I am trying to solve the following problem- $$ $$ Given $∈H^2(^2)$, show that it satisfies:
$|(+2)−2(+)+()|≤||$
$ \forall x,y \in R^2$
for some constant $c$. $$ $$ Here, $$ H^{2}(R^2)=\{f:R^{2} \to R: D^{\alpha}f \in L^{2}(R^{2}) \; \forall |\alpha| \leq 2 \} $$ With norm $$ ||f||_{H^2}= ( \sum_{|\alpha| \leq 2} ||D^{\alpha}f||_{2}^{2})^{\frac{1}{2}}. $$ Where $\alpha \in (N \cup \{0\})^{2}$ multi-index and $D^{\alpha}f$ is the weak $\alpha$-derivative of $f$. $$ $$ So far, taking the definition of Fourier transform and it's inverse to be: $$ \hat{f}(\epsilon)=\int_{R^n}f(x)e^{i \epsilon \cdot x}dx \; \text{and} \; \check{f}(x)=\frac{1}{(2 \pi)^n} \int_{R^n}f(\epsilon)e^{-i \epsilon \cdot x} d \epsilon, $$ I have that $$ |f(x+2y)-2f(x+y)+f(x)| = \frac{1}{(2 \pi)^2}\int_{R^2} \hat{f}(\epsilon) \; (e^{-2i \epsilon \cdot y}- 2e^{-i\epsilon \cdot y}+1)\; e^{-i \epsilon \cdot x} \;d \epsilon . $$ By Cauchy-Schwarz inequality, $$ \leq\frac{1}{(2\pi)^2} \; (\int_{R^2} |\hat{f}|^2(1+|\epsilon_{1}|^2+|\epsilon_{2}|^2)^{2} d \epsilon)^{\frac{1}{2}} \;(\int_{R^2}|e^{-2i \epsilon \cdot y}- 2e^{-i\epsilon \cdot y}+1|^{2} \; (1+|\epsilon_{1}|^2+|\epsilon_{2}|^2)^{-2} d\epsilon)^{\frac{1}{2}}. $$ I am aware that $$ (\int_{R^2} |\hat{f}|^2(1+|\epsilon_{1}|^2+|\epsilon_{2}|^2)^{2} d \epsilon) $$ can be bounded by a constant multiple of $||f||_{H^{2}}^{2}$ using Plancherel's formula in $L^{2}$ and the Fourier transforms of derivatives. $$ $$ I am not sure what to about $$ \int_{R^2}|e^{-2i \epsilon \cdot y}- 2e^{-i\epsilon \cdot y}+1|^{2} \; (1+|\epsilon_{1}|^2+|\epsilon_{2}|^2)^{-2} d\epsilon.$$ Any help would be greatly appreciated. Thank you.