Pointwise convergence and uniformly convergence. Is this correct? Let $F$ be a $C^1$ function defined in $\mathbb R$ and satisfies $F(0)>0$ and if $|x|>1$ then $F(x)=0$
Then, I think $g_n (x):=F(x-n)$ can converge pointwisely to $g(x)=0$ but don't think it can converge uniformly. Is this correct? Anyway I don't know how to prove it.
And also how about $h_n (x):=F(x-\frac{1}{n})$? Does this function converge uniformly? or pointwisely?
 A: Yes, the sequence $(g_n)$ converges pointwisely, but not uniformly, to $g$.
Namely, for every fixed $x\in\mathbb{R}$, as soon an $n > |x| + 1$ you have that
$$
g_n(x) = F(x-n) = 0
\qquad (n > |x| + 1),
$$
since $|x-n| > n - |x| > 1$, so that the point $x-n$ does not belong to the support of $F$.
This proves the pointwise convergence.
On the other hand,
$$
\sup_{x\in\mathbb{R}} |g_n(x) - g(x)| = 
\sup_{x\in\mathbb{R}} |F(x-n)| \geq F(0) > 0,
$$
hence the convergence cannot be uniform.
For the sequence $(h_n)$, you can prove that $(h_n)$ converges uniformly to $F$.
The pointwise convergence is almost trivial: since $F$ is continuous, you have that
$$
\lim_{n\to +\infty} h_n(x) = \lim_{n\to+\infty} F\left(x - \frac{1}{n}\right) = F(x),
\qquad \forall x\in\mathbb{R}.
$$
To prove the uniform convergence it is enough to observe that $F$ is uniformly continuous. Hence, for every $\epsilon>0$, there exists $\delta > 0$ such that
$$
|F(x) - F(y)| < \epsilon
\qquad \forall x,y\in\mathbb{R}, \ |x-y| < \delta.
$$
Hence, if $n > 1/\delta$, for every $x\in\mathbb{R}$ one has
$|(x-1/n) - x| = 1/n < \delta$, so that
$$
\sup_{x\in\mathbb{R}} |h_n(x) - F(x)|
= \sup_{x\in\mathbb{R}} \left|F\left(x - \frac{1}{n}\right) - F(x)\right|
\leq \epsilon
\qquad
(n > 1/\delta).
$$
Since $\epsilon > 0$ is arbitrary, this proves the uniform convergence of the sequence.
Remark: in your assumptions you have $F\in C^1$ (which is not needed: continuity suffices).
This extra regularity can be used to prove the last estimate without the explicit use of uniform continuity.
Indeed, since $F'$ is continuous and $F'(x) = 0$ for $|x| \geq 1$, there exists a constant $L>0$ such that $|F'(x)|\leq L$ for every $x\in\mathbb{R}$.
Hence, using the mean value theorem,
$$
|F(x-1/n) - F(x)| \leq L \frac{1}{n}\,,
\qquad \forall x\in\mathbb{R}.
$$
