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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?

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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 $0$.

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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 $x<0$.

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  • $\begingroup$ Impressive! But it's not clear to me why $f^\prime$ is integrible. $\endgroup$ – Mohsen Shahriari Oct 26 '15 at 18:59
  • $\begingroup$ @MohsenShahriari You are right. I assumed that $f$ is the integral of $f'$. But since $f$ is differentiable everywhere, $f'$ is Lebesgue integrable, and the argument works (if I remember well my measure theory). $\endgroup$ – Julián Aguirre Oct 26 '15 at 19:33
  • $\begingroup$ It isn't true that the derivative of an everywhere-differentiable function is Lebesgue integrable. A counterexample is $f(x)=x^2 \sin(x^{-2})$, extended by continuity at $x=0$. $\endgroup$ – Julian Rosen Oct 26 '15 at 19:47
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    $\begingroup$ I forgot to add the condition $f'$ is (locally) bounded. $\endgroup$ – Julián Aguirre Oct 26 '15 at 20:42
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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 righthand 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 $x$.

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