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.