On Sobolev space $W^{2,2}$, is the inner product of the derivative of a function and the function itself zero?

Question

Let $\;f\in W^{2,2}(\mathbb R;\mathbb R^2)\;$, then is it always true that $\;{\langle f,f' \rangle}_{L^2}=0\;$?

I'm dealing with some integrals and I observed that in my notes the term $\;\int_{\mathbb R} (ff')'\;dx\;$ always vanishes without further explanation.

Hence I thought that the inner product $\;{\langle f,f' \rangle}_{L^2}\;$ should be zero whenever $\;f \in W^{2,2}\;$ but I can't see how to prove it.

Answer 1

You appear to be asking about two different integrals, $\int_{\mathbb{R}}ff'$ and $\int_{\mathbb{R}}(ff')'$. The answer is yes for both: such integrals are zero for all $f\in W^{2,2}(\mathbb{R})$. (The codomain being two-dimensional does not make a difference.)

Since smooth compactly supported functions are dense in $W^{2,2}(\mathbb{R})$, the questions to ask are:

1. Is this integral expression zero for smooth compactly supported functions?
2. Is it continuous with respect to the $W^{2,2}$ norm?

The answer is "yes" to both, in both cases. For smooth compactly supported functions, we have an integral of the derivative of some expression (using $ff' = (f^2/2)'$), and such an integral vanishes.

For the second question, it's easier to think in terms of the bilinear form $B(f, g) = \int fg'$ or $B(f, g) = \int (fg')'$. Is this bilinear form bounded on $W^{2, 2}$, meaning $|B(f, g)|\le C\|f\|\|g\|$? Yes in both cases (using Cauchy-Schwarz): only derivatives up to order 2 are involved, and they are all controlled by the $W^{2,2}$ norm in the suitable way.