Let $f\in C^0(S^1\backslash\{p\})\cap L^\infty(S^1)$ for some $p\in S^1$. I guess $$\lim_{\epsilon\rightarrow 0}f_\epsilon(p)=\frac{1}{2}\lim_{x\rightarrow p-}f(x)+\frac{1}{2}\lim_{x\rightarrow p+}f(x)$$ when $f$ is nice enough, where $f_\epsilon$ is $\epsilon-$mollification of $f$.
Question : Is my guess true? If so, what gives sufficient condition for $f$ so the guess is satisfied?
Let $$f_-(x):=\bigg\{\begin{array}{cc} f(x)\quad \text{on}\quad [p-1/2,p]\\ f(-x) \quad\text{on}\quad[p,p+1/2]\end{array},$$$$f_+(x):=\bigg\{\begin{array}{cc} f(-x)\quad \text{on}\quad [p-1/2,p]\\ f(x) \quad\text{on}\quad[p,p+1/2]\end{array}.$$ Then $f_-$ and $f_+$ are even at $p$, and continuous. Therefore, $$f_\epsilon(p)=\int_{(p-\epsilon/2,p+\epsilon/2)}\eta_\epsilon(x-p)f(x)dx=\int_{(p-\epsilon/2,p)}\eta_\epsilon(x-p)f(x)dx+\int_{(p,p+\epsilon/2)}\eta_\epsilon(x-p)f(x)dx=\frac{1}{2}\int_{(p-\epsilon/2,p)}\eta_\epsilon(x-p)f_-(x)dx+\frac{1}{2}\int_{(p,p+\epsilon/2)}\eta_\epsilon(x-p)f_+(x)dx=\frac{1}{2}{f_-}_\epsilon(p)+\frac{1}{2}{f_+}_\epsilon(p)\xrightarrow{\epsilon\rightarrow 0}\frac{1}{2}(f_-(p)+f_+(p))=\frac{1}{2}(\lim_{x\rightarrow p-}f(x)+\lim_{x\rightarrow p+}f(x)).$$