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I am looking at the following exercise:

Let $f:[a,b] \to \mathbb{R}$ integrable. Show that $\exists \xi \in [a,b]$ such that $\int_a^{\xi} f(u) du= \int_{\xi}^{b} f(u) du$

We define $G(x)=\int_a^{x} f(u) du+ \int_{b}^{x} f(u) du$

$$\int_a^{x} f(u) du \text{ and } \int_{b}^{x} f(u) du \text{ are uniformly continuous at} [a,b],\text{ so } G \text{ is also uniformly continuous }$$

  • $G(a)=- \int_a^b f(u) du$
  • $G(b)=\int_a^b f(u) du$

So, $G(a) \cdot G(b) \leq 0$

  • $G(a) \cdot G(b) <0$: $$G \text{ is uniformly continuous at } [a,b],\text{ so it is also continuous at this interval, so from the Bolzano theorem } \exists \xi \in (a,b) \text{ such that } \int_a^{\xi} f(u)du= \int_{\xi}^b f(u)du$$
  • At the case $G(a) \cdot G(b)=0$, are the only possible roots $\xi=a \text{ and } \xi=b$,are there also other ones?
evinda
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3 Answers3

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It may be more fruitful to define $$H(x) = \int_a^x f(u) \, du - \int_x^b f(u) \, du.$$ If $H(a) = 0$ you are done. Otherwise note that $H(a) = - H(b)$ so that one of these numbers is positive and the other is negative. There exists $\xi$ with $H(\xi) = 0$ because $H$ is continuous on $[a,b]$.

Umberto P.
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The answer to "are there no other roots?" is not necessarily. For example, set $f(x)=0$, then every value of $x$ is a root, i.e. $H(x)=0$ for all $x$ (as defined in @Umberto's solution).

vadim123
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There can be more roots. Take for example $f(x)=\sin x$, $x\in[0,4\pi]$.

Then, $$\int_0^{x}\sin udu=1-\cos x$$ and $$\int_x^{4\pi}\sin udu=\cos x-1$$

The equation $$1-\cos x=\cos x-1$$ has three solutions in the intrerval, namely $0$,$2\pi$ and $4\pi$.

To guarantee uniqueness you need $f$ to be monotonic. (Maybe with slightly weaker conditions, such monotonic almost always it can also be guaranteed).

ajotatxe
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