Let $\eta_{\epsilon}$ be a standard mollifier. For $f\in L^{\infty}(\mathbb{R})$ we have $|f^{\epsilon}(x)|\leq \int\limits_{R}|f(x-y)|\eta_{\epsilon}(y)dy \leq ||f||_{\infty},$ which implies $||f^{\epsilon}||_{\infty} \leq ||f||_{\infty}.$
Thus convolution does not increase the $L^{\infty}$ norm.
Similarly, if $TV(f)<\infty$ do we have $TV(f^{\epsilon}) \leq TV(f)?$ If not, can we get $TV(f^{\epsilon}) \leq C TV(f)?$ for some $C$ independent of $\epsilon?$
Defintion: Total variaton and Functions of bounded variation
Let $f: \mathbb{R} \rightarrow \mathbb{R}$ and Let $ \mathcal{P} =\left\{P=\{ x_0, \dots , x_{n_P}\} \mid P\text{ is a partition of } \mathbb{R} \text{ satisfying } x_i\leq x_{i+1}\text{ for } 0\leq i\leq n_P-1 \right\} $
Total variation of $f$ is defined by $TV(f)=\sup_{P \in \mathcal{P}} \sum_{i=0}^{n_{P}-1} | f(x_{i+1})-f(x_i) |.$ A function $f$ is called function of bounded variation if $TV(f)<\infty.$
This is a matter of chasing definitions: Fix $x_0<x_1<\ldots<x_n$ a "partition" of $\mathbb{R}$, then \begin{equation} \begin{split} \sum_{j=1}^n |f_\varepsilon(x_j)-f_\varepsilon(x_{j-1})| & = \sum_{j=1}^n \left| \int_{\mathbb{R}} [f(x_j-y)-f(x_{j-1}-y)]\eta_\varepsilon(y)\, dy\right| \\ & \leq \int_\mathbb{R} \left(\sum_{j=1}^n |f(x_j-y) - f(x_{j-1}-y)|\right)\eta_\varepsilon(y)\, dy. \end{split} \end{equation} Now notice that, for fixed $y\in \mathbb{R}$, we can consider the partition $x_0-y<x_1-y<\ldots<x_n-y$ and so $$ \sum_{j=1}^n |f(x_j-y) - f(x_{j-1}-y)| \leq TV(f). $$ Since $\int_\mathbb{R}\eta_\varepsilon\, dx =1$ the inequality follows.