The Fourier Uncertainty Principle:
If $f\in \mathscr L^2(\mathbb R)$ and $xf(x),\xi\hat f(\xi)\in \mathscr L^2(\mathbb R)$ then
\begin{equation}\left\|xf(x)\right\|_2\big\|\xi\hat f(\xi)\big\|_2\ge \frac 12\left\|f\right\|_2\big\|\hat f\big\|_2\end{equation}
The proof I found here (page 9) begins with finding $f'$. But why is $f$ even differentiable?
I decided to argue as follows: If $f$ is in the Schwartz space it is infinitely many times differentiable and the inequality follows as in the proof I linked. If it is in $\mathscr L^2$ then by density, we have a sequence of Schwartz functions $f_n$ that converge point-wise a.e. and in the 2 norm to $f$ satisfying
\begin{equation}\left\|xf_n(x)\right\|_2\big\|\xi\hat f_n(\xi)\big\|_2\ge \frac 12\left\|f_n\right\|_2\big\|\hat f_n\big\|_2\end{equation}
Now the right hand side tends to the desired
\begin{equation}\frac 12\left\|f\right\|_2\big\|\hat f\big\|_2\end{equation}
but where does the left hand side converge to? If $x^2f_n^2(x)$ was bounded by an integrable function I could use the Dominated Convergence Theorem, but I don't have that. Any ideas?
EDIT: As User98130 said we can for $f\in \mathscr L^2(\mathbb R)$ we can consider $g(x)=f(x)(1+x^2)^{1/2}\in \mathscr L^2$ and approximate it by functions $g_n\in \mathscr C^{\infty}_c$ in the $\mathscr L^2$ norm. If $f_n(x)=g_n(x)(1+x^2)^{-1/2}$, then $f_n\in \mathscr C^{\infty}_c\subseteq \mathcal S$ and $f_n\to f$, $xf_n(x)\to xf(x)$ in the 2-norm. This is because
\begin{equation}\left\|f_n-f\right\|_2,\left\|xf_n(x)-xf(x)\right\|_2\le\big\| (1+x^2)^{1/2}(f_n(x)-f(x))\big\|_2=\left\|g_n-g\right\|_2\to 0\end{equation}
But how do we prove that $\xi\hat f_n(\xi)\to \xi\hat f(\xi)$ in the $\mathscr L^2$ norm?
Best Answer
Edit: the approximation suggested below works only in the $x$ variable.
I suggest working with $f$ and its $\mathscr L^2$ derivative $f'$ directly, without an approximation. The $\mathscr L^2$ derivative comes from the inverse Fourier transform of $\xi \hat f(\xi)$. There is a step-by-step guide to the proof in Real Analysis by Folland, page 255.
Not only $f$, but also $g:=(1+x^2)^{1/2}f$ is in $L^2$. Approximate $g$ by smooth functions $g_n$ in $L^2$. Define $f_n=g_n/(1+x^2)^{1/2}$. Observe that $f_n\to f$ in $L^2$ and $xf_n\to xf$ in $L^2$. (Indeed, multiplication by an $L^\infty$ function is a continuous linear operator on $L^2$.)
By the way, you are right that $f$ does not have to be classically differentiable. Under the stated conditions, it belongs to the Sobolev space $W^{1,2}$, meaning it has a weak derivative which is in $L^2$. Perhaps the author had the weak derivative in mind when writing that proof.