Problem of the Week # 233 - Sep 15, 2016

[Ackbach]

Here is this week's POTW:

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

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[Ackbach]

Re: Problem Of The Week # 233 - Sep 15, 2016

This was Problem B-2 in the 1997 William Lowell Putnam Mathematical Competition.

Congrats to Opalg for his correct solution, which follows:

Spoiler

Multiply both sides by $f'(x)$, getting $f(x)f'(x) + f'(x)f''(x) = -xg(x)(f'(x))^2$. Now change the variable from $x$ to $t$, and integrate from $0$ to $x$: $$\int_0^x\frac d{dt}\bigl((f(t))^2 + (f'(t))^2\bigr)\,dt = -\int_0^x tg(t)(f(t))^2dt.$$
If $x>0$ then $tg(t)(f(t))^2 \geqslant 0$ for $0\leqslant t\leqslant x$. So the right side of that equation is negative.

If $x<0$, write the right side of the equation as \(\displaystyle \int_x^0 tg(t)(f(t))^2dt.\) This time, the integrand is negative for $x\leqslant t\leqslant 0$. So again the right side of the equation is negative.

Therefore, for all values of $x$, $$\tfrac12(f(x))^2 + \tfrac12(f'(x))^2 - \tfrac12(f(0))^2 - \tfrac12(f'(0))^2 = \int _0 ^x \frac d{dt}\bigl((f(t))^2 + (f'(t))^2\bigr)\,dt \leqslant 0.$$ It follows that $(f(x))^2 + (f'(x))^2 \leqslant (f(0))^2 + (f'(0))^2$, and hence $|f(x)| \leqslant \sqrt{(f(0))^2 + (f'(0))^2}.$ Thus $|f(x)|$ is bounded.