6

Assume that $f$ is continuous and differentiable. Futher, let $g(r,t)\geq0$ be such that for $t>x$ $$ \lim_{r \to \infty} \frac{g(r,t)}{g(r,x)} = 0. $$

Show that $$ \lim_{r \to \infty} \frac{\int_x^{\infty} (f(x)-f(t)) g(r,t)dt}{\int_x^{\infty} (x-t) g(r,t)dt} = f'(x), $$ where all the functions are also sufficiently integrable for the claim to make sense.

Some notes:

From the assumptions it follows that for $t\geq x$ $$ \lim_{r \to \infty} \frac{g(r,t)}{g(r,x)} = \delta_x, $$ where $\delta_x$ is the Dirac delta. Then we can divide the denominator and the numerator by $g(r,x)$. But I can't immediately see why that would reduce things to be the $f'(x)$ since the integrals are taken separately in the denominator and the numerator. Are some additional assumptions needed?

NPHA
  • 339
  • Additional assumptions are definitely needed, because the integrals may not even converge for any value of $r$. No information is given about the behavior of $g(r,t)$ as $t\to\infty$. Also I suggest the substitution $t=x+h$ and multiplying top and bottom by $-1$ to make it more clear where the derivative is supposed to come from. – SmileyCraft Nov 10 '23 at 12:26
  • @SmileyCraft Well I just forgot the state that all the functions in the expressions are sufficiently integrable. Otherwise the whole problem makes no sense at all. But I don’t think that gives any hints how to approach the problem. – NPHA Nov 10 '23 at 22:24
  • For anyone looking for an example of this: Take $g(r,t)=e^{-rt}$, $r,t>0$. In this case the above fraction reduces to $r^2\int_x^\infty(f(t)-f(x))e^{r(x-t)}\mathrm{d}t=r\int_x^\infty f'(t)e^{r(x-t)}\mathrm{d}t=\int_0^\infty f'(x+t)re^{-rt}\mathrm{d}t$ which converges to $f'(x)$ as $r\to\infty$ for well-behaved $f$. – junjios Nov 11 '23 at 00:20
  • 1
    We have $$\frac{\int_{x}^{\infty}(f(x)-f(t))g(r,t),\mathrm{d}t}{\int_{x}^{\infty}(x-t)g(r,t),\mathrm{d}t}=\int_{x}^{\infty}f'(s)k(r,s),\mathrm{d}s $$ where $$k(r,s) = \frac{\int_{s}^{\infty}g(r,t),\mathrm{d}t}{\int_{x}^{\infty}\left(\int_{s'}^{\infty}g(r,t),\mathrm{d}t\right),\mathrm{d}s'}.$$ Hence, the problem boils down to finding the condition on $g$ so that $k(r,s)$ tends to $\delta_x(s)$ in weak sense as $r\to\infty$. – Sangchul Lee Nov 13 '23 at 01:14
  • Do you assume that $g$ is non-negative? If yes, it should be specified. – Christophe Leuridan Nov 13 '23 at 21:20
  • @ChristopheLeuridan I don't immediately see why this is important, but it can be assumed to be non-negative. I added it to the question. – NPHA Nov 16 '23 at 13:05
  • @SangchulLee Thank you. I this shows what is essential in the problem. However, it would be interesting to get some conditions for $g$ such that the limit holds. – NPHA Nov 16 '23 at 13:06
  • @NPHA If $g$ takes positive and negative values, the denominator may vanish, or may be very small, providing big variations of the ratio. – Christophe Leuridan Nov 16 '23 at 19:35
  • I recommend setting $x=0$. It then follows from hopital rule that the clame is correct for any infinitesimal positive $t$. I notice division by an integral , so maybe this helps too : https://math.stackexchange.com/questions/2991661/understanding-when-the-multiplication-of-two-definite-integrals-gives-unity

    Just my first ideas

     – mick Nov 17 '23 at 00:19

0 Answers0