Property of smooth functions on the adeles Let $k$ be a number field,  $\mathbb A$ the ring of adeles of $k$, $\mathbb A_f$ the finite adeles, and $\mathbb A_{\infty}$ the infinite adeles.  
Let $\phi: \mathbb A = \mathbb A_{\infty} \times \mathbb A_f \rightarrow \mathbb C$ be a continuous function which is smooth in the first variable and locally constant in the second variable.  Some authors call such a function smooth.  
Is it the case that for every $x \in \mathbb A$, there exists an open neighborhood $U$ of $x$, and an open compact subgroup $H$ of $\mathbb A_f$, such that
$$\phi(x'+h) = \phi(x')$$
for all $x' \in U, h \in H$?
This seems to be the claim in an answer to one of my questions on Mathoverflow, which I am trying to understand.
My attempt:
Lemma 1: For every $x \in \mathbb A$, there exists an open compact subgroup $H$ of $\mathbb A_f$ such that $\phi(x+h) = \phi(x)$ for all $h \in H$.
The problem becomes whether in the Lemma we can choose $H$ uniformly for $x'$ in a sufficiently small neighborhood of $x$.  
Proof of Lemma 1: Let $x = (x_1,x_2) \in \mathbb A = \mathbb A_{\infty} \times \mathbb A_f$.  There exists an open neighborhood $V \subset \mathbb A_f$ of $x_2$ such that $$\phi(x_1,y') = \phi(x_1,x_2)$$ for all $y' \in V$.  Let $H$ be an open compact subgroup of $\mathbb A_f$ which is contained in the neighborhood $V - x_2$ of the identity.  Then for all $h \in H$, we have $h + x_2 \in V$, and so 
$$\phi(x + h) = \phi((x_1,x_2) + (0,h)) = \phi(x_1,x_2+h) = \phi(x_1,x_2) = \phi(x)$$
$\blacksquare$
To prove what I want, I will definitely need to make use of the fact that $\phi$ is continuous.  So far I haven't been successful.
 A: Say a function $f:X\times Y\to Z$ is [locally] [constant in the second variable] if each $(x,y)$ has a neighborhood $U\times V\subset X\times Y$ of $(x,y)$ such that $f(u,v)=f(u,v')$ for all $u\in U$ and $v,v'$ in $V.$
Say a function $f:X\times Y\to Z$ is [locally constant] [in the second variable] if for each $(x,y)$ there is a neighborhood $V\subset Y$ of $y$ such that $f(x,v)=f(x,v')$ for all $v,v'$ in $V.$
It sounds like you're using the second meaning to define smooth functions. I believe the first definition is right and the second definition is wrong. At least from a quick Google of "smooth adelic function". ☺
Here's a counterexample that I hope answers your question.
Let $\psi:\mathbb R\to\mathbb R$ be a bump function such that $\psi(x)=1$ for $|x|\leq 1$ and $\psi(x)=0$ for $|x|\geq 2.$ Define a function $\phi:\mathbb A_\mathbb{Q}\to\mathbb R$ by
$$\phi(x_\infty,x_2,\dots)=\begin{cases}
x_\infty\cdot \psi(x_\infty/|x_2|_2)&(x_2\neq 0)\\
x_\infty&(x_2=0)
\end{cases}$$
where $|x_2|_2$ is the usual $2$-adic norm.
To see this is continuous, define $N=\{0\}\cup\{2^{-v}\mid v\in\mathbb Z\}.$
If you believe that the map $\mathbb A_\mathbb{Q}\to\mathbb R\times N$ given by $(x_\infty,x_2,\dots)\mapsto(x_\infty,|x_2|_2)$ is continuous, and that the map $\mathbb R\times N\to\mathbb R$ defined by $(x,y)\mapsto \begin{cases}x\cdot \psi(x/y)&(y\neq 0)\\x&(y=0)\end{cases}$ is continuous, then you should believe that their composition $\phi$ is also continuous.
This function is [locally constant] [in the second variable]: this is hopefully clear for $x_2\neq 0,$ while to check the property at $x_2=0$ we need that $\phi(0,x_2,\dots)=0$ for all $x_2,$ and for each $x_\infty\neq 0$ we have $\phi(x_\infty,x_2,\dots)=0$ whenever $|x_2|_2\leq \tfrac12 |x_\infty|.$
But there are no $U,H$ with $U$ an open neighborhood of $0$ and $H$ an open subgroup of $\mathbb Q_2$ such that $\phi(u,h,\dots)=\phi(u,0,\dots)$ for all $u\in U$ and $h\in H.$ Just pick a non-zero $h\in H$ and take $0<u\leq |h|_2$ to get $\phi(u,h,\dots)=u\neq 0=\phi(u,0,\dots).$
