Can we write $\int f(x)g(x)dx \leq C\int g(x)dx$ given that $\sup_x|f(x)|\leq C<+\infty$? My answer is yes, but I'm confused with the following $\int f(x)dx \leq C\int dx$. What $\int dx$ means, is it just a length of integration, i.e. $\int_a^bdx = b -a$?
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Yes, $\int , dx$ just means $\int 1 , dx$. Your inequality is trie as long as $g \geq 0$. For a counterexample in the general case, consider $f \equiv 0$ and $g \equiv -1$. – PhoemueX Jun 11 '14 at 14:26
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... and $a \le b$. – Robert Israel Jun 11 '14 at 14:27
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This is only true for all the given $f$ if $g(x)\ge0$. – robjohn Jun 11 '14 at 14:27
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The expression $\int f(x)g(x)dx$ is in general a function (or strictly speaking, a set of functions). It is not a real number and thus it does not make sense to say that $\int f(x)g(x)dx\leq C\int g(x)dx$ as two sets of functions are generally non-comparable. You may consider rewriting your question using definite integrals.
zuriel
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You are correct on both counts.
For your first question, you are using the fact that $p(x) \leq q(x) \Rightarrow \int_a^bp(x)dx \leq \int_a^bq(x)dx$. [As commented by PhoemueX, we need $g(x) \geq 0$, otherwise we cannot conclude $f(x)g(x) \leq Cg(x)$ from $f(x) \leq C$]
Your second question is just an application of the first where $g(x) = 1$.
Michael Albanese
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