techniques to know whether given function is in $H^s(\Omega)$ or not?

Question

Suppose $f$ be a function defined as $f_m(x)=x^m(1-x)^m$ if $x\in [0,1]$ and then we extend it by zero in $[0,3]$. Since $f_1(x)$ is not differntiable at $x=1$. I would like to know how to calculate the value of $s$ such that $f\in H^s([0,3])$. Further if we increase $m$ how does the value of $s$ changes? Kindly provide some hint that in general, what are the techniques to know that given function lies in which $H^s([0,3])$? Thanks

Answer 1

Firstly, $f_1$ is not discontinuous at $x=1$, it is $C^{0,1}$.

Secondly, $f_m=g_m \circ u^+$ where $u(x)=x(1-x)$, $g_m(x) = x^m$, and $u^+(x)=\max\{u(x),0\}$ is the positive part of $u$. Since $u\in C^\infty([0,3])$, it follows that $u^+ \in H^1((0,3))$. Moreover, if $m\geqslant 1$ then $g_m$ is locally Lipschitz so by the chain rule, $f_m \in H^1((0,3))$.

If $0<m<1$ then I believe that you should have $f_m \in H^m((0,3))$ but you should double check this.

Answer 2

As indicated by JackT, $f_m$ is not discontinuous when $m\geq 0$. Notice also that you can also extend your function to $0$ on the whole space $\Bbb R$ as the singularity is the same at $x=0$ and $x=1$ (the function is symmetric with respect to $1/2$

Then to know in what $H^s$ space it belongs, you can:

• compute $m$ derivatives and look if it gives a $L^2$ function when $m$ is an integer.
• Use the Fourier transform $\widehat f_m$ and compute $\int_{\Bbb R} |\widehat f_m(y)|^2 \,(1+|y|^2)^{s}$
• Use the integral formula $$ |f|_{H^s}^2 = \iint_{\Bbb R^2} \frac{|f(x)-f(y)|^2}{|x-y|^{d+2s}} \,\mathrm d x\,\mathrm d y $$ and use the fact that if $s\in(0,1)$, for a compactly supported function, $f\in H^s$ if and only if $|f|_{H^s} <\infty$. If $s$ is larger, then you can write $s = n + \alpha$ with $n = \lfloor s\rfloor$, and then $f\in H^s$ if and only if $|\nabla^n f|_{H^\alpha} <\infty$

In your case, the last formula might be more efficient than the second since the Fourier transform might be difficult to compute.