12
$\begingroup$

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 $h(c) = 0 \implies f(x) = g(x).$

$\endgroup$
3
  • $\begingroup$ 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
  • 7
    $\begingroup$ 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
  • 9
    $\begingroup$ 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
3
$\begingroup$

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 $b$.

$\endgroup$
-1
$\begingroup$

Let

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

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

$\endgroup$
1
  • 2
    $\begingroup$ Nice example. Continuous? $\endgroup$ – Did Jan 25 '13 at 17:25
-2
$\begingroup$

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).

$\endgroup$
2
  • 2
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.