Maximum Number of modes of $V=U+Z$ where $Z$ standard normal and $|U|\le a$ Let $f_V$ be a pdf of  random variable $V$ where
\begin{align}
V=U+Z
\end{align}
and where $U$ and $Z$ are independent and $Z$ is Gaussian. Moreover, suppose that $|U| \le A$.
Can we find the upper bound on the number of modes of $V$?   
Note that the the pdf $V$ is given by 
\begin{align}
f_V(v)=E\left[\frac{1}{\sqrt{2\pi}}e^{-\frac{(v-U)^2}{2}}\right]
\end{align}
I am actually not interested in a very tight bound. I more interested in the methods that can be used to produce such bounds. 
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The pdf $f_V$ is a mixture of the family $\F_A:=(f_a)_{a\in[-A,A]}$ of the normal pdfs $f_a$ with mean $a$ and variance $1$. By Theorem 6 on p. 2129 of Kemperman, for each natural $s$, the pdf $f_V$ will have at most $s$ modes (or, more precisely, modal intervals) iff any mixture of any $2s$ members of the family $\F_A$ has at most $s$ modal intervals. 
This theorem is a generalization of Theorem 4 on p. 2128 of Kemperman, which implies that all mixtures of a family of pdfs are unimodal iff any mixture of any two members of the family is unimodal. 
Further, by Remark 1 on p. 2133 of Kemperman, our pdf $f_V$ will be unimodal for all admissible $U$ iff for all $a,b,x$ such that $-A\le a<x<b\le A$ we have
\begin{equation}
 f'_a(x)f''_b(x)\ge f'_b(x)f''_a(x), 
\end{equation}
which can be rewritten as $(a-x)(b-x)\ge-1$ for all such $a,b,x$, which is equivalent to $A\le1$. 
The case of more modes than $1$ appears much more difficult analytically. 
