Does the maximum value of the following integral exist?

[Tspirit]

Suppose ##\intop_{-\infty}^{+\infty}(f(x))^{2}dx=1##, and ##a=\intop_{-\infty}^{+\infty}(\frac{df(x)}{dx})^{2}dx##, does a maximum value of ##a## exist? If it exists, what's the corresponding ##f(x)##?

 

[andrewkirk]

No it doesn't exist. Consider the function ##f:\mathbb R\to\mathbb R## that is equal to ##\frac{\sin nx}{\sqrt\pi}## on the interval ##[0,2\pi]## and zero outside it. The first integral is 1 regardless of the value of ##n## but the second integral increases without limit as ##n## increases.

 

[Tspirit]

No it doesn't exist. Consider the function ##f:\mathbb R\to\mathbb R## that is equal to ##\frac{\sin nx}{\sqrt\pi}## on the interval ##[0,2\pi]## and zero outside it. The first integral is 1 regardless of the value of ##n## but the second integral increases without limit as ##n## increases.

Yes, you are right. Thanks.

 

[zinq]

Nice example, andrewkirk ! (Now I wonder what if the original problem were changed only so that the second integral used the 4th power of the derivative instead of its square.)

 

[Tspirit]

Nice example, andrewkirk ! (Now I wonder what if the original problem were changed only so that the second integral used the 4th power of the derivative instead of its square.)

I think it is like this: ##\intop_{+\infty}^{-\infty}f(x)dx=1##,and ##a=\intop_{-\infty}^{+\infty}(\frac{df(x)}{dx})^{4}dx##, does a maximum value of a exist?
If we use the example andrewkirk said ##“\frac{sin(nx)}{\sqrt{\pi}}”##, we have $$a=\intop_{-\infty}^{+\infty}(\frac{df(x)}{dx})^{4}dx=\intop_{-\infty}^{+\infty}(\frac{ncos(nx)}{\sqrt{\pi}})^{4}dx,$$ $$(ncos(nx))^{4}=n^{4}\left(\frac{1+cos2nx}{2}\right)^{2}=n^{4}[\frac{1}{4}+\frac{1}{2}cos2nx+\frac{1}{8}(1+4con4nx)],$$ so when ##n## is infinite, the ##a## is also infinite.