When does convolution erase non-monotonicities? Suppose $\phi:\Bbb R\to[0,\beta]$ is a bounded continuous function such that $\phi(-\infty)=0$ and $\phi(\infty)=\beta$.
Assume $\phi$ is non-decreasing except near zero, i.e. there exists $r>0$ such that $\phi$ is non-decreasing on $\Bbb R\setminus[-r,r]$, but it  may be decreasing on subsets of $[-r,r]$. (Assume also that $r$ is minimal.)
Suppose $k$ is a probability density, and I want to take the convolution
$$ (k*\phi)(x) = \int_{-\infty}^{\infty}k(x-y)\phi(y)dy $$
I think that if $r$ is small and the variance of $k$ is large, then $k*\phi$ should be monotone, i.e. the non-monotonicities in $\phi$ were "eroded away" by convolution with $k$. Are there any results in terms of $r$ and the variance of $k$?
 A: Suppose that for all real $x$
$$\phi(x):=f(x):=\begin{cases}
0&\text{ if }x\le-1, \\
 1-|x| &\text{ if } -1<x\le 1, \\
 1-1/x &\text{ if }  x>1  
\end{cases}
$$
and 
$$k(x):=k_b(x):=\frac1{2b}\,1_{|x|<b},
$$
where $b>1/2$. Then for any such $b$ and any real $x$
$$(f*k_b)'(x)=\frac1{2b}\,\int_{-b}^b f'(x-t)\,dt=\frac{f(x+b)-f(x-b)}{2b},
$$
where $f'$ is the almost-everywhere derivative of the absolutely continuous function $f$. So, 
$$(f*k_b)'(b)=\frac{f(2b)-f(0)}{2b}=-\frac1{4b^2}<0. 
$$ 
So, here the convolution does not fully erase the non-monotonicity on the interval $[-1,1]$, however large the variance of $k$ is. 
Here are graphs of $f$ (left) and $f*k_b$ (right, for $b=5$): 

A: Iosif Pinelis' post answers your question as stated.  In case you're thinking about modifying your assumptions to guarantee monotonicity, here are a few observations.
Let $\beta=1$.  Let $\psi=(k * \phi)$ and suppose that $\phi$ is non-decreasing.  Suppose that $\phi$ and $k$ are differentiable (and hence $\psi$ is continuous and differentiable).
Then we can interpret $\psi$ as the CDF of some continuous random variable $X_{\psi}$, $k$ the pdf of some continuous r.v. $X_k$, and $\phi$ the CDF of a r.v. with negative probability (see, e.g., Wikipedia on that).  Then
$$X_k + X_{\phi} = X_{\psi}$$
i.e.,
$$k * \phi = \psi$$
However, we aren't given $k$ or $\psi$.  Suppose we are given only $\phi$.  Then, per Ruzsa and Szekely, we can always find some $k$ and $\psi$ that satisfy the equations above, i.e., for any $\phi$, there is some kernel $k$ that will make $\phi$ non-decreasing.
Of course, this doesn't provide a single $k$ for all $\phi$, but perhaps it's enough for your purposes.
