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Let $ g:[a,b]\to\mathbb{R} $ and $ f:[a,b]\to\mathbb{R} $ be continuous functions, such that $ f\left(x\right)\leq g\left(x\right) $ for any $ x\in[a,b] $.

Assume $ \intop_{a}^{\boldsymbol{b}}g\left(x\right)dx=\intop_{a}^{\boldsymbol{b}}f\left(x\right)dx $.

Is it true that $ f=g $?

I think that its true. I'll explain why:

we know that $ g\left(x\right)\geq f\left(x\right) $ and thus $ \intop_{a}^{\boldsymbol{b}}g\left(x\right)dx\geq\intop_{a}^{\boldsymbol{b}}f\left(x\right)dx $ so

$ \intop_{a}^{\boldsymbol{b}}(g\left(x\right)-f\left(x\right))dx\geq0 $

If we will assume by contradiction that $ g\neq f $ then we could find $ x_{0}\in[a,b] $ such that $ g\left(x_{0}\right)-f\left(x_{0}\right)>0 $.

But $ f $ and $ g $ are continuous, and thus $ g-f $ is continuous, and we found $ x_0 $ such that $ (g-f)(x_0 $ is positive, thus the integral $ \intop_{a}^{b}\left(g-f\right) $ should also be positive.

So that's why it seems correct to me, but I wanted to make sure because that's a strong assumption to make.

Thanks in advance.

Ralph Clausen
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FreeZe
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  • Can you add $dx$ in your integrals? – Vincenzo Tibullo Aug 25 '20 at 11:48
  • Your notation needs to be cleaned up, but your statement is correct. – Benjamin Aug 25 '20 at 11:49
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    Yes, I'll edit the post – FreeZe Aug 25 '20 at 11:53
  • FreeZe, you are right, if both functions are continuous, then they are equals. – Angelo Aug 25 '20 at 11:53
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    You are correct. You could also say: if $g - f$ is positive at a single point, then you could find a small ball (an interval) around that point in which $g-f$ is positive by continuity and the integral of $g - f$ over that ball would be positive. But this will be a contradiction to the hypothesis that the integral of g - f is identically zero on the whole interval [a, b]. – Daniel Akech Thiong Aug 25 '20 at 11:54
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    "But and are continuous, and thus − is continuous, and we found 0 such that (−)(0 is positive, thus the integral ∫(−) should also be positive." : This is exactely what you have to prove. – Walace Aug 25 '20 at 12:02
  • Which definition of integral do you know ? This is fairly trivial if you know the formal Darboux/Riemann definition of integralibility (show that the lower sums are bounded by a positive constant). – Lelouch Aug 25 '20 at 12:11

2 Answers2

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You have the right idea, having $(g-f)(x_0)>0$ is not enough to conclude that the integral is positive. Instead of considering $h=g-f$, we can suppose $f=0$.

If $g\neq 0$, there is $x_0\in [a,b]$ s.t. $g(x_0)>0$. We can suppose WLOG that $x_0\in (a,b)$. Since $g$ is continuous, there is $\delta >0$ s.t. $$(x_0-\delta ,x_0+\delta )\subset (a,b)\quad \text{and}\quad g|_{(x_0-\delta ,x_0+\delta )}>0.$$ Let $M>0$ s.t. $g|_{(x_0-\delta ,x_0+\delta )}\geq M$ (such $M$ exist by continuity of $g$). In particular, $$\int_a^b g\geq \int_{x_0-\delta }^{x_0+\delta }g\geq 2M\delta >0,$$ which is a contradiction.

Surb
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    having $ (g-f)(x_0) >0 $ and $ (g-f)(x) \geq 0 $ for any x $ \in [a,b] $ and $ g-f $ is continious is not enough to conclude that the integral is positive? – FreeZe Aug 25 '20 at 11:55
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    Yes, but you have to justify why it is enough. Which is not clear in your proof (you just state the proposition, but you ave to provide a proof). In particular, continuity is crucial here. @FreeZe – Surb Aug 25 '20 at 11:58
  • FreeZe, you are right, it is enough. Actually it is not necessary to suppose $f=0$, it is enough to consider the difference $(g-f)(x)$ and continuity. – Angelo Aug 25 '20 at 11:58
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another method consider $$h(x)=\int_{a}^{x}(f(x)-g(x))dx$$

clearly $h'(x) \le0$ or it is non increasing and $h(a)=0$ (given) therefore $h(x)\le 0$ for all $x>a$.

given $h(b)=0$ and as $b>a$ it leaves us no choice but to accept $h(x)=0$

To explain the last point suppose there exists some $c$ in $(a,b)$ such that $h(c)$ is not zero. Then $ h(c)<0$ as mentioned above.(h(x) is decreasing.)Now as h(x) is decreasing we have $h(b)<h(c)<0$ thus a contradiction!

thus $h(x)=0$ for all x in $[a,b]$ or it is a constant function . $h'(x)=0$ for all x in $[a,b]$. $f(x)=g(x)$

Albus Dumbledore
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