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$f:[0,1] \to \mathbb{R}$ is continuous.

$\int_{0}^{x} f(t) dt = \int_{x}^{1} f(t) dt \implies f(x) = 0 \forall x \in [0,1]$

I don't understand why I can say $f(1)=f(0)=0$. My attempt is using FTC:

$\int_{0}^{x} f(t) dt = \int_{x}^{1} f(t) dt \implies \int_{0}^{x} f(t) dt= \int_{1}^{x} -f(t) dt \implies f(x) = -f(x) \forall x \in (0,1) \implies f(x) = 0 \forall x \in (0,1)$. However, the statement is about $[0,1]$. How can I conclude for $f(0)$ and $f(1)$?

Thanks

  • you didn't conclude that $f(x)=-f(x)$. You only know that $\int_0^x f(t)dt = \int_0^x -f(t)dt$ – Yanko Sep 26 '18 at 21:04
  • @Yanko now differentiate. – Umberto P. Sep 26 '18 at 21:04
  • @UmbertoP. Yes but this in fact solves the problem, because you get that $\int_0^x f(t) dt = 0$ hence differentiation leads to $f(x)=0$. However I don't know if we can differentiate because the OP didn't specify whether $f$ is continuous or not. – Yanko Sep 26 '18 at 21:07
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    Surely you need more from $f$? Say, continuity? – Theoretical Economist Sep 26 '18 at 21:08
  • I think the statment is false, because $f(x) =0$ for $(0,1), f(1)=1,f(0)=1$ satisfy the hypothesis but is not the null function on $[0,1]$... –  Sep 26 '18 at 21:09
  • @TheoreticalEconomist, yes, sorry, I forgot. $f$ is continuous –  Sep 26 '18 at 21:10
  • @dude3221 Your example works but it is not continuous. The claim is true once you assume the continuity of $f$. Did you learn about the fundamental Theorem of calculus? – Yanko Sep 26 '18 at 21:11
  • ohh, of course! my example was not continuous indeed, I must be tired. I know about TFC. –  Sep 26 '18 at 21:15
  • Use the continuity definition involving sequences. For instance, can you show $\lim_{n\rightarrow \infty }f\left ( \frac{1}{n} \right ) \neq f\left ( \lim_{n\rightarrow \infty }\left ( \frac{1}{n} \right ) \right )$ if $f(0)\neq0$? – LPenguin Sep 26 '18 at 22:54
  • @dude3221 Or, use that $,\int_{0}^{1} f(t) dt = \int_{1}^{1} f(t) dt = 0,$ and $,\int_{0}^{x} f(t) dt + \int_{x}^{1} f(t) dt = \int_{0}^{1} f(t) dt = 0,$, then see If $\int_0^x f \ dm$ is zero everywhere then $f$ is zero almost everywhere. – dxiv Sep 26 '18 at 23:33

2 Answers2

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Assume $f$ is continuous on $[0,1]$. If $\int_{0}^{x}f(t)dt=\int_{x}^{1}f(t)dt$ for all $0 \le x \le 1$, then the Fundamental Theorem of Calculus gives $$ f(x)=\frac{d}{dx}\int_0^x f(t)dt=\frac{d}{dx}\int_x^1f(t)dt = -f(x) $$

So $f(x)=0$ for $0 \le x \le 1$. (You take a right-hand derivative at $x=0$ and a left-hand derivative at $x=1$.)

Disintegrating By Parts
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I agree that the Fundamental Theorem of Calculus is the weapon to use here. However, if you want to get your hands a little bit dirtier, notice that $$\int_0^x f(t)dt = \int_x^1f(t)dt = \int^1_0f(t)dt - \int_0^xf(t)dt$$ implies that $$\int_0^x f(t)dt = \frac 12 \int_0^1 f(t)dt.$$ In particular, this also means (write the previous identity for $x=a$ and $x=b$, and substract them) that $$\int_a^bf(t)dt = 0,\quad\forall a,b\in(0,1).$$ Finally, assume there exists $x_0\in(0,1)$ such that $f(x_0)>0$. Since $f$ is continuous, there exists $\delta>0$ such that $f(x)>\frac 12 f(x_0) > 0$ for each $|x-x_0|<\delta$ (definition of continuity with $\epsilon = \frac 12f(x_0)$). This implies that the integral of $f$ over $(x_0-\delta,x_0+\delta)$ must be strictly positive, contradicting the previous equation. If you assume $f(x_0)<0$, an analogous reasoning follows.

foreignvol
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