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Let $f\in L^1([a,b])$, and extend $f$ to be $0$ outside $[a,b]$. Let $$ \phi(x)=\frac{1}{2h}\int_{x-h}^{x+h}f $$

How to prove $$ \int_a^b\left | \phi\right | \leqslant\int_a^b\left|f\right| $$

To @martini:

I asked this question because I find something strange: if we let $f(x)=x^2$, then $\phi(x)=x^2+\frac{1}{3}h^2$, so the inequality doesn't hold apparently.

Clearly, the problem is I do not extend $f$ to be $0$ outside $[a,b]$, so $\phi$ should be smaller near $a$ and $b$. However, when I put this example into your proof, I cannot figure out what goes wrong. It seems you have proved $$ \int_a^b(x^2+\frac{1}{3}h^2) \leqslant\int_a^b x^2 $$

I think the problem lies in your exchange of $\chi$, but still not clear.

hxhxhx88
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  • Expand the expression for $\phi$, and then interchange the order of integration. – Willie Wong Jun 25 '13 at 11:53
  • @WillieWong, please refer to my update in the post. – hxhxhx88 Jun 25 '13 at 15:03
  • The corrections due to the edge terms are quite large compared to $h^2$. For $k < h$ and $x = b-k$ you have $$\phi(x) = \frac{k + h}{2h} x^2 - \frac{h^2 - k^2}{2h} x + \frac{1}{6h}(h^3 + k^3) $$ As you wrote, what you plugged in as "$\phi$" is not really the correct function. If you plug in the correct function you will not see a contradiction. – Willie Wong Jun 25 '13 at 15:23
  • @WillieWong, my point is exactly where the proof given by martini shows the condition $f$ equals 0 outside [a,b]? If I still let $f(x)=x^2$ outside $[a,b]$, it seems the proof can pass as well. i.e., consider two function $\tilde f$ and $\hat f$, the former is extended to be zero and the latter is extended to still be $x^2$, then put them into the proof, replace $f$ there, what makes the situation of $\hat f$ fails? – hxhxhx88 Jun 25 '13 at 15:43
  • @WillieWong, in the situation of $\hat f$, it should fail, for the reason shown in my update. But the proof seems to work fine. – hxhxhx88 Jun 25 '13 at 15:54
  • Ah, the step where it is used is in the third line. If $[x-h,x+h] \subset [a,b]$ then you can change the bounds of integration and add a characteristic function. But if $f$ doesn't vanish outside $[a,b]$ and if either $x-h < a$ or $x+h > b$, that line is false. (In fact, in such a case the third line is going to be less than the second line.) – Willie Wong Jun 25 '13 at 15:54
  • @WillieWong, you helped me a lot!!!! I have been confused for a long time....Thank you!!! – hxhxhx88 Jun 25 '13 at 15:56

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We have where $\chi_I$ denotes the indicator function of $I$$\def\abs#1{\left|#1\right|}$ \begin{align*} \int_a^b \abs{\phi(x)}\, dx &= \frac 1{2h}\int_a^b \abs{\int_{x-h}^{x+h} f(\xi)\, d\xi}\, dx\\ &\le \frac 1{2h}\int_a^b \int_{x-h}^{x+h} \abs{f(\xi)}\, d\xi\, dx\\ &= \frac 1{2h}\int_a^b \int_a^b \abs{f(\xi)}\chi_{[x-h,x+h]}(\xi)\, dx\, d\xi\\ &= \frac 1{2h}\int_a^b\abs{f(\xi)} \int_a^b \chi_{[\xi-h,\xi+h]}(x)\, dx\,d\xi\\ & \text{note that $x-h \le \xi\le x+h \iff \xi-h\le x \le \xi+h$}\\ &= \frac 1{2h}\int_a^b\abs{f(\xi)}\int_{\xi-h}^{\xi+h}\, dx\, d\xi\\ &\le \frac {2h}{2h} \int_a^b \abs{f(\xi)}\, d\xi. \end{align*}

martini
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