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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 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 $x$.

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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 $x$.

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    $\begingroup$ This intuition is misleading. The same way of thought would go if the problem was stated as $ | f ' ( x ) | \le 2 \sqrt { | f ( t ) | } $, which is a completely different situation, as there is the nonzero function $ f ( x ) = x ^ 2 $ satisfying it. $\endgroup$ – Mohsen Shahriari Jun 27 at 22:00
  • $\begingroup$ You are right! Thanks for pointing it out, should I delete my answer? $\endgroup$ – Bhaswat Jun 27 at 22:04
  • $\begingroup$ I don't think so. It's good to have such things pointed out. $\endgroup$ – Mohsen Shahriari Jun 27 at 22:05
  • $\begingroup$ I'm thinking when would that kind of thinking work. I'm trying to search the other solutions for insight, and for example find out that if the problem was stated as my above comment, which parts would not work. $\endgroup$ – Mohsen Shahriari Jun 27 at 22:07
  • $\begingroup$ I am also thinking about the same! I think the first problem is that such a $dt$ doesn't exist only in reality. Some positive quantity less than any other positive quantity. So probably need to reformulate it in terms of formal limit. $\endgroup$ – Bhaswat Jun 27 at 22:09

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