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Assuming Cauchy Schwarz inquality as follows... $$\left|\int_a^b{f(x)g(x)dx} \right|\le \left(\int_a^b{|f(x)|^2}dx\right)^{1/2}\left(\int_a^b{|g(x)|^2}dx\right)^{1/2} $$ Where $g(x)=0$ and $f(x)=f'(x)$. Show that...$$\left(\int_a^b{|f(x)|^2dx}\right)^{1/2}\le C\left(\int_a^b{|f'(x)|^2}dx\right)^{1/2} $$

You can suppose that $f$ is a smooth function supported on [a,b] implying f is absolutely integrable and square integrable such that $\mathscr F f$ is well defined. C will depend on $b-a$. The problem is outlined in three steps.

a) To begin, use the fundamental theorem of calculus to write formula for f(x) in terms of integral of its derivative, you must use the fact that f is zero at the endpoints. b) Invoke Cauchy Schwarz as explained above... c) Integrate the expression.

One possibility is to use integration by parts such that $\int_a^bf(x)dx=-\int_a^bxf(x)dx$ where x might be replaced with a specific x-C.

However, I have had more luck with the following, though it concerns me because it does not follow the steps given... $$\int_a^b f(x)^2 dx = \int_b^a \left(\int_a^x f'(y)dy\right)^2dx$$ $$\int_a^x f'(y)dy\le\left(\int_a^x|f'(y)|^2dy\right)^{1/2}\left(\int_a^x1^2dy\right)^{1/2}=\left(\int_a^x|f'(y)|^2dy\right)^{1/2}\left(\sqrt{x-a}\right)$$ Using the Cauchy Schwarz inequality as describe above and that $f(a)=0$. Then substituting this... $$\int_a^b f(x)^2 dx \le\int_a^b\left(\left(\sqrt{x-a}\right)\left(\int_a^x|f'(y)|^2dy\right)^{1/2}\right)^2dx$$ Taking the square root of both sides gives ALMOST the answer, but there are obvious problems: I do not know if it is legal to move the square root through the first integral. I am not sure how to evaluate the double integral to obtain C from $\sqrt{x-a}$ and obtain the correct bounds for what should only be one integral (but is a double integral).

$$\left(\int_a^b f(x)^2 dx\right)^{1/2} \le\int_a^b\left(\sqrt{x-a}\right)\left(\int_a^x|f'(y)|^2dy\right)^{1/2}dx$$

I have certainly tried some other alternatives, but this one has gotten me the closest.

Any help is much appreciated!

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  • $\begingroup$ Minor variant of math.stackexchange.com/questions/371965/… $\endgroup$ – stochasticboy321 Dec 4 '15 at 1:18
  • $\begingroup$ @stochasticboy321 Hm, I have looked at that problem and tried something similar, but it didn't seem to give me what I wanted. Especially because I did not know what the constant should be and because they show the integral of simply f(x) is less than... instead of f(x)^2. I made some more edits above to show improved work. $\endgroup$ – mentorship Dec 4 '15 at 2:25
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https://math.stackexchange.com/a/499219/214892

Yes I believe this is in fact already correct, after obtaining... $$\int_a^b f(x)^2 dx \le\int_a^b\left(\left(\sqrt{x-a}\right)\left(\int_a^x|f'(y)|^2dy\right)^{1/2}\right)^2dx$$

Let the squared and square root terms cancel. Then make the observation that the maximum will occur when x = b such that$\int_a^bx-a\left(\int_a^x|f'(y)|^2dy\right)dx\le\int_a^b\left(b-a\right)\left(\int_a^b|f'(y)|^2dy\right)dx=(b-a)^2\int_a^b|f'(y)|^2dy$ The integral can then be made in terms of x since y is a dummy variable and the entire inequality can be taken to the $\frac{1}{2}$ power.

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