Suppose $f \geq 0$, if continuous on [a,b] and $\int_{a}^{b} f(x)dx=0$. Prove that $f(x)=0$ for all $x\in [a,b]$.

Question

My attempt: Let P the partition of $[a,b]$. Let $x_{i}^{*} \in [x_{i-1},x_{i}]$ and f is non negative in $[x_{i-1},x_{i}]$.

Since f is continuous on $[a,b]$, then $f$ is R-integrable with $\int_{a}^{b} f(x)dx=0$.

$$\displaystyle \lim_{n\rightarrow \infty} \sum_{i=1}^{n}f(x_{i}^{*}) \delta x_{i}=0$$

This implies $f(x_{i})=0$ for all i. i.e $f(x)=0$.

Is this proof is correct?

If you can use lebesgue integration Show that the integral is 0 on any open set using the fact that every open set is a disjoint union of open intervals. — Sorfosh, Dec 08 '19 at 19:59
Actually, this problem is from the Riemann integral chapter. — User124356, Dec 08 '19 at 20:06
Can you kindly specify what is $\delta x_i$? Is $\delta x_i = x_i - x_{i-1}$? — the_candyman, Dec 08 '19 at 20:19
Yes. I choose the partition of n interval and $\delta x_{i}=x_{i}-x_{i-1}$ — User124356, Dec 08 '19 at 20:21
You are saying that, since $\delta x_i > 0$, then the limit is true if and only if $f(x_i^*) = 0$? — the_candyman, Dec 08 '19 at 20:24
Does this answer your question? Prove the integral of $f$ is positive if $f ≥ 0$, $f$ continuous at $x_0$ and $f(x_0)>0$ — Maximilian Janisch, Dec 08 '19 at 22:19

score 2 · Answer 1 · answered Dec 08 '19 at 20:03

I would have used the Mean value theorem to solve this.

Take a generic point $c \in (a ,b).$ You know that:

$$\int_{a}^{b} f(x) dx = \int_{a}^{c} f(x) dx + \int_{c}^{b} f(x) dx = 0.$$

Since $f$ is continuous in $[a, b]$ (with $b > a$), then there exist $d \in (a, c)$ and $e \in (c, b)$ such that:

$$\int_{a}^{c} f(x) dx + \int_{c}^{b} f(x) d = f(d)(c-a) + f(e)(b-c) = 0.$$

This can be rewritten as follows:

$$f(d)(c-a) = -f(e)(b-c).$$

We know that $a < c < b$. As a consequence, $(c-a) > 0$ and $(b-c) > 0$.

Therefore:

$$f(d) = -f(e) \frac{c-a}{b-c}.$$

Since $f \geq 0$ in $[a, b]$, the previous equation is only satisfied by $f(d) = f(e) = 0.$

This means that $f(x) = 0$ for all $x \in [a, b].$

Yes. This is a nice proof. Can you suggest mistakes in my proof? Thanks for your help. — User124356, Dec 08 '19 at 20:09
@kapil I would, but your notation is incomplete. Please, take a look to my comment to your question. — the_candyman, Dec 08 '19 at 20:20

score 2 · Answer 2 · answered Dec 08 '19 at 20:15

Assume by contradiction that $f(x_0)>0$ for some $x\in [a,b]$. Since $f$ is continuous, then there is a closed ball$^*$ around $x_0$ of radius $\epsilon$ s.t $f(x)>0$ for all points inside this ball. Since the closed ball is compact, $f$ has a (positive) minimum value inside the ball - $m$. What does this tell you about the Lower Darboux sum of $f$ in a partition which involves this ball? Try reaching a contradiction to the fact that the integral of $f$ is $0$.

$*$Note that the assumption about the closed ball is not accurate if $x_0$ is on the boundary of $[a,b]$, but this can easily be fixed - try to think how.

Peter Szilas · Answer 3 · 2019-12-09T14:44:46.293

Overkill?

$f$ is continuous on $[a,b]$.

$F(x):=\displaystyle{\int_{a}^{x}}f(x)dx$, $x \in [a,b]$; and

$F$ is differentiable: $F'(x)=f(x) \ge 0$,

i.e. $F$ is an increasing function of $x$.

$\rightarrow F(x)\equiv 0$ , for $x \in [a,b]$.

(Since $0 \le F(x) \le F(a)=0$)

Then $F'(x)=f(x)=0$, for $x \in (a,b)$.

Since $f$ is continuous on [a,b] : $f(x)=0$ , $x \in [a,b].$ (Consider $\lim_{x \rightarrow a^+}f(x)=f(a)$, similarly for $b$)

Suppose $f \geq 0$, if continuous on [a,b] and $\int_{a}^{b} f(x)dx=0$. Prove that $f(x)=0$ for all $x\in [a,b]$.

3 Answers3