Finding $\lambda$ for which $v(x)=|x|^\lambda\in L_2(\Omega), H^1(\Omega)$

Question

Let $\Omega$ be the unit ball in $\mathbb{R}^d,d=1,2,3$,i.e.,$\Omega=\{x\in\mathbb{R}^d:|x|<1\}$. For which values of $\lambda\in\mathbb{R}$ does the function $v(x)=|x|^\lambda$ belong to $(a) L_2(\Omega),(b) H^1(\Omega)$?

So far I did:

For $d=1$ we have $\Omega=\{x\in\mathbb{R}:|x|<1\}$. Then \begin{align*} \int_\Omega|v(x)|^2\,dx=&\int_{-1}^1|x|^\lambda\,dx\\ =&\int_{-1}^0(-x)^\lambda\,dx+\int_0^1\,x^\lambda\,dx\\ =&\frac{2}{\lambda+1} \end{align*}

So for $\lambda\neq -1$ $v\in L_2(\Omega)$. Now \begin{align*} \int_\Omega|v'(x)|^2\,dx=&\int_{-1}^1(|x|^\lambda)'\,dx\\ =&\int_{-1}^0(-\lambda)(-x)^{\lambda-1}\,dx+\int_0^1\,\lambda x^{\lambda-1}\,dx\\ =&0 \end{align*} So no restriction on $\lambda$.

Therefore for $\lambda\neq -1$ $v\in H^1(\Omega)$.

I am not quite sure whether I am right or wrong? Also I don't know how to proceed with $d=2$ and $d=3$. Any help would be greatly appreciated. Thanks in advance.

Answer 1

I think you forgot to square your function when taking the $L^2$-norm.

Generally, in $\mathbb R^d \,\,\, (d \ge 2),$ by the transformation formula $$\int_{B_1(0)} \lvert x \rvert^{2\lambda}dx = \int_{0}^1 \int_{\partial B_1(0)} r^{2\lambda} r^{d-1} d\sigma\, dr = \omega_d \int^1_0 r^{2\lambda + d - 1} dr$$ where $\omega_d$ is a dimensional constant $($the surface measure of the sphere $\mathbb S^{d-1})$. For convergence of this integral, you need $2\lambda + d -1 > -1$ or $\lambda > -d/2$ (and similarly for the derivatives, though the power will change because of the differentiation so we require $\lambda > -d/2+1$).

Your one-dimensional case is also a bit wrong (regardless of the fact that you forgot to square the function). For the integral $$\int_{-1}^{1} \lvert x \rvert^{2\lambda} dx$$ to converge we need integrability near the origin, so we need $2\lambda > -1$ or $\lambda > -1/2$. Likewise for the derivative to be in $L^2$, we need $$\int_0^1 \big(\lvert x \rvert^{\lambda-1}\big)^2 dx$$ to converge which required $2\lambda - 2 > -1$ or $\lambda > 1/2$. Thus the function is in $L^2$ for $\lambda > -1/2$ and $H^1$ for $\lambda > 1/2$.