I want to compute the Sobolev norm $H^2([0,1]^2)$ for a 2d function such as $f(x,y)=x^{10}y^5$. I am confused whether it is,
$$\|f\|_{H^2}^2=\|f\|_{L^2}^2+\|\partial_x f\|_{L^2}^2+\|\partial_y f\|_{L^2}^2+\|\partial^2_{xx} f\|_{L^2}^2+\|\partial^2_{yy} f\|_{L^2}^2+\|\partial^2_{xy} f\|_{L^2}^2$$
or is it,
$$\|f\|_{H^2}^2=\|f\|_{L^2}^2+\|\partial_x f+\partial_y f\|_{L^2}^2+\|\partial^2_{xx} f+\partial^2_{xy} f+\partial^2_{yy} f\|_{L^2}^2$$
Thank you for any help. If both of them are wrong, then what is the right answer? I am confused with the notation.
Actually I think its neither, one usually has all elements of the Hessian, and you're missing $ \partial_y\partial_x f$. In snappy multi-indices notation , one writes $$\|f\|^2_{H^2} = \sum_{0\le \alpha\le 2} \|D^\alpha f\|^2_{L^2}$$ though at times it is convenient to use the equivalence of norms on finite dimensional vector spaces to write e.g. $$ \pmb|f\pmb|_{H^2} = \sum_{0\le \alpha\le 2} \|D^\alpha f\|_{L^2}$$ Here $\alpha$ runs through the multiindices $(0,0),(1,0),(0,1),(1,1),(2,0),(0,2)$. One easily checks that $\pmb| {\pmb|}{}_{H^2}$ and $\| \|_{H^2}$ are equivalent norms on $ H^2$.
Regarding your second choice for the norm, which I'll write $|\!|\!|f|\!|\!|$, if you add this remaining Hessian term and consider a function $g\in C^2(\mathbb R)$ and let $f$ be defined by $$f(x,y) = g(x-y)$$ then $f\in L^2([0,1]^2)$, and $$ \partial_x f + \partial_y f = 0$$ $$ \partial_{x}^2 f + 2\partial_x\partial_y f + \partial_y^2 f = 0$$ I think it should be possible to use this to create a function $f\notin H^2$ such that $|\!|\!|f|\!|\!| = \|f\|_{L^2}$.