Approximation by convolution with mollifiers

Remarks

Part 1 follows by differentiating under the integral sign, which is allowed because $\varphi_\varepsilon$ has bounded derivatives and compact support. Part 2 holds because $\varphi_\varepsilon(x - y) = 0$ when $\abs{x - y} > \varepsilon$. For parts 3 and 4, write $f_\varepsilon(x) - f(x) = \int \varphi_\varepsilon(y) \big( f(x - y) - f(x) \big) dy$, an average of differences over $\abs{y} \le \varepsilon$. For continuous $f$, uniform continuity on compact sets makes it small. For $f \in L^p$, Minkowski's integral inequality bounds its norm by $\sup_{\abs{y} \le \varepsilon} \norm{f(\cdot - y) - f}_{L^p}$. This tends to $0$ by the continuity of translation in $L^p$. The bound $\norm{f_\varepsilon}_{L^p} \le \norm{f}_{L^p}$ is Young's convolution inequality with $\norm{\varphi_\varepsilon}_{L^1} = 1$.

Consequence: for $1 \le p < \infty$ the bump functions are dense in $L^p(\R^n)$. Indeed, first cut $f$ off to a large ball, then mollify, and use parts 1, 2 and 4.

Part 4 fails for $p = \infty$. If $f$ is the indicator function of $[0, \infty)$ on $\R$, then each $f_\varepsilon$ is continuous, so $\norm{f_\varepsilon - f}_{L^\infty} \ge 1/2$ for all $\varepsilon$.

Going deeper

For a student

This theorem is the workhorse behind density arguments in analysis. To prove an identity or an inequality for all $f \in L^p$ or all $f$ in a Sobolev space, one proves it for smooth functions and passes to the limit. Mollification commutes with weak derivatives, $\partial^\alpha (\varphi_\varepsilon * f) = \varphi_\varepsilon * \partial^\alpha f$, so the same argument shows that smooth functions are dense in $W^{k,p}(\R^n)$. At a Lebesgue point of $f$, $f_\varepsilon(x) \to f(x)$, so the convergence also holds almost everywhere.