I have a pretty basic doubt regarding Sobolev Space.
Suppose $v \in H^k(\mathbb{R}^n) := \{ v \in L^2(\mathbb{R}^n), D^{\alpha}(v) \in L^2(\mathbb{R}^n) \text{ for } \alpha \leq k \}$, where the derivative is taken in the distribution sense, so $v$ needn't be differentiable. Then how can we conclude using Plancherel that $||D^\alpha v||_{L^2} = || \zeta^\alpha \hat{v}||_{L^2}$?
I am aware that for $C^k \cap L^2$ functions we have $\widehat{D^\alpha v} = (2 \pi i \zeta)^\alpha \hat{v}$, and using Plancherel then we have $||\hat{v}||_{L^2} = ||v||_{L^2}$ since the Fourier transformation is an isometry on $L^2$, but my confusion stems from the fact that $v$ is not necessarily differentiable.
Best Answer
Let $v\in H^k$ and let $v_n = \phi_n*v$ where $\phi_t(x)=t^{-n}\phi(x/t)$ is a $C^\infty_c$ mollifier with $\int \phi=\int \phi_t = 1$. Then it is "standard" that $v_t \in C^\infty \cap H^k$ , and $v_t \to v $ in $H^k$ as $t\to 0$, i.e. $D^\alpha v_t \to D^\alpha v$ in $L^2$ norm.
Now by what you know, $$ \|D^\alpha v_n \|_{L^2} = \|\zeta^\alpha \hat v_n\|_{L^2}= \|\zeta^\alpha \widehat{\phi_t}\hat v \|_{L^2}$$ Observe that $\widehat{\phi_t}(\zeta) = \hat \phi(t\zeta)\to \hat \phi(0) = 1 $ pointwise, with the dominating function $$|\zeta^\alpha \widehat{\phi_t}\hat v |^2 \le \|\widehat{\phi}\|^2_{L^\infty} |\zeta|^{2\alpha} |\hat v|^2 \in L^1.$$ So DCT allows us to send $t\to 0$ and obtain the result.