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If $f: \mathbb{R} \rightarrow \mathbb{R}$ is a differentiable function, $f(0)=0$ and $f' = f^2$, then $f = 0$. Any help?

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    $\begingroup$ Don't worry, if this is your first time going through an analysis sequence it will be rough absorbing all the theorems and methods of solving problems. Keep going! $\endgroup$
    – abnry
    Feb 17, 2015 at 1:59
  • $\begingroup$ Thanks! Yeah, it's my first time. I can understand the book, but coming up with my own proofs for the exercises is really challenging and it's screwing with my self esteem! $\endgroup$
    – violeta
    Feb 17, 2015 at 3:13

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Note that your function is increasing due to $f' = f^2 \geq 0$. So if you have such a function which is not identically zero, the set of $x$ for which $f(x) = 0$ is some closed interval $I$ (possibly unbounded) which is not all of ${\mathbb R}$. Let $x_0$ be an endpoint of $I$. Replacing $f(x)$ by $-f(-x)$ if necessary, we can assume $x_0$ is the right endpoint of $I$.

On $(x_0,\infty)$ the equation $f' = f^2$ is separable, and standard calculus techniques give that $f(x) = -{1 \over x + c}$ for $x > x_0$ for some constant $c$. But then taking limits as $x \rightarrow x_0$ gives $f(x) = -{1 \over x_0 + c}$, contradicting that $f(x_0) = 0$ since $x_0 \in I$.

This contradiction shows that $f = 0$ is the only solution.

By the way, solving $f' = f^2$ here only uses basic real analysis techniques... it's the same as ${d \over dx} ({1 \over f(x)} + x) = 0$, which by the mean value theorem for example says that ${1 \over f(x)} + x = c$ for some constant $c$. This implies that $f(x) = -{1 \over x + c}$.

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  • $\begingroup$ Nice! I really like this one, it's very pretty. Thanks everyone else for the answers anyways! $\endgroup$
    – violeta
    Feb 17, 2015 at 14:28
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Here's a proof just from basic theorems of real analysis and a little topology.

Since $f'=f^2\ge0$, $f$ is nonnegative for $x\ge0$. We have $f(x)=0$ for all $x\in[0,x^*]$; by shifting the function $x\mapsto x-x^*$ we can assume $x^*=0$, and then if $f\ne0$ we have arbitrarily small $x>0$ with $f(x)>0$. Now assume for a contradiction that $f(x)>0$ for some $x\in(0,1)$, and let $M=\min(1,f(x)/x)$. Since the set $[M,\infty)$ is closed and $f(x)/x$ is continuous with limit $0$ at $0$ (since $f'(0)=0$), there is a least $y\in(0,1)$ such that $f(y)/y\ge M$.

By the Mean Value Theorem, $f^2(z)=f'(z)=f(y)/y$ for some $0<z<y$. But if $f(y)/y\ge 1$ then $f(z)=\sqrt{f(y)/y}\ge1$, and if $1\ge f(y)/y\ge f(x)/x$ then $f(z)\ge f(y)/y$. In either case $z<y$ also has $f(z)/z\ge f(z)\ge M$, a contradiction. Thus $f=0$ on $[0,\infty)$, and by applying this to $g(x)=-f(-x)$, we have $f=0$ on $\Bbb R$.

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  • $\begingroup$ I think you left out that $y$ is to be the infimum of the set $f^{-1}([M, \infty)) \cup [0, \infty)$. $\endgroup$
    – abnry
    Feb 17, 2015 at 2:44
  • $\begingroup$ @nayrb (It's possible that your comment is from an old version since I made many edits, but) I wrote that "there is a least $y>0$ such that $f(y)\ge M$", which is intended to be interpreted as "Let $y$ be the infimum of $f^{-1}([M,\infty))\cap(0,\infty)$, which is in the set since it is closed". $\endgroup$ Feb 17, 2015 at 2:55
  • $\begingroup$ Oh! I read it quickly as "there is at least $y > 0$... Got it! $\endgroup$
    – abnry
    Feb 17, 2015 at 3:04
  • $\begingroup$ By the Mean Value Theorem, shouldn't it be $(y - 0)f'(z) = f(y)$? $\endgroup$
    – violeta
    Feb 17, 2015 at 3:09
  • $\begingroup$ @raffa Wow, don't know how I messed that up. Well the argument still works with a little modification. $\endgroup$ Feb 17, 2015 at 3:41
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$f' = f^2$ is a differential equation. What do you know about differential equations? If you know the Existence and Uniqueness Theorem, this question is easy.

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  • $\begingroup$ Nothing! The question is from an introductory real analysis book. It's actually on the section about Riemann integrals. $\endgroup$
    – violeta
    Feb 17, 2015 at 1:32
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Here's an idea that doesn't "use" differential equations:

Assume $f \geq 0$ for $x \geq 0$ and consider the interval $(0, \epsilon)$, $\epsilon<1$ where $0 \leq f < 1/2$, which exists by continuity of $f$.

First, pick an $x_0 \in (0, \epsilon)$. By the mean value theorem, $$f(x_0) = f(x_0)-f(0) = x_0 f'(x_1) \leq f(x_1)^2$$ for some $0 < x_1 < x_0$. Define $x_2$ and in general $x_n$ in a similar manner so that $f(x_n) \leq f(x_{n+1})^2$. Then $$f(x_0) \leq f(x_n)^{2^n} \leq (1/2)^{2^n}.$$ Taking the limit we get $f(x_0) = 0$.

Now fill in the gaps and deal with the other cases (or combine them) and figure out a way to show $f = 0$ beyond $(0, \epsilon)$. If you use this idea in your homework just write down you got the idea here, no biggie.

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    $\begingroup$ I find it interesting how our answers both use essentially the same method but in different ways. $\endgroup$ Feb 17, 2015 at 2:23
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    $\begingroup$ I think raffa's point that MVT gives $f(x_0)=x_0f'(x_1)$, not $f(x_0)=f'(x_1)$, also applies to this answer. $\endgroup$ Feb 17, 2015 at 3:45
  • $\begingroup$ @MarioCarneiro Good point, what a silly error! It's corrected now. $\endgroup$
    – abnry
    Feb 17, 2015 at 14:00
  • $\begingroup$ Probably should put in the assumption that $\epsilon<1$, or else your inequality doesn't hold. $\endgroup$ Feb 17, 2015 at 14:07
  • $\begingroup$ Fixed it, thanks! $\endgroup$
    – abnry
    Feb 17, 2015 at 14:15
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If $f$ is differentiable it is continuous.

In particular, $Z=f^{-1}(\{0\})$ is closed and $0 \in Z$.

Choose some $t_0 \in Z$. Define $M(\delta)= \sup_{x \in [t_0-\delta,t_0+\delta]} |f(x)|$. It is straightforward to check that $M$ is continuous and non decreasing.

Then the function $\delta \mapsto |\delta| M(\delta)$ is continuous and zero at zero.

Now choose some $\delta>0$ such that $\delta M(\delta) <1$.

Then for $t \in [t_0-\delta,t_0+\delta]$, $|f(t)| \le | \int_{t_0}^t f^2(s) ds | \le |\delta| M(\delta)^2 $. Taking the $\sup$ of the left hand side over $[0, \delta]$ gives $M(\delta) \le \delta M(\delta)^2 $, which implies $M(\delta) = 0$, and in particular, $f(t) = 0 $ for all $t \in [t_0-\delta,t_0+\delta]$.

This shows that $Z$ is open.

Since $Z$ is a non-empty open and closed subset of $\mathbb{R}$, we must have $Z= \mathbb{R}$.

Alternative approach: The key here is that $f$ is defined for all of $\mathbb{R}$.

Suppose $f(t_0) >0$ for some $t_0$. A quick computation shows that if $\phi(t) = {1 \over f(t)}$, then $\phi'(t) = -1$ and so $\phi(t) = \phi(t_0) + t_0 - t$. However, this would give that $\lim_{t \uparrow (t_0+\phi(t_0) )} f(t) = \infty$, which contradicts the fact that $f$ is defined for all of $\mathbb{R}$.

A similar approach works for $f(t_0) <0$. Hence $f$ is identically zero.

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