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$f:[0,1] \to \mathbf R$ is continuous. If $$\int_0^{x/3} f(t)dt =\int_0^xf(t)dt$$ for all $x$ in $[0,1]$, prove that $f$ is identically $0$.

My thought is to prove that the maximum and minimum of $f$ are equal then $f$ is constant and this constant can only be zero. But I can't think of a way to do that. Can somebody help and give me some hints. Thanks.

Caleb Stanford
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user236626
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    My thought is to look at $F(x) = \int_0^x f(t) \mathrm{dt}$ as a continuous and differentiable function, satisfying $F(x) = F(x/3)$. – hardmath Aug 26 '15 at 03:33
  • Are you sure $x\in[0,1]$ ? I think it will be $x>0$ – Empty Aug 26 '15 at 03:34
  • @S.Panja-1729: I believe in this case $f(t)$ need only be defined on $[0,1]$ and thus only on that interval do we need to argue it is identically zero. – hardmath Aug 26 '15 at 03:35
  • It must be $x\in (0,1]$..Otherwise the statement is FALSE – Empty Aug 26 '15 at 03:37
  • @S.Panja-1729: Why would the statement be false? Do you have a counterexample where it is true for all $x\in (0,1)$ but false if we require the condition as well for endpoints $x=0,1$? – hardmath Aug 26 '15 at 03:40
  • @S.Panja-1729 No. It also holds for $f(0)=0$. – Vim Aug 26 '15 at 03:40
  • If we take $x=0$ and $f(x)=5$ then the relation holds but $f\not \equiv 0$ – Empty Aug 26 '15 at 03:41
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    @S.Panja-1729: be careful that the supposition is on all $x \in [0,1]$, not just a particular point. Your $f$ certainly doesn't satisfy that. –  Aug 26 '15 at 03:47
  • @6005: Often the extensive use of $\LaTeX$ in titles will cause formatting problems, esp. when a post gets on the Hot Questions list and is displayed on SE sites which do not support MathJax. – hardmath Aug 26 '15 at 03:50
  • @hardmath The title should be descriptive of the actual problem to be solved. Otherwise, you have to click on it to find out. Which is a big pain in the behind in pretty much any context--looking for a problem to answer, finding duplicates, locating a question that you previously answered or referenced, etc. I know about LaTeX having some formatting problems in titles, but I think having a descriptive title is a bigger concern. Otherwise we have ten questions all named "prove $f$ is identically zero" and they're all different. (In fact, I see four or five such questions in the side bar.) – Caleb Stanford Aug 26 '15 at 03:57
  • I also don't much care for the hot networks question list, I think it sucks, and I hope we get less clicks from it. Anyway it is not our problem if people don't display our titles correctly. – Caleb Stanford Aug 26 '15 at 03:57
  • It's easy to show F is constant, see my answer –  Aug 26 '15 at 04:34

4 Answers4

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Hint:

FTC $\implies \frac13 f(x/3) = f(x)$ for all $x \in [0,1]$

Added:

Which in turn gives:

$$\frac1{3^n} f\left(\frac1{3^n}x\right) = f(x), \ \forall \ x \in [0,1], \ n \in \mathbb N$$

Cameron Williams
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Let $F(x) = \int_0^x f(t) \mathrm{dt}, x \in [0,1]$ be a primitive of $f$.

Taking $x \in [0, 1]$, from $F(x)=F(x/3)$ we get (by recursively changing $x \rightarrow x/3$) $F(x)=F(x/3^{n}), \forall n \in \mathbf N$. Because F continous and $(x/3^n) \rightarrow 0$ when $n \rightarrow \infty$ we have $F(x)=F(0)$. Therefore F is constant and $f = F'= 0$

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Expand @BolzWeir's answer.

First I propose to show that $f(0)=0$:

Due to the equality we have $$\int_{x/3}^x f =0$$ for all $x\in[0,1]$. Note that $f$ is uniformly continuous on $[0,1]$, which is to say, for an $\epsilon>0$ we can find a uniform $\delta>0$ that suits all $x\in [0,1]$. Suppose $f(0)=a> 0$, then let $\epsilon=a/2$, and take an arbitrary $x\in(0,\delta)$, we would have $$\int_{x/3}^x f >a/2\cdot 2x/3>0$$ a contradiction. The case where $a<0$ is handled similarly.

By continuity, it means $f(x)\to 0$ as $x\to 0^+$. And for all $x\in[0,1]$, based on @BolzWeir's answer $$3^nf(x)=f(\frac x{3^n})\to 0\quad \text{as}\, n\to\infty$$ which indicates $f(x)=0$.

Vim
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Here is another way:

It is straightforward to see from the hypotheses that $\int_0^x f = \int_0^{x \over 3^n} f$ for any $n$ and, taking limits, that $\int_0^x f = 0$ for any $x$. In particular, it follows that $\int_a^b f = \int_0^b f -\int_0^a f = 0$ and hence (reasoning by contradiction on a suitably small interval) that $f = 0$.

If $f$ is just integrable, the above reasoning can be extended to show that $f(x) = 0$ for ae. $x$.

copper.hat
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