Does $f(0)=0$ and $\left|f^\prime(x)\right|\leq\left|f(x)\right|$ imply $f(x)=0$?

Question

Let $f:\mathbb{R}\to\mathbb{R}$ be a function such that $f(0)=0$ for all real numbers $x$, $\left|f^\prime(x)\right|\leq\left|f(x)\right|$. Can $f$ be a function other than the constant zero function?

I coudn't find any other function satisfying the property. The bound on $f^\prime(x)$ may mean that $f(x)$ may not change too much but does it mean that $f$ is constant?

I thought for a while and found that $f^\prime(0)=0$ and by using mean value theorem, if $x\neq0$ then there's a real number $y$ between $0$ and $x$ such that $\left|f(x)\right|=\left|xf^\prime(y)\right|\leq\left|xf(y)\right|$. Anything further?

Answer 1

Suppose $x$ is some real number with $f(x)=0$, and let $y\in [x-1/2,x+1/2]$ be such that $|f(y)|$ is maximized. Then the mean value theorem implies there is some $z$ between $x$ and $y$ such that $$ 2\cdot |f(y)|\leq \frac{|f(y)|}{|y-x|}=\left|\frac{f(y)-f(x)}{y-x}\right|=|f'(z)|\leq |f(z)|\leq |f(y)|, $$ which implies $f(y)=0$. We have shown that $f(x)$ implies $f$ is $0$ on $[x-1/2,x+1/2]$, so $f$ must be identically $0$.

Answer 2

Let $F(x)=|f(x)|$. Then if $x>0$ $$ F(x)=\Bigl|\int_0^xf'(t)\,dt\Bigr|\le\int_0^x|f'(t)|\,dt\le\int_0^x|f(t)|\,dt=\int_0^xF(t)\,dt. $$ From here we get $$ \Bigl(e^{-x}\int_0^xF(t)\,dt\Bigr)'=e^{-x}\Bigl(F(x)-\int_0^xF(t)\,dt\Bigr)\le0. $$ Thus, $e^{-x}\int_0^xF(t)\,dt$ is decreasing. Since its value at $x=0$ is $0$ and $e^{-x}>0$, we see that $\int_0^xF(t)\,dt\le0$ for all $x>0$. Since $F\ge0$, this implies that $F(x)=0$ for all $x>0$.

A similar reasoning applies if $x<0$.

Answer 3

Continuing the idea that I mentioned after proposing the question: Let's define $y_1:=y$. By the mean value theorem there's a real number $y_2$ between $0$ and $y_1$ such that $\left|f(y_1)\right|\leq\left|y_1f^\prime(y_2)\right|\leq\left|y_1f(y_2)\right|$ so $\left|f(x)\right|\leq\left|xy_1f(y_2)\right|\leq\left|x^2f(y_2)\right|$. Continuing this way, we inductively conclude that for every positive integer $n$, there's a real number $y_n$ between $0$ and $x$ such that $\left|f(x)\right|\leq\left|x^nf(y_n)\right|$. So if $0<x<1$, using the fact that $f$ is bounded on a bounded interval, we take the limit of the right-hand side of the last equation as $n$ tends to infinity and conclude that $f(x)=0$. Since $f$ is continuous, we have $f(x)=0$ for $0\leq x\leq1$. Now, if $m$ is a positive integer, the function $g(x)=f(x+m)$ has the property $\left|g^\prime(x)\right|\leq\left|g(x)\right|$. This lets us to prove that $f(x)=0$ for $m\leq x\leq m+1$ inductively. The function $h(x)=f(-m+1-x)$ can be treated in the same manner and that allows us to prove $f(x)=0$ for every real number $x$.

Answer 4

I am just outlining an alternate approach to visualize what is actually going on. I am aware that it is not completely rigorous as it depends on a infinitesimal quantity. $\space $Suppose $x$ is time and $f(x)$ is the position of a body on the real line. It is given that the magnitude of velocity i.e. speed is less than equal to its distance from origin in magnitude. At time $0$, the speed is $0$ as the particle is at $0$ distance from the origin, so in next $\mathrm d t$ times, where $\mathrm d t \to 0$, it will move $0 \cdot \mathrm d t=0$ distance. In next step of length $\mathrm d t$ also, it will not move because it still has zero speed because it is at zero distance from the origin. So the body is stuck at the origin. So for all times, the position is zero. So, $f(x) = 0$ for all $x$.