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Assume that a function $f$ is integrable on $[0, x]$ for every $x > 0$.

Prove that for any $x > 0$, $\displaystyle\left (\int_{0}^{x}fdx \right )^2\leq x\int_{0}^{x}f^2dx$.

I have no idea how to even start this... What concept should I be using?

EDIT:

So upon the hint of using C-S inequality/using a dummy variable for clarity, I have come up with the following proof:

Let $g$ be a constant function s.t. $g=1$ for any $x >0$.

Note that $\displaystyle\ x \cdot \int_{0}^{x}f^2(t)dt = \left(\int_{0}^{x}g(t)dt \right) \cdot \left( \int_{0}^{x}f^2(t)dt \right)$.

Since we know that f and g is integrable, we can apply the Cauchy-Schwarz Inequality for integrals. The inequality states that (integral of $f \cdot g$,... etc..).

Thus, $$ \left (\int_{0}^{x}f(t)\cdot g(t)dt \right )^2 = \left (\int_{0}^{x}f(t)dt \right )^2 \leq \int_{0}^{x}g(t)dt \cdot \int_0^x f^2(t)dt=x\int_0^x f^2dt .$$

Q.E.D.

//I don't know if I should be using $t$ or $x$ here though... As a matter of fact, shouldn't the statement change to

Prove that for any $t,x>0$, [inequality] holds.

now that we use $t$? Or am I misunderstanding the use of a dummy variable?//

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    @DavideGiraudo Ah yes, I do and I think I know how to approach this!2011-11-21
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    @AustinMohr Would it be wrong to integrate with respect to x - or would it just be confusing? Also, do I need to define t before hand, or can I just use it right away?2011-11-21
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    @TravisLex It's confusing, and considered bad form. You don't have to define it before hand if you use it as a dummy variable.2011-11-21
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    As process91 says elsewhere, you seem to have $dx$ inside the integral and $x$ as a limit of integration2011-11-22
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    @DavideGiraudo Could you check the proof I have put up there? (or is it common here to upload a new question for that?)2011-11-22
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    @Henry Yes, the problem was stated like this... Could you check my proof to see if I used the dummy variable properly?2011-11-22
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    @process91 Thank you for the prompt response. Have I used it correctly in my edited proof? I am a little confused as you can see by the end note...2011-11-22
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    @TravisLex Almost, just the last $dx$ should be $dt$.2011-11-22
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    @TravisLex Also, the way that you presented your proof (as an addition to your question) was fine. It is also fine to answer your own question, if it is sufficiently different from the others.2011-11-22

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