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Suppose we have two continuous functions $f,g:[a,b)\to\mathbb{R}$ such that the improper integrals $\int_a^bf(x)\,dx$ and $\int_a^bg(x)\,dx$ are convergent. Does this imply that the limit $\lim_{y\to b^-}\int_a^yf(x)g(x)\,dx$ exists - finite or infinite? What about continuous functions $f,g:[a,\infty)\to\mathbb{R}$ and analogous improper integrals $\int_a^\infty \dots\, dx?$

Looks like my textbook uses this to prove some version of Holder's inequality and I'm having problems proving it. I know that it is true for nonnegative functions $f,g,$ since then the function $I(y)=\int_a^yf(x)g(x)\,dx$ is nondecreasing. I also found example of functions $f,g$ such that $\int_a^\infty f(x)\,dx$ and $\int_a^\infty g(x)\,dx$ are convergent and $\int_a^\infty f(x)g(x)\,dx=\infty,$ therefore the latter may not be convergent, but that's all I have.

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Stan Tendijck answer doesn't really answer the question completely. In his example, the limit $\lim\limits_{y\to b^-} \int_a^y f(x)g(x)dx = \infty$ exists.

For an actual counterexample, consider $$ f(x) = \sum_{n=0}^\infty 2^{n/2} \cdot 2^{n+2}\left(\left| x - \frac{2^{-n-1}+2^{-n}}2 \right| -2^{-n-2} \right) I_{[2^{-n-1},2^{-n}]} $$ and $$ g(x) = \sum_{n=0}^\infty (-1)^n 2^{n/2} \cdot 2^{n+2}\left(\left| x - \frac{2^{-n-1}+2^{-n}}2 \right| -2^{-n-2} \right) I_{[2^{-n-1},2^{-n}]}. $$

Then $$ \int_0^1 f(x)dx \simeq \sum_{n=0}^\infty 2^{n/2} \cdot 1 \cdot 2^{-n-1} < \infty, $$ thus also $g$ is absolutely integrable on $[0,1]$, but the improper integral $\lim\limits_{y\to0^+} \int_y^1 f(x)g(x)dx$ does not exist because $$ \int_{2^{-n-1}}^{2^{-n}} f(x)g(x)dx \simeq (-1)^n. $$

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  • $\begingroup$ It depends on what do you mean by the existence of a limit. Your counter example is indeed better than mine, although both are proper counter examples in my opinion :) $\endgroup$ – Stan Tendijck Jun 26 '18 at 18:34
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I don't think this is actually true.

Take for example $f(x)=g(x)=1/(x-b)^{1/2}$. Then, it is obvious that both functions are continuous on $[a,b)$. Moreover, they are both integrable over $[a,b)$ but the product isn't. A similar example can be taken in the second case.

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