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Let $f$ be a twice differentiable real valued function such that $f(x)+f''(x)=-xg(x)f'(x)$ Where $g(x)\geq 0$ for all real $x$ Show that $|f(x) |$ is a bounded function.

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    $\begingroup$ Put $h(x)=f(x)^2+f^{\prime}(x)^2$, and show (compute the derivative) that $h(x)\leq h(0)$ for all $x$. $\endgroup$ – Kelenner Sep 7 '17 at 16:31
  • $\begingroup$ Is this a Putnam problem? I am sure that I have seen this problem in a math contest. $\endgroup$ – Batominovski Jul 20 '20 at 20:30
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Multiplying both sides by $f'(x)$, one has $$ \frac12[(f(x))^2+(f'(x))^2]'=-xg(x)(f'(x))^2\le 0 $$ for $x>0$. Then integrating both sides from $0$ to $x$, one has $$ (f(x))^2+(f'(x))^2\le(f(0))^2+(f'(0))^2$$ from which one obtains $$ (f(x))^2\le (f(x))^2+(f'(x))^2\le (f(0))^2+(f'(0))^2. $$ So $|f(x)|$ is bounded. If $x<0$, one can use the same way to establish.

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Start with the observation that

$\dfrac{d}{dx}((f'(x))^2 + (f(x))^2) = 2f'(x)f''(x) + 2 f(x) f'(x); \tag 1$

since we are given that

$f''(x) + f(x) = -x g(x) f'(x), \tag 2$

we find, upon multiplying (2) through by $f'(x)$, that

$f'(x)f''(x) + f(x)f'(x) = -x g(x) (f'(x))^2; \tag 3$

thus (1) yields

$\dfrac{d}{dx}((f'(x))^2 + (f(x))^2) = -2x g(x) (f'(x))^2. \tag 4$

We may integrate (4) 'twixt $0$ and any $x \ge 0$:

$((f'(x))^2 + (f(x))^2) - ((f'(0))^2 + (f(0))^2)$ $= \displaystyle \int_0^x \dfrac{d}{ds}((f'(s))^2 + (f(s))^2)ds = -2\int_0^x s g(s)(f'(s))^2 ds. \tag 5$

If we set

$M_+(x) = \displaystyle -2\int_0^x s g(s)(f'(s))^2 ds, \tag 6$

we that

$M_+(x) \le 0 \tag 7$

since $s$, $g(s)$, and $(f'(s))^2$ are all non-negative on $[0, x]$. Therefore

$(f'(x))^2 + (f(x))^2 = (f'(0))^2 + (f(0))^2 + M_+(x) \le (f'(0))^2 + (f(0))^2); \tag 8$

also,

$\vert f(x) \vert^2 = (f(x))^2 \le (f'(x))^2 + (f(x))^2 \le (f'(0))^2 + (f(0))^2, \tag 9$

whence

$\vert f(x) \vert \le \sqrt{(f'(0))^2 + (f(0))^2} \tag {10}$

for all $x \ge 0$.

In the event that $x < 0$, noting that (4) binds for all $x \in \Bbb R$, we may write

$((f'(0))^2 + (f(0))^2) - ((f'(x))^2 + (f(x))^2)$ $= \displaystyle \int_x^0 \dfrac{d}{ds}((f'(s))^2 + (f(s))^2)ds = -2\int_x^0 s g(s)(f'(s))^2 ds; \tag {11}$

now setting

$M_-(x) = \displaystyle -2\int_x^0 s g(s)(f'(s))^2 ds, \tag{12}$

we see that, since $xg(x)(f'(x))^2 \le 0$ for $x \le 0$,

$M_-(x) \ge 0, \tag{13}$

and re-arranging (11) we obtain

$(f'(x))^2 + (f(x))^2 = (f'(0))^2 + (f(0))^2 - M_-(x) \le (f'(0))^2 + (f(0))^2); \tag{14}$

which again leads to (9) and hence (10). We see that $\vert f(x) \vert$ is bounded for all $x \in \Bbb R$.

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