Lower bound on $L^2$-norm of derivative

Question

Suppose I want to approximate the indicator function $f:= \chi_{[0,1]}$ on $\mathbb{R}$ with mollifications (choose the standard mollifier $\phi$). Then we know that with respect to $L^2$ we have $\phi_\epsilon * f \to f$ as $\epsilon\to 0$. What should happen for the derivatives $\phi'_\epsilon * f$ is that they do not converge $L^2$ (the closer we get, the steeper the slope should get at say $x=0$). However I am having difficulties proving this with an explicit lower bound on $\lvert\lvert\phi'_\epsilon * f\rvert\rvert_2$, that explodes as $\epsilon\to 0$. Since we get a factor of $1/\epsilon$ from $\phi'_\epsilon$, i suppose this is the divergence rate but I do not have a precise argument for that. Can someone help out with a reference here? I am sure this was done before.

Answer 1

Heuristically speaking, the derivative of $f$ is a sum of Dirac deltas, the regularization of which stays bounded in $L^1$ and grows like $\epsilon^{-1}$ in $L^\infty$. Thus, by Hölder’s inequality, you expect the $L^2$ norm of $\phi_\epsilon’ *f$ to grow like $1/\sqrt \epsilon$.

Explicitly, assume that $\phi_1$ has has support in $[-1,1]$. Then by algebraic manipulations it holds $$ [\phi_\epsilon’ *f](x)= \phi_\epsilon(x)-\phi_\epsilon(x-1),$$ where the two addends on the right-hand side have disjoint supports assuming $\epsilon<1$. This means that $$ || \phi_\epsilon’ *f||_{L^2}^2=|| \phi_\epsilon||_{L^2}^2+|| \phi_\epsilon(\cdot-1)||_{L^2}^2=2|| \phi_\epsilon||_{L^2}^2. $$ Then one has to compute the quantity on the right hand side: $$ || \phi_\epsilon||_{L^2}^2=\int_\mathbb R|\epsilon^{-1}\phi(\epsilon^{-1}t)|^2dt= \epsilon^{-1}\int_\mathbb R|\phi(s)|^2ds\sim \epsilon^{-1}, $$ so that the $L^2$ norm of $\phi_\epsilon’ *f$ grows like $1/\sqrt \epsilon$.

Edit. If you remember my answer to this question, you can also think of this on the Fourier side and have very precise information: the Fourier transform of $f$ is bounded by $1/|\xi|$, so that the Fourier transform of $\frac{d}{dx}f$ is bounded. By Plancherel’s theorem, $||f_\epsilon’||_{L^2}=|||\xi|\widehat {f_\epsilon}||_{L^2}=|||\xi|\widehat f ( \cdot)\widehat \phi(\epsilon\cdot) ||_{L^2}\leq|||\xi|\widehat f||_{L^\infty}||\widehat\phi(\epsilon\cdot)||_{L^2},$ and you can compute the latter norm explicitly.