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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 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?

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    I edited your question. Please double check I did not change anything (apparently, Yuval caught one mistake already).2012-08-06
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    Your proof sounds fine, just remember to invoke explicitly the fundamental theorem of calculus.2012-08-06
  • 0
    Well "dx" is missing...2012-08-06

2 Answers 2