Let $\{f_n\}_{n\in\mathbb{N}}$ be a sequence of functions with $f_n:\mathbb{R}\to\mathbb{R}$ defined as: $$f_n(x)=e^{-(x-n)^2}.$$
a) Show that the sequence is pointwise convergent towards zero on $\mathbb{R}$.
b) Show that the sequence is uniformly convergent towards zero on $[a,b]$ where $a<b$.
c) Show that the sequence is not uniformly convergent towards zero on $\mathbb{R}$.
My solutions for a) and c):
a) For all $x\in\mathbb{R}$ we have that: $$f_n(x)=e^{-(x-n)^2}=\frac{1}{e^{(x-n)^2}}\to\frac{1}{e^\infty}=\frac{1}{\infty}=0 \quad\text{for }n\to\infty.$$ Hence $\{f_n\}_{n\in\mathbb{N}}$ is pointwise convergent towards $f(x)=0$.
c) We know that $e^x\geq 1$ for $x\geq 0$, and that $e^x=1\Leftrightarrow x=0$. Since $(x-n)^2\geq 0$ we have that: $$f_n(x)=e^{-(x-n)^2}=\frac{1}{e^{(x-n)^2}}\leq 1,\qquad e^{-(x-n)^2}=1 \quad\Leftrightarrow\quad x=n.$$ This means that $f_n(x)$ reaches its maximum in $f_n(n)=1$. Since $e^x>0$ we have that: \begin{align*} &d_\mathbb{R}(f,f_n)=\sup|f(x)-f_n(x)|=\sup|0-e^{-(x-n)^2}|=\sup (e^{-(x-n)^2})\\ &=\max f_n(x)=1. \end{align*} We then have that: \begin{align*} &\lim_{n\to\infty}d_\mathbb{R}(f,f_n)=\lim_{n\to\infty}1=1\neq 0 \quad\text{for }x\in\mathbb{R}. \end{align*} Since $d_\mathbb{R}(f,f_n)$ does not converge towards zero, the sequence is not uniformly convergent on $\mathbb{R}$.
Can anybody help me with b)?
For part $\text{(b)}$, consider what happens when $a\le x\le b$ and $n\gg b:$ $$ e^{-(x-n)^2} = e^{-((x-b)-(n-b))^2} \le e^{-(n-b)^2} \to 0 \text{ as } n\to\infty. \tag 1 $$ So think about
The way I thought of the inequality in $(1)$ is by thinking about what the graph of $x\mapsto e^{-(x-n)^2}$ looks like. It's "bell curve" centered at $n.$ When $n\gg b$ then $b$ is in one of its very thin tails and the value of the function gets smaller as $x$ gets farther from $n;$ thus is smaller everywhere else in $[a,b]$ than at $b.$