0

Suppose that $f$ is continuous on $[a, b]$ and $\int_{a}^{b} fg = 0$ for every continuous function $g$ on $[a, b]$. Prove that $f = 0$.

My attempt: since $\int_{a}^{b} fg = 0$ for every continuous function $g$ on $[a,b]$, in particular taking $g(x) = 1$ for all $x \in [a, b]$ we get $$ 0 = \int_{a}^{b}f(x)dx. $$ Now to finish the proof I assume that I have to use the hypothesis that $f$ is continuous. What comes to my mind is the mean value theorem for integrals, to get a point $c \in (a, b)$ such that $$ 0 = \int_{a}^{b} f(x)dx = f(c)(b-a), $$ from what I can conclude that $f(c) = 0$, but this holds for some $c$, not for all of them. How should I proceed?

J. W. Tanner
  • 63,683
  • 4
  • 43
  • 88
MrGran
  • 775
  • 3
    $0 = \int_{a}^{b}f(x)dx$ does not imply that $f=0$. Better choose $g=f$. – Martin R May 02 '24 at 16:49
  • 1
    This has been asked and answered before: https://math.stackexchange.com/q/2538728/42969, https://math.stackexchange.com/q/2009922/42969, https://math.stackexchange.com/q/253157/42969 – all found with Approach0 – Martin R May 02 '24 at 16:50
  • I see, thank you. Now it is straightforward, since for $f=g$ one gets $\int_{a}^{b} f^{2} = 0$ and since the map $(f,g) \mapsto \int_{a}^{b}fg$ is an inner product it is positive definite, so $f=0$ for the equality to make sense. – MrGran May 02 '24 at 16:59

0 Answers0