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Putnam Problem of the Day

http://math.harvard.edu/putnam/ Mon 25 Mar 06:58:41 2019 No.10495275 View Reply Original Report

Quoted By: >>10495943 >>10496078

$ \text{Let }f\text{ be a twice-differentiable real-valued function satisfying} \\ ~\qquad \qquad \qquad \qquad \quad \quad f(x)+f''(x)=-xg(x)f'(x), \\ \text{where }g(x)\geq 0\text{ for all real }x\text{. Prove that }|f(x)|\text{ is bounded.} $

http://math.harvard.edu/putnam/

http://math.harvard.edu/putnam/ Mon 25 Mar 2019 07:01:56 No.10495277 Report

Quoted By:

Previous Thread >>10490281

Anonymous

Anonymous Mon 25 Mar 2019 09:26:17 No.10495445 Report

Quoted By: >>10495459 >>10495466

Multiplying by f', we get:
$f(x)f'(x) + f'(x)f''(x) = -xg(x)f'(x)^2$
By FTA, we get:
$\frac{f(x)^2 + f'(x)^2}{2} = \frac{f(0)^2+f'(0)^2}{2} - \int_0^x tg(t)f'(t)^2 dt$
Now note that, for each x, $\int_0^x tg(t)f'(t)^2 dt \ge 0$ (consider the cases x >= 0 and x < 0 separately).
It follows that
$\frac{f(x)^2}{2} \le \frac{f(x)^2 + f'(x)^2}{2} \le \frac{f(0)^2 + f'(0)^2}{2}$

Anonymous

Anonymous Mon 25 Mar 2019 09:35:54 No.10495459 Report

Quoted By:

>>10495445
Woops, forgot that g is not a priori continuous. Obviously this was too easy

Anonymous

Anonymous Mon 25 Mar 2019 09:42:13 No.10495466 Report

Quoted By:

>>10495445
Woops, forgot that the RHS is not necessarily integrable, but either way you can replace the argument with the following:
$xg(x)f'(x)^2$ has the sign of x, hence $x \mapsto f(x)^2 + f'(x)^2$ is nondecreasing for negative x and nonincreasing for positive x, hence, for each x, $f(x)^2 + f'(x)^2 \le f(0)^2 + f'(0)^2$ and conclude as above.

Anonymous

Anonymous Mon 25 Mar 2019 12:05:30 No.10495644 Report

Quoted By:

bump

Anonymous

Anonymous Mon 25 Mar 2019 15:09:01 No.10495943 Report

Quoted By:

>>10495275
thinly veiled homework thread

Anonymous

Anonymous Mon 25 Mar 2019 16:02:18 No.10496078 Report

Quoted By:

>>10495275
You should be able to put it in normal form and use the Sturm theorems to prove it's bounded

Anonymous

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Anonymous Mon 25 Mar 2019 16:10:52 No.10496103 Report

Quoted By:

I'm too much of a brainlet to figure these out but I like the images in these threads :3

Anonymous

Anonymous Tue 26 Mar 2019 03:39:49 No.10497476 Report

Quoted By: >>10497505

source?

Anonymous

Anonymous Tue 26 Mar 2019 03:59:04 No.10497505 Report

Quoted By:

>>10497476
good will hunting

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