# Show that $f$ is bounded if $f(x)+f''(x)=-xg(x)f'(x)$ where $g\ge0$

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.

• Put $h(x)=f(x)^2+f^{\prime}(x)^2$, and show (compute the derivative) that $h(x)\leq h(0)$ for all $x$. – Kelenner Sep 7 '17 at 16:31
• Is this a Putnam problem? I am sure that I have seen this problem in a math contest. – Batominovski Jul 20 '20 at 20:30

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.

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

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 $$x \in \Bbb R$$.