Let $f$ be a real valued function and $$\|f\|_{\alpha, \beta, \infty} = \|x^\alpha D^\beta f\|_\infty\\ \|f\|_{\alpha, \beta, 2} = \|x^\alpha D^\beta f\|_{L^2}$$ where $\alpha$ and $\beta$ are multi-indices. I would like to show that $$\|f\|_2 \leq \|(1+x^2)^{-1}\|_2~\|(1+x^2)f\|_\infty \tag{1}$$ and use this result to prove $$\|f\|_{\alpha, \beta, 2} \leq C\big(\|f\|_{\alpha, \beta, \infty} + \|f\|_{\alpha+2, \beta, \infty}\big) \tag{2}$$ where $C$ is some positive constant.
So far, using $f = f(1+x^2)(1+x^2)^{-1}$, I have tried using Holder's inequality and the Cauchy-Schwartz inequality, but in both cases I get extra powers of two, and I cannot figure out how to bound the $L^2$ norm by an $L^\infty$ norm.
Any tips on establishing these inequalities?
We have
\begin{align*} \Vert f \Vert_2 &= \Big( \int \vert f(x) \vert^2 dx\Big)^{1/2} =\Big( \int (1+x^2)^{-2}\vert (1+x^2) f(x) \vert^2 dx\Big)^{1/2} \\ &\leq \Big( \int (1+x^2)^{-2} \Vert (1+y^2) f\Vert_\infty^2 dx\Big)^{1/2} = \Vert (1+y^2) f\Vert_\infty \Big( \int \vert (1+x^2)^{-1} \vert^2 dx\Big)^{1/2} \\ &=\Vert (1+y^2) f\Vert_\infty \Vert (1+x^2)^{-1} \Vert_2. \end{align*} Let's write $h=x^\alpha D^\beta f$, then we have (using the previous computation for $h$, instead of $f$ and changing $y$ back to $x$) \begin{align*} \Vert f\Vert_{\alpha,\beta,2} &= \Vert h \Vert_2 \leq \Vert (1+x^2)^{-1}\Vert_2 \Vert (1+x^2) h\Vert_\infty \\ &\leq \Vert (1+x^2)^{-1}\Vert_2 (\Vert h\Vert_\infty + \Vert x^2 h\Vert_\infty ) \\ &=\Vert (1+x^2)^{-1}\Vert_2 (\Vert x^\alpha D^\beta f \Vert_\infty + \Vert x^{\alpha +2} D^\beta f \Vert_\infty) \\ &= \Vert (1+x^2)^{-1}\Vert_2 (\Vert f \Vert_{\alpha, \beta, \infty}+ \Vert f \Vert_{\alpha +2, \beta, \infty}). \end{align*} Now, if we are integrating over a domain in $\mathbb{R}$, we have $C=\Vert (1+x^2)^{-1}\Vert_2<\infty$ (if we are in higher dimensions, then we also need to be a bit more careful about our indices $\alpha$).