Proof of integrals being equal to zero

Question

Alright hello, the proof on paper is really simple yet I'm having a lot of trouble with it. The proof is this: Let $f(x)$ be continuous on $[a,b]$, and $\forall x\in [a,b], f(x) \geq 0$. If $\int_a^b f(x)\ dx = 0$, then prove $∀x\in [a,b], f(x) = 0$.

Now I've messed around with it and I figure it's probably some simple use of squeeze theorem or something but I'm not entirely sure.

The method I devised was using the fundamental theorem of calculus along with the mean value theorem which I won't get into typing the whole thing, BUT I got this far with it.

1.) There exists a $c\in [a,b]$ such that $f(c)=0$.

2.) I can keep manually proving there exists values between the already calculated values or between $a$ and the next greatest point and $b$ with the highest point that isn't equal to $b$.

However, even if I could find a way to extend that to show that there's infinitely many points in the interval $[a,b]$ that produce an output of zero, I am not entirely sure if the cardinality of that set of points would be the same as the reals which would mean that the function may not necessarily equal zero, if approaching it that way.

Just to add a little diagram to the logic I'm explaining (sorry I'm awful with LaTex so this will have to do.)

NO FTC: [a------b] FTC once: [a----c----b] where f(c)=0 FTC twice: [a--d---c---g--b] where f(d)=f(c)=f(g)=0

I just keep seeming to miss the obvious solution to the problem and could really use a point in the right direction. Thanks much! Sorry again for my lackluster formatting

Answer 1

Suppose that exists $\xi\in [a,b]$ such that $f(\xi)>0$. Because $f$ is continuous, the sign permanence theorem ensures that exists $\delta>0$ such that $f(x)>0, \hspace{3pt}\forall x\in (\xi-\delta, \xi+\delta)$. So, it follows that: $$\int_{\max\{a,\xi-\delta\}}^{\min\{b,\xi+\delta\}}f(x)\ dx>0$$ that is a contradiction with the hypothesis assumed.

Answer 2

Can we assume that if $f(x) \ge 0$ for all $x\in [a,b]$ then for any $[\alpha, \beta]$ where $a \le \alpha < \beta \le b$ then $\int_{\alpha}^{\beta}f(x) dx \le \int_a^b f(x)dx$?

Can we assume that if for all $x \in [\alpha, \beta], g(x) > 0$ then $\int_{\alpha}^\beta g(x)dx > 0$?

If $f$ is continuous on all $[a,b]$ and there is a $w\in [a,b]$ where $f(w)\ne 0$ (and as $f(w)\ge 0$ then $f(w) > 0$, can we show there is $\alpha, \beta$ where $a \le \alpha \le w \le \beta \le b$ where $f(x) > 0$ for all $x \in [\alpha, \beta]$.

If so w have proven by contradiction that if there is an $f(w)> 0$ then there is an $\alpha,\beta$ where $0 < \int_\alpha^\beta f(x)dx \le \int_a^b f(x)dx$. We are done.

I imagine it is the third paragraph:

If $f$ is continuous on all $[a,b]$ and there is a $w\in [a,b]$ where $f(w)\ne 0$ (and as $f(w)\ge 0$ then $f(w) > 0$, can we show there is $\alpha, \beta$ where $a \le \alpha \le w \le \beta \le b$ where $f(x) > 0$ for all $x \in [\alpha, \beta]$.

That is not intuitive.

But this follows directly from the definition of continuous.

If $f(w) = c > 0$ and $f$ is continuous then for every $\epsilon > 0$ there is $\delta > 0$ so that if $|x-w| < \delta$ then $|f(x)-f(w)|< \epsilon$. Let $\epsilon = c$ and let $\delta$ be the resulting $\delta$. Let $\alpha = \max(a, w-\delta)$ and $\beta = \min(b, w+\delta)$ and .... voila!