Suppose $f$ and $k$ are Lebesgue measurable functions $\mathbf R\to\mathbf R$ and satisfy $k(x)\geq 0$ for all $x$, and $\int_{\mathbf R} k(x)\,dx=1$. Also suppose $f$ has bounded variation, and $f(-\infty)=0$ and $f(\infty)=1$. Let $$(k*f)(x) = \int_{-\infty}^{\infty}k(x-y)f(y)dy$$ denote the convolution of $k$ and $f$.
Let $TP(f)$ denote the number of turning points of a measurable function. We can define turning points like follows: let $P=\{a=x_0,x_1,\dots,x_n=b\}$ be a partition of $\mathbf R$. Then we can define $$ D(f; P) = \left\{ f(x_1)-f(x_0),\dots, f(x_n)-f(x_{n-1}) \right\}$$
$$ TP(f) = \sup_P (\# \text{sign changes of }D(f; P))$$
My question: When does $k*f$ have more turning points than $f$? Equivalently, under what conditions will $k*f$ have at most as many turnings points as $f$?
Since $f$ has bounded variation we know that $TP(f)$ is at most countably infinite. But consider an example like the Gaussian kernel, especially in the context of heat flow. It is intuitively obvious that heat flow will not create new turning points. I cannot seem to prove this in the language of functional analysis.
It is easy to show $(k*f)(-\infty)=0$ and $(k*f)(\infty)=1$, and also that the variation of $k*f$ is not greater than the variation of $f$.
ALso, if $f$ has no turning points then neither does $k*f$, because if $f$ is monotone then the variation of $f$ is $1$, and if $k*f$ has a turning point then its variation is strictly greater than $1$, a contradiction.