Let $(f_n)$ be a sequence of funtions defined by $f_n(x)=e^{-(x-n)^2}$ for $x\in \mathbb R$. Show that $f_n\to 0$ pointwise on $\mathbb R$, uniformly on $[a,b]$ where $a<b$, and not uniformly on $\mathbb R$.
i) Observe that $(x-n)^2=x^2+n^2-2xn$. If $x\geq 0$, then $0<f_{n}(x)\leq e^{-( n^2 - 2xn)}$. If $x < 0$, then $0 < f_n(x) <e^{-(x^2 + n^2)}$. Each case show that $f_n(x)\to 0$ as $n\to \infty$ for all $x\in \mathbb{R}$ by Squeeze theorem.
ii) If $x\in [a,b]$, then $a-n<x-n<b-n$. Looks like I have to check the cases $1\leq a<b$ and $a<b\leq 1$. If $a\geq 1$, then $(a-n)^2<(x-n)^2<(b-n)^2$, and so $e^{-(b-n)^2}<f_n(x)<e^{-(a-n)^2}$. If $b\leq 1$, then $(b-n)^2<(x-n)^2<(a-n)^2$, and so $e^{-(a-n)^2}<f_n(x)<e^{-(b-n)^2}$. Taking $n\to \infty$ in both cases show that $f_n(x)\to 0$ for all $x$ in $[a,b]$ by Squeeze theorem. I am not sure about this approach.
iii) Where do I start?
For each real $x $,
$$\lim_{n\to+\infty}e^{-n^2+2nx+x^2} $$ $$=e^{-\infty}=0$$
on the other hand,
$$f_n(n)=1 \implies $$ $$\sup_{\mathbb R}|f_n (x)-0|\geq 1 \implies $$ the convergence is not uniform at $\mathbb R $.
At $[a,b] $, we have normal convergence because
$\forall x\in [a,b]\;\;\;n-x\geq n-b>0$ for enough great $n $, thus
$$e^{-(x-n)^2}\leq e^{-(n-b)^2}$$ $$\implies \sup_{[a,b]}|f_n(x)-0|\leq e^{-(n-b)^2} $$
and goes to zero.