation removed. It seems to me that you used formula of $\delta(f(t))$, which introduced multiplication by $k''(t)$, but the already existing factor $H(-k''(y))k(y)$, wouldn't that hinder? Request you to exapnd a bit for my understanding. Also it would be nice, if you could comment on expresion for $\Gamma(t)$. $\endgroup$ – Rajesh Dachiraju Dec 15 '15 at 4:20
  • $\begingroup$ @RajeshDachiraju, I updated my answer a bit. $\endgroup$ – Joonas Ilmavirta Dec 15 '15 at 8:12
  • $\begingroup$ Thank you very much @Joonas. That was very helpful. $\endgroup$ – Rajesh Dachiraju Dec 15 '15 at 8:26

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.