When does the convergence of the regularization of a function is decreasing?

Question

Hi everyone: Let $\theta(x)$ equal $k\exp\left(-\frac{1}{1-\|x\|^2} \right)$ if $\|x\|<1$, and equal $0$ if $\|x\|\geq1.$ Here $\|\cdot\|$ designates the Euclidean norm in $\mathbb{R}^n$, and the constant $k$ is chosen such that $\int_{\mathbb{R}^n}\theta(t)\,d\lambda(t)=1$. If we set $$\theta_\varepsilon(x)=\frac{1}{\varepsilon^N}\theta\left(\frac{x}{\varepsilon}\right),$$ we obtain a $\mathcal{C}^\infty$ function whose support is the Euclidean ball $B(x,\varepsilon)$, for all $\varepsilon>0$. Then we can define by convolution $$f_\varepsilon(x):= (f\ast \theta_\varepsilon)(x).$$

My question is: if $f$ is only locally integrable, is the convergence of $f_\varepsilon$ to $f$ decreasing, as $\varepsilon$ approaches $0$? Thanks for your reply.

Answer 1

No. If $f_\epsilon(x)>f(x)$, then $-f_\epsilon(x)<-f(x)$, so for any $f$, either $f$ or $-f$ is a counterexample.

The convergence is decreasing when $f$ is subharmonic. This is because $\frac{d}{d\epsilon}f_\epsilon$ is the convolution of $f$ against a radially symmetric function that is negative on small radii and positive for large ones. The fact that the mean of $f$ on a sphere of radius $r$ increases with $r$ implies that $ \frac{d}{d\epsilon}f_\epsilon \ge 0 $.

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I think I caused this confusion by a sloppy answer here; will edit.