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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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    @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

4 Answers 4

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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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    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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    I think this is clearer after changing the dummy variables.2011-11-22