# Prove that $f$ is identically equal to $0$ on a closed interval

I was given $f$ is continuous on $[a,b]$ and $\int_a^bf(x)g(x)dx=0$ for all continuous function $g$ on $[a,b]$.

How do I prove that $f$ is identically equal to $0$ on $[a,b]$?

I tried to assume $f$ does not equal to $0$ to get a contradiction, but got stuck. Can I reach this problem with a different approach?

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See what you can deduce from taking $g=f$. – David Mitra Dec 7 '12 at 16:45

Hint: If $\int_a^b fg=0$ for all $g$ continuous on $[a,b]$ then

$$\int_a^b f(x)^2dx=0.$$

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Can I let $\epsilon={|f(t)|\over 2}$ and then do it by cases, and get $\int_a^bf(x)^2dx$ either $<0$ or $>0$, which is a contradiction? – Akaichan Dec 11 '12 at 15:42
$f(x)^2=|f(x)|^2$ since $f$ is continuous if its non-zero you can bound it below by $\varepsilon>0$ in some small interval, then this gives that the integral must be positive. – JSchlather Dec 11 '12 at 15:47
So, it's clear that $\int_a^bf(x)^2dx$ can never be negative,, right? Do I need to state that? – Akaichan Dec 11 '12 at 16:29
If $g\geq 0$ then $\int_a^b g(x)dx \geq 0$ assuming $a<b$ of course. You should be able to prove it, but you can most likely state it. – JSchlather Dec 11 '12 at 16:38
and from $\int_a^bf(x)^2dx=0$, I can just infer that $f(x)^2=0$? – Akaichan Dec 11 '12 at 18:07

If you want to go the contradiction route, here's one. Assume there is some $c\in (a, b)$ s.t. $f(c) \neq 0$. Just for simplicity's sake, say $f(c)\gt 0$.

By definition of continuity, there is some $\epsilon >0$ such that for any $x \in (c-\epsilon, c+\epsilon)$, we have $f(x) > 0$. Let $g$ be a function which is $0$ outside that interval, and positive within the interval. Then $$\int_a^bf(x)g(x)dx$$ is strictly positive. Thus we have a contradiction. (The case $f(c)< 0$ is completely analogous, only the integral is strictly negative. Still, we have a contradiction.)

An example of a $g$ with the properties we want (just to convince you that it exists) is the funtion which is equal to $0$ outside the interval $(c-\epsilon, c+\epsilon)$, and within the interval is equal to $\epsilon^2 - (x-c)^2$

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Then $\int_{a}^{b}f(x)^2dx = 0$. If $f$ is not null, there exist $x \in [a,b]: f(x)^2 >0$ by continuidty $f>c$ in some interval $I \subset \Omega$. Thus $\int_{a}^{b}f(x)^2dx \ge \int_{a}^{b}c^2 = c^2 |I| > 0$. Contradiction.

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Another option is for each $x \in (a,b)$, construct a sequence of continuous functions (say piecewise linear) $\{p_n\}$ with the property that $\int_a^b f(y) p_n(y) dy \to f(x)$ as $n \to \infty$. Basically, we want a sequence of functions of total integral 1 so that the majority of the mass of these functions is contained in increasingly smaller intervals around $x$.

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