Approximating piecewise linear function

Try a convolution with a function of small width and great height (: Richard P. Feynman in 'Space-Time Approach to Quantum Electrodynamics'). Name this function $\delta(x)$ (not quite by coincidence). The simplest one is $\,\delta(x) = B(x/\sigma)/\sigma\,$ where $\sigma \rightarrow 0$ and: $$ B(t) = \left\{ \begin{array}{lll} 0 & \mbox{for} & t \le -1/2 \\ 1 & \mbox{for} & -1/2 \le t \le +1/2 \\ 0 & \mbox{for} & +1/2 \le t \end{array} \right. $$ The geometry of $\,\delta(x)\,$ is a rectangle with height $1/\sigma$ and width $\sigma$ , resulting in an area $1$, thus establishing that the function $\,\delta(x)\,$ is normed. Now define: $$ \overline{f}(x) = \int_{-\infty}^{+\infty} \delta(x-t)\,f(t)\,dt = \frac{1}{\sigma} \int_{x-\sigma/2}^{x+\sigma/2} f(t)\,dt $$ This is what we find: $$ \overline{f}(x) = \left\{ \begin{array}{lll} x/x_s & \mbox{for} & x \le x_s-\sigma/2 \\ P(x) & \mbox{for} & x_s-\sigma/2 \le x \le x_s+\sigma/2 \\ (1-x)/(1-x_s) & \mbox{for} & x_s+\sigma/2 < x \end{array} \right. $$ Where the (piece of a) parabola $\,P(x)\,$ is defined as: $$ P(x) = \frac{1}{\sigma}\int_{x-\sigma/2}^{x_s} \frac{t}{x_s}\,dt + \frac{1}{\sigma}\int_{x_s}^{x+\sigma/2} \frac{1-t}{1-x_s}\,dt = \\ \frac {4\,{{\it x_s}}^{2}+4\,{x}^{2}-4\,x\sigma+8\,x\sigma\,{ \it x_s}+{\sigma}^{2}-8\,{\it x_s}\,x-4\,\sigma\,{\it x_s}}{8\sigma \,{\it x_s}\, \left( -1+{\it x_s} \right) } $$ Thus, with the simplest convolution kernel, the result is a piecewise analytic function. The original $\,f(x)\,$ is in black, the convoluted one $\,\overline{f}(x)\,$ is in $\color{red}{\mbox{red}}$, $\,x_s\,$ is chosen at random and $\,\sigma = 1/10\,$ in the renditions depicted here:

A nice property of the convolution is that $\overline{f}(x)$ inherits the smoothness of the kernel. One might alternatively define that kernel as a Gauss function. Thus: $$ \overline{f}(x) = \frac{1}{\sigma\sqrt{2\pi}}\int_{-\infty}^{+\infty} e^{-\frac{1}{2}(x-t)^2/\sigma^2} f(t)\,dt $$ This results in a not so simple expression, but the advantage is that $\overline{f}(x)$ is truly analytic. I leave this as an exercise to the reader. Visualization does not differ much from the ones given.