Suppose $f$ is a continuous function on $[a,b]$ and $$ \int_a^b f(x)g(x) = 0$$ for every integrable function. Show that $f(x) = 0$ on $[a,b]. $

Here is what I have so far:

Consider any $x \in [a,b].$ Consider any $y >0.$ Say $g(u) = 1$ for $x<u<x+y,$ and $g(u) = 0$ otherwise. Hence $\int_x^{x+y} f(u) du = 0.$ Hence $\frac{\int_x^{x+y} f(u) du}{y} = 0.$ tend $y = 0$ we get $f(x) = 0.$ Hence proved.

Is this correct?

  • $\begingroup$ I edited your question. Please double check I did not change anything (apparently, Yuval caught one mistake already). $\endgroup$
    – user2468
    Aug 6 '12 at 20:21
  • $\begingroup$ Your proof sounds fine, just remember to invoke explicitly the fundamental theorem of calculus. $\endgroup$ Aug 6 '12 at 20:21
  • $\begingroup$ Well "dx" is missing... $\endgroup$
    – Hawk
    Aug 6 '12 at 20:27

Hint: Let $x_0$ be a point at which $f(x_0) \ne 0$; Since $f$ is continuous there is an interval about $x_0$ in which $f$ is non-zero (can you prove this?). Now, can you find a $g(x)$ for which $\int_a^b f(x) g(x) dx \ne 0$ given this information?


Note that $\int_{[a,b]}f^2=0$. Then, $f\ge 0$. If there exist $x_0$ such that $f(x_0)>0$ Then, by continuity there exist an interval such that $f^2>0$ there and $\int_{[a,b]}f^2>0$, contradiction.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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