in an application I have the following vector field $$\beta(x,y)=\frac{1}{\sqrt{x^2+y^2}}(-y,x)$$
and I need to check if this lives in the Sobolev space $[W^{1,\infty}([0,1]^2)]^2$
Looking at the first component $\frac{y}{\sqrt{x^2+y^2}}$, I should check that:
- it is in $L^\infty([0,1]^2)$
- its gradient is in $L^\infty([0,1]^2)$
I think it's not in $L^\infty$, as it blows up at the origin. At the same time, the $L^\infty$ norm doesn't "see" measure zero sets like the origin $(0,0)$, so I'm wondering if it's in $L^\infty$ or not. Thanks for your help.
Your function is in $L^\infty ([0,1]^2 ; \mathbb R^2)$ but it is not in $W^{1,\infty}([0,1]^2 ; \mathbb R^2)$. Indeed, for $(x,y) \neq (0,0)$, $$ \vert \beta(x,y) \vert= \frac{\sqrt{x^2+y^2}}{\sqrt{x^2+y^2}} =1,$$ so $$ \| \beta \|_{L^\infty ([0,1]^2 ; \mathbb R^2)} = \underset{(x,y) \in [0,1]^2}{\mathrm{ess} \, \sup} \vert \beta(x,y) \vert =1.$$
However, you can check that the gradient is not in $L^\infty$. We have that for $(x,y) \neq (0,0)$, $$ \frac{\partial \beta}{\partial x} =\bigg ( \frac{xy}{(x^2+y^2)^{\frac32}}, \frac{y^2}{(x^2+y^2)^{\frac32}}\bigg ).$$ Hence, $$ \bigg \vert \frac{\partial \beta}{\partial x} \bigg \vert =\frac1{(x^2+y^2)^{\frac32}} \sqrt{x^2y^2+y^4} = \frac{\vert y \vert}{x^2+y^2}.$$ Approaching $(0,0)$ along the line $x=0$ we find $$ \bigg \vert \frac{\partial \beta}{\partial x} \bigg \vert = \frac{\vert y \vert}{y^2}=\frac1{\vert y \vert} \to \infty$$ as $y \to 0$.