I am trying to prove the following:

Take $f, g:[a,b] \to \mathbb{R}$ such that $f$ and $g$ are continuous. If $$\int_a^b f(x) \ \mathrm{d}x = \int_a^b g(x) \ \mathrm{d}x,$$

then there exists some $c \in [a,b]$ such that $f(c) = g(c).$

Here's my current proof. I'd welcome any feedback regarding correctness and clarity.

Current Proof

Assume that there exists no such $c$. There are then three possibilities. First, it is possible that $f(x) > g(x)$ $\forall$ $x \in [a,b]$. However, this cannot be, since then we would have $$\int_a^b f(x) \ \mathrm{d}x > \int_a^b g(x) \ \mathrm{d}.x$$

Similarly, we cannot have $g(x) > f(x)$ $\forall$ $x \in [a,b]$, since then $$\int_a^b f(x) \ \mathrm{d}x < \int_a^b g(x) \ \mathrm{d}x.$$ Thus, there exists some $x \in [a,b]$ such that $f(x) > g(x)$ and some $y \neq x$ such that $g(y) > f(y).$ Assume without loss of generality that $x < y.$ Consider a new function $h:[a,b] \to \mathbb{R}$ defined by $$h(x) = f(x) - g(x).$$ Clearly, $h$ is continuous, as the difference of two continuous functions. From the above, we have that $h(x) > 0$ and $h(y) < 0.$ Apply the Intermediate Value Theorem to $h$ on the interval $(x,y).$ Thus, there exists some $c \in (x, y)$ such that $f(c) = 0$. Since $(x, y) \subset [a, b]$, we have found an element of $[a,b]$ such that $h(c) = 0 \implies f(x) = g(x).$

  • $\begingroup$ Since you're invoking the IVT on $h$ anyway, just stick with that. You don't really need the other stuff. $\endgroup$ – Josephine Moeller Jan 21 '13 at 1:47
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    $\begingroup$ I'm sure you're aware, but your proof essentially shows that if $\int_a^b h(x) dx = 0$ and $h$ is continuous on $[a,b]$, then $\exists x \in [a,b]$ such that $h(x) = 0$. $\endgroup$ – JavaMan Jan 21 '13 at 1:50
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    $\begingroup$ Your proof is very correct and clear but not very concise. If you already know about Rolle’s theorem, just apply it to $$ F(x)=\int_{a}^x f(t)-g(t) dt $$ $\endgroup$ – Ewan Delanoy Jan 21 '13 at 2:23

While your proof is correct, there are ways you could simplify it. For example, as noted in the comments, you could apply Rolle's theorem to $\int_a^x f(x') - g(x') dx'$ since you know it will be zero at the endpoints $a$ and $b$.



$$\begin{align*} &f(x)=0, x\in [0,1]\\\\ &g(x)=\begin{cases}-1,&x\in\left[0,\frac12\right]\\\\1,&x\in\left(\frac12,1\right]\;.\end{cases} \end{align*}$$

Then $$\int_0^1f(x)dx=\int_0^1g(x)dx$$ and $f(x)\neq g(x)$, for every $x\in [0,1]$.

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    $\begingroup$ Nice example. Continuous? $\endgroup$ – Did Jan 25 '13 at 17:25

It can be proved: ∫{a}^{b}f(x)dx=∫{a}^{b}g(x)dx and f(x)≥g(x), for all x∈[a,b], then there is c∈[a,b] such that f(c)=g(c).

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    $\begingroup$ Well, this is the question, isn't it? $\endgroup$ – Did Jan 25 '13 at 17:25
  • $\begingroup$ Who asked this question before? It is essential f>=g... $\endgroup$ – CriMo Jan 25 '13 at 17:59

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