Real function such that $|f'(x)|\le|f(x)|$ has fixed sign

Question

Let a differentiable real function $f$ be such that $\left|f'(x)\right|\le|f(x)|$. Prove that $f(x)$ has the same sign for all $x$. (Let 0ドル$ and any real number have the same sign)

1. If $f$ has no zeros then we are done because $f$ is continuous.

2. If $f$ has zeros, WLOG let $f(0)=0$. One way is to consider $F(x)=\mathrm e^{-2x}f^2(x)$, where $$F'(x)=2\mathrm e^{-2x}\left(f(x)f'(x)-f^2(x)\right)\le0$$ due to $\left|f'(x)\right|\le|f(x)|$. The rest is simple.

But I did not think of such $F(x)$. I wanted to know whether my idea as follows can give a complete proof. (I don't want to see answers that continue the approach above!!!)

Let $f$ have zeros. If the zeros are dense, then by continuity $f\equiv0$. Otherwise, I think we can find $f(x_0)=0$ such that both $f$ and $f'$ are positive/negative on $(x_0,x_0+\delta)$ (I did not prove this!). WLOG positive, consider $g(x)=\mathrm e^{-x}f(x)$, then $g'(x)=\mathrm e^{-x}(f'(x)-f(x))\le0$ so $g(x)$ is a decreasing function which gives a contradiction.

• MathIsAlwaysRight: Why does g(x) being decreasing give a contradiction?
• youthdoo: Because $g$ and $f$ have the same sign. @MathIsAlwaysRight
• MathIsAlwaysRight: sorry if I'm misunderstanding something, but what if g just decreases towards some asymptote? It doesn't necessarily have to change its sign, does it?
• youthdoo: @MathIsAlwaysRight But $g(x_0)=0$.
• Christophe Boilley: Why $f'$ would be positive/negative on $(x_0,x_0+\delta)$? It is false in general (and useless for your conclusion).

Answer 1

Let $I=\{x:\ f(x)=0\}$. If $I=\emptyset$, we're done since $f$ is continuous. Because of continuity, $I$ consists of closed intervals, hence $\mathbb R\setminus I$ is open and thus is a (at most countable) collection of open intervals. Let $b$ belong to some such interval $(a,c)$ where $c\in(a,+\infty]$ and assume w.l.o.g. that $f(b)>0$; hence $f(x)>0$ for all $x\in(a,b]$, $f(a)=0$.

Now, $$ \int_a^b \frac{f'(x)}{f(x)}dx=\ln(f(b))-\ln(f(a))=+\infty $$ and at the same time $$ \int_a^b \frac{f'(x)}{f(x)}dx\le \int_a^b \frac{|f'(x)|}{f(x)}dx \le \int_a^b 1 dx=b-a $$ Contradiction.

• youthdoo: Why is the second inequality?
• Gary: There is a $\le$ missing in front of $\int_a^b 1 dx$.
• Gary: Should be $\le$ and not $=,ドル right?
• van der Wolf: Thanks, @Gary !