Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I'm trying to show that if f is continuous real-valued function on [a,b] satisfying integral from 1 to b (f(x)*g(x)) dx = 0 for every confinuous function g on [a,b] then f(x) = 0 for all x an element in [a,b]

What I tried is assuming that f(x) >= 0 for all x an element in [a,b] but I need help in breaking the cases and how many cases are there?

share|cite|improve this question


What about using $g(x)=f(x)$ ...

share|cite|improve this answer
This is ONE case, but what if f <0, f > 0, etc? – mary Apr 29 '11 at 5:15
Using $g=f$ works for every function $f$. User: you seem to be under the impression that either $f\ge0$ on $[a,b]$ or $f\le0$ on $[a,b]$. This is not true, think about $f$ defined by $f(x)=\sin(\omega x)$ for every $x$, for a large $\omega$. – Did Apr 29 '11 at 5:27
Fabian means that the problem is really not that hard. However, more general statements are true $f$ may be chosen from larger classes of functions than that of continuous ones (e.g. $f\in L^1$) and in the same manner the annihilating family, the $g$'s, may be chosen more restrictive (e.g. trigonometric polynomials $g$). – AD. Apr 29 '11 at 6:40

If you want to prove this by contraposition or contradiction, you don't want to assume that $f(x)\geq 0$ for all $x\in [a,b]$.

The negation of the conclusion of the stated theorem is that $f(x) \neq 0$ for at least one $x\in [a,b]$.

Assuming this, there is has to be some $c\neq 0$ and $x_c \in [a,b]$ such that $f(x_c)=c$. Now, a quick argument using continuity of $f$ lets us conclude that $f$ is actually bounded away from $0$ on some interval around $x_c$.

Once you have that, you can think about how to construct a continuous function $g$ such that $\int fg$ is nonzero.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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