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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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First note: $$\int_0^x f(x) \, \textrm{d}x = \int 1_{[0,x]}(t) f(t) \, \textrm{d}t.$$

Then:

$$\left (\int 1_{[0,x]}(t) f(t) \, \textrm{d}t \right )^2 \leq \int 1_{[0,x]}(t)^2 \, \textrm{d}t \cdot \int_0^x f(t)^2 \, \textrm{d}t.$$

By Cauchy-Bunyakovski-Schwarz.

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    This is how I first thought about it (+1). However, I would write $$\int_0^x1_{[0,x]}(t)^2\mathrm{d}t$$ to make it look more like C-B-S.2011-11-21
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    @robjohn Done. ${}$2011-11-21
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    @robjohn I have never seen the notation 1_[0,x] before... What does it indicate?2011-11-22
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    @Travis: I don't know if I have either, but it appears to be a function which is $1$ on $[0,x]$ and $0$ elsewhere.2011-11-22
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    To my knowledge that is just standard notation for the indicator function! How would you write it? $\chi$?2011-11-22
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    FWIW, I see the notation $1_A$ for the indicator function of set $A$ more often in the context of probability theory, measure theory, and convex analysis texts, and $\chi_A$ in other analysis texts. (The $\chi$, of course, stands for *characteristic*.) Unfortunately, the term "characteristic function" has a different meaning in the context of probability distributions, and in the context of convex analysis, $\chi_A$ is the characteristic function defined to be $(+\infty)(1 - 1_A)$ (it takes value $+\infty$ away from $A$ and $0$ at $A$).2011-11-22
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    Habitually I use $\chi_A$, but I do agree in principle that $1_A$ is better notation.2011-11-22
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    Maybe a bold $\mathbf 1$ or blackboard bold $1$ is better to distinguish between the number $1$. But I'm pretty sure there will not be much confusion.2011-11-22
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It might be worth mentioning that this is really just a scaled version of the $x = 1$ case. If you change variables to $u$ where $t = xu$, then your inequality reduces to $$\bigg(\int_0^1 f(xu)\,du\bigg)^2 \leq \int_0^1 f(xu)^2\,du$$ This is for example Jensen's inequality for $\phi(x) = x^2$.

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Your notation with the $x$ in the limits and the variable of integration is a bit non-standard. However, if we fix that and divide both sides by $x^2$, we get $$ \left(\int_0^xf(t)\frac{\mathrm{d}t}{x}\right)^2\le\int_0^xf(t)^2\frac{\mathrm{d}t}{x} $$ Which is Jensen's inequality since $\dfrac{\mathrm{d}t}{x}$ is a unit measure on $[0,x]$.

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    Now that I look at Zarrax's answer, this is just a scaled version of that.2012-01-27
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You can use Cauchy-Schwarz inequality as suggested by Davide. Or more elementary, we can prove it in this way: Consider the function $g(y)=f(y)-\frac{1}{x}\int_0^xf(t)dt$ where $x>0$. Clearly we have $$\int_0^xg^2(y)dy\geq 0.$$ On the other thand, $$\int_0^xg^2(y)dy=\int_0^x\Big(f(y)-\frac{1}{x}\int_0^xf(t)dt\Big)^2dy=$$

$$=\int_0^x\Big[f^2(y)-f(y)\Big(\frac{1}{x}\int_0^xf(t)dt\Big)+\Big(\frac{1}{x}\int_0^xf(t)dt\Big)^2\Big]dy=$$ $$= \int_0^xf^2(y)dy-\frac{1}{x}\Big(\int_0^xf(t)dt\Big)^2\ .$$ Now the result follows by combining the above inequality and equality.

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    The expansion in your last step doesn't seem right... or perhaps I'm just reading it wrong, it's confusing with the $dx$. Could you switch to different dummy variables?2011-11-21
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    What I find slightly confusing is that $f(\cdots)$ is a multiplication. But I think process91 is correct and you have $\int_0^xg^2dx=\int_0^xf^2dx-\Big(\frac{2}{x}-\frac{1}{x^2}\Big)\Big(\int_0^xfdx\Big)^2$2011-11-22
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    I think this is clearer after changing the dummy variables.2011-11-22