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This question already has an answer here:

Let $f:\mathbb{R}\rightarrow\mathbb{R}$ be a continuous and differentiable function in all $\mathbb{R}$. If $f(0)=0$ and $|f'(x)|\leq |f(x)|$ for all $x\in\mathbb{R}$, then $f\equiv 0$.

I've been trying to prove this using the Mean Value Theorem, but I can't get to the result. Can someone help?

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marked as duplicate by user299912, Arnaldo, Davide Giraudo real-analysis Jul 10 '17 at 20:36

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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Consider $S := \{ x \in \mathbb{R} : f(x) = 0 \}$. This set is closed by the continuity of $f$. We now claim $S$ is also open. To show this, suppose $x_0 \in S$, and let $A := \sup \{ |f(x)| : x \in (x_0 - \frac{1}{2}, x_0 + \frac{1}{2}) \}$. Then, for $x \in (x_0 - \frac{1}{2}, x_0 + \frac{1}{2})$, we have $f(x) = \int_{x_0}^x f'(t)\,dt$, and $|f'(t)| \le |f(t)| \le A$ for $t$ between $x$ and $x_0$, so $|f(x)| \le |x - x_0| A \le \frac{1}{2} A$. Thus, by the definition of $A$ as a supremum, $0 \le A \le \frac{1}{2} A$, which implies $A = 0$. This implies that $f(x) \equiv 0$ for $x \in (x_0 - \frac{1}{2}, x_0 + \frac{1}{2})$, establishing that $S$ is a neighborhood of $x_0$.

Now, we have shown that $S$ is a clopen subset of $\mathbb{R}$, and we are given that $0 \in S$ so in particular, $S$ is nonempty. By the connectedness of $\mathbb{R}$, this implies $S$ is all of $\mathbb{R}$, which is equivalent to the desired conclusion.

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  • $\begingroup$ This proof was inspired by the proof of Picard's theorem on existence and particularly uniqueness of solutions to a first-order differential equation. $\endgroup$ – Daniel Schepler Jul 7 '17 at 18:57
  • $\begingroup$ This solution should also be fairly straightforward to adapt to the similar question if you replace $|f'(x)| \le |f(x)|$ with $|f'(x)| \le |e^x f(x)|$, for example - whereas I think Severin Schraven's answer would require more work to adapt to such a problem. $\endgroup$ – Daniel Schepler Jul 7 '17 at 18:59
  • $\begingroup$ You are absolutely right. I decided not to use the connectedness, as in this case one can get away with an easy induction argument and I wasn't sure whether the OP was familiar with this topological fact. $\endgroup$ – Severin Schraven Jul 7 '17 at 19:10
  • $\begingroup$ One has to be careful with the definition of integration here. Differentiable with a bounded derivative does not imply that the derivative is Riemann integrable for example. $\endgroup$ – WimC Jul 7 '17 at 20:18
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    $\begingroup$ @WimC Oh, right... If you don't want to use Lebesgue integration, then the Mean Value Theorem should do just as well to show $|f(x)| \le |x - x_0| A$. $\endgroup$ – Daniel Schepler Jul 7 '17 at 20:40
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Let $\vert x \vert \leq \frac{1}{2}$ and let $x_0\in \mathbb{R}$ such that $f(x_0)=0$.

$$ \vert f(x_0+x) \vert = \left\vert f(x_0) + \int_0^x f'(x_0+t) dt \right\vert = \left\vert \int_0^x f'(x_0+t) dt \right\vert $$

using the triangular inequality for integrals yields

$$ \leq \left\vert \int_0^x \vert f'(x_0+t) \vert dt \right\vert \leq \left\vert \int_0^x \vert f(x_0+t) \vert dt \right\vert \leq \left\vert \int_0^x \max_{\vert s\vert \leq \frac{1}{2}}\vert f(x_0+s) \vert dt \right\vert \leq \vert x \vert \cdot \max_{\vert s\vert \leq \frac{1}{2}}\vert f(x_0+s) \vert.$$

Now we take the maximum over all $\vert x \vert \leq \frac{1}{2}$ and obtain

$$ \max_{\vert s\vert \leq \frac{1}{2}}\vert f(x_0+ s) \vert \leq \frac{1}{2} \max_{\vert s\vert \leq \frac{1}{2}}\vert f(x_0+s) \vert. $$

Thus, $\max_{\vert s\vert \leq \frac{1}{2}}\vert f(x_0+ s) \vert=0$ and hence

$$ \forall \vert t \vert \leq \frac{1}{2}: \ f(x_0+t)= 0.$$

By induction we prove that $f(t)=0$ for all $t\in \left[ -\frac{n}{2}, \frac{n}{2} \right]$ for all $n\in \mathbb{N}$ and therefore $f$ is identicially zero.

Added: If one does not want to use integrals, one can use the mean value theorem in the following way: For $\vert x \vert \leq \frac{1}{2}$ there exists $\vert \xi(x) \vert \leq \vert x \vert$ such that

$$ \vert f(x) \vert = \vert f(x) - f(0) \vert = \vert f'(\xi(x)) \vert \cdot \vert x \vert \leq \vert f(\xi(x)) \vert \cdot \vert x \vert \leq \max_{\vert s \vert \leq \frac{1}{2}} \vert f(s) \vert \cdot \vert x\vert .$$

Taking again the maximum over all $\vert x \vert \leq \frac{1}{2}$ one arrives at $\max_{\vert s \vert \leq \frac{1}{2}} \vert f(s) \vert \leq \frac{1}{2} \max_{\vert s \vert \leq \frac{1}{2}} \vert f(s) \vert$ and proceeds as above.

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  • $\begingroup$ While I think the intended solution was to just use derivatives, using the integral works just as well. $\endgroup$ – Sean Roberson Jul 7 '17 at 18:51
  • $\begingroup$ @SeanRoberson Did you have something like the added argument in mind? Or what did you mean? $\endgroup$ – Severin Schraven Jul 7 '17 at 19:12
  • $\begingroup$ Almost - I didn't quite piece it together but I had an idea of using a supremum in some fashion. $\endgroup$ – Sean Roberson Jul 7 '17 at 19:13
  • $\begingroup$ @SeanRoberson If you have a proof that neither relies on connectedness nor doing everything on different intervalls, then I'd really like to see this proof :) that would be really cool $\endgroup$ – Severin Schraven Jul 7 '17 at 19:18
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Intuitively, the solution to $|f'| \leq |f|$ with $f(0) = c$ can not grow out of the region bounded by the solutions to $f' = +f$ and $f' = -f$ with the same initial condition $f(0) = c$. The boundary solutions are $f_\pm(x) = c e^{\pm x} = 0$ with $c = 0$ we have $f_\pm(x) \equiv 0$.

It's so obvious intuitively, I think, but was not at all as easy to prove as I thought, but finally I came up with a proof. But first a lemma:

Lemma
Let $h : \mathbb [0, \infty) \to \mathbb R$ be differentiable and satisfy

  1. $h(x_0) > 0$,
  2. $h'(x) > 0$ when $x > x_0$ and $h(x) > 0$.

Then $h(x) > 0$ for all $x \geq x_0$.

Proof of lemma
Assume that $h(a) \leq 0$ for some $a > x_0$. Since $h$ is continuous, by the intermediate value theorem, $h$ takes the value $0$ in at least one point between $x_0$ and $a$. Let $x_1 = \inf \{ t \in (x_0, a) \mid h(t) = 0 \}$. Since $h$ is continuous and $h(x_0) > 0$ we have $x_1 > x_0$ and $h(x) > 0$ when $x_0 < x < x_1$. Then $h(x_1) - h(x_0) < 0$ and by the mean value theorem there exists some $\xi \in (x_0, x_1)$ such that $h'(\xi) = (h(x_1) - h(x_0))/(x_1 - x_0) < 0$. But this contradicts that $h'(x) > 0$ when $x > x_0$ and $h(x) > 0$. Thus $h(x) > 0$ for $x \geq x_0$.

Proof of statement in question
Take $\lambda>0$.

Let $g(x) = \lambda e^x$. Then $g-f$ satisfies the conditions of the lemma with $x_0=0$. Thus, for all $x > 0$ we have $(g-f)(x) > 0$, i.e. $f(x) < \lambda e^ x$.

In the same way, taking $g(x) = -\lambda e^x$ we have $f-g$ satisfying the conditions of the lemma so $f(x) > -\lambda e^x$ for $x > 0$.

Thus, for $x > 0$ we have $-\lambda e^x < f(x) < \lambda e^x$. Since $\lambda>0$ was arbitrary we must have $f(x) \equiv 0$ for $x > 0$.

Reversing the function, i.e. letting $f(x) \to f(-x)$ in the above, we also get that $f(x) \equiv 0$ for $x < 0$.

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