What conditions are necessary for approximation of identity to converge to the function?

Question

Let $\Phi:\mathbb{R}^n \times (0,\infty) \rightarrow (0,\infty)$ be defined as: \begin{align*} \Phi(x,t) = \frac{1}{(\sqrt{4\pi t})^n}e^{-||x||^2/4t} \end{align*} Then denoting for a fixed $t$, $\Phi_t: \mathbb{R}^n \rightarrow (0,\infty) $ as $\Phi_t(x) = \Phi(x,t)$ we have the following properties: \begin{align*} \int_{\mathbb{R}^n}\Phi_t\ dx &= 1\\ \lim_{t\rightarrow0}\Phi_t(x) &= 0, x\not=0 \end{align*} If $f \in L^1(\mathbb{R}^n) \cap L^{\infty}(\mathbb{R}^n)$, then suppose we have: \begin{align*} \lim_{t\rightarrow 0}f*\Phi_t(x) = f(x), \forall x \in \mathbb{R}^n \end{align*} (I have proved that if $f$ is continuous at $x$, then the limit holds true) What can we say about $f$? Is it continuous $a.e$?

Answer 1

A function in $L^{1}(\mathbb{R}^{n})$ is not continuous a.e in general. For example, take $n=1$ and the function $f = \mathbb{1}_{\mathbb{Q} \cap [0,1]}$ the function $f$ is nowhere continuous. Despite this fact, $\varphi_{t}*f \to f$ in $L^{1}([0,1])$ as $t \to 0.$ Here is a reminder of the proof $$ \int_{\mathbb{R}}\left\vert \int_\mathbb{R} f(x-y)\,\varphi_{t}(y)\,dy - f(x) \right\vert \,dx = \int_{\mathbb{R}}\left\vert \int_\mathbb{R} \bigl( f(x-y)-f(x) \bigr)\,\varphi_{t}(y)\,dy \right\vert \,dx \\ \leq \int_{\mathbb{R}} \int_\mathbb{R} \left\vert \bigl( f(x-y)-f(x) \bigr)\,\varphi_{t}(y) \right\vert \,dy \,dx = \int_{\mathbb{R}} \,\varphi_{t}(y) \, \int_\mathbb{R} \left\vert \bigl( f(x-y)-f(x) \bigr) \right\vert \,dx \,dy\\ =\int_{\varphi} \frac{1}{\sqrt{t}}\varphi_{1}(\frac{y}{\sqrt{t}})\, \int_\mathbb{R} \left\vert \bigl( f(x-y)-f(x) \bigr) \right\vert \,dx \,dy \\ = \int_{\varphi} \varphi_{1}(y)\, \int_\mathbb{R} \left\vert \bigl( f(x-y\cdot \sqrt{t})-f(x) \bigr) \right\vert \,dx \,dy. $$ But the translation operator (for further clarifications look at this post Translation operator and continuity ) is continuous in $L^{1}$; i.e, $$ \lim_{a \to 0} \int_{\mathbb{R}} \vert f(x-a)-f(x) \vert = 0. $$ You then apply this result with $a= y\sqrt{t}$ and you use the dominated convergence theorem. As a dominating function you can take $$ \varphi_{1}(y)\cdot 2\Vert f \Vert_{L^{1}} $$ because $$ \varphi_{1}(y)\, \int_\mathbb{R} \left\vert \bigl( f(x-y\cdot \sqrt{t})-f(x) \bigr) \right\vert \,dx \leq \varphi_{1}(y)\, \bigl( \int_\mathbb{R} \left\vert f(x-\sqrt{t}\,y) \right\vert dx + \int_\mathbb{R} \left\vert f(x) \right\vert dx \bigr) \\ = 2 \cdot \varphi_{1}(y)\cdot \Vert f \Vert_{L^{1}} $$ Moreover, if you take a sequence of functions from $\{ \varphi_{t} \}_{t \geq 0};$ for example, $\{ \varphi_{\frac{1}{n}} \}_{n \in \mathbb{N}}$ then you have a subsequence that converge almost everywhere to $f.$ In fact, every sequence $\{ f_{n} \}_{n \in \mathbb{N}}$ which converge in $L^{1}$ to a function $f$ also admits a subsquence convergent almost everywhere to $f.$ Despite these two types of convergence we do not have that $f$ is continuous almost everywhere.