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?


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


//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?//

  • $\begingroup$ @DavideGiraudo Ah yes, I do and I think I know how to approach this! $\endgroup$ – Travis Lex Nov 21 '11 at 22:13
  • $\begingroup$ @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? $\endgroup$ – Travis Lex Nov 21 '11 at 22:14
  • $\begingroup$ @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. $\endgroup$ – process91 Nov 21 '11 at 22:30
  • $\begingroup$ As process91 says elsewhere, you seem to have $dx$ inside the integral and $x$ as a limit of integration $\endgroup$ – Henry Nov 22 '11 at 1:24
  • $\begingroup$ @DavideGiraudo Could you check the proof I have put up there? (or is it common here to upload a new question for that?) $\endgroup$ – Travis Lex Nov 22 '11 at 3:02

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


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

By Cauchy-Bunyakovski-Schwarz.

  • $\begingroup$ 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. $\endgroup$ – robjohn Nov 21 '11 at 22:25
  • $\begingroup$ @robjohn Done. ${}$ $\endgroup$ – Jonas Teuwen Nov 21 '11 at 22:31
  • $\begingroup$ @robjohn I have never seen the notation 1_[0,x] before... What does it indicate? $\endgroup$ – Travis Lex Nov 22 '11 at 2:42
  • $\begingroup$ @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. $\endgroup$ – robjohn Nov 22 '11 at 6:19
  • 1
    $\begingroup$ To my knowledge that is just standard notation for the indicator function! How would you write it? $\chi$? $\endgroup$ – Jonas Teuwen Nov 22 '11 at 11:38

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 $\phi(x) = x^2$.


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 $[0,x]$.

  • $\begingroup$ Now that I look at Zarrax's answer, this is just a scaled version of that. $\endgroup$ – robjohn Jan 27 '12 at 22:46

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.

  • $\begingroup$ 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? $\endgroup$ – process91 Nov 21 '11 at 22:28
  • $\begingroup$ What I find slightly confusing is that $f(\cdots)$ is a multiplication. But I think process91 is correct and you have $\int_0^xg^2dx=\int_0^xf^2dx-\Big(\frac{2}{x}-\frac{1}{x^2}\Big)\Big(\int_0^xfdx\Big)^2$ $\endgroup$ – Henry Nov 22 '11 at 1:21
  • $\begingroup$ I think this is clearer after changing the dummy variables. $\endgroup$ – Paul Nov 22 '11 at 9:15

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