For $n \ge 1$, define functions $f_n$ on $[0,\infty)$ by
$$f_n (x) =
\begin{cases}
e^{-x} &\quad\text{for}\quad 0 \le x \le n\\
e^{-2n} (e^n + n – x) &\quad\text{for}\quad n \le x \le n + e^n \\
0 &\quad\text{for}\quad x\ge n + e^n.
\end{cases} $$
Find the pointwise limit $f$ of $f_n$. Show that the convergence is uniform on $[0,\infty)$.
Would the $\lim_{n\to\infty}f_n=e^{-x}$ and it is uniform since $f=e^{-x} \forall x\in[0,\infty).$
Best Answer
For any $x\in[0,\infty)$, there is always $N_x\in\mathbb{N}$ s.t. $N_x>x$, thus we have $f_n(x)=e^{-x},\forall n\geq N_x$, which means
$$\lim_{n\to\infty}f_n(x)=e^{-x},\quad\forall x\in[0,\infty)$$
To show that the convergence is uniform, it suffices to show that
$$\lim_{n\to\infty}\sup_{x\geq 0}|f_n(x)-f(x)|=0$$
For any $n\in\mathbb{N}$, there are 3 cases
$$|f_n(x)-f(x)|=|e^{-x}-e^{-x}|=0,\quad x\leq n$$
Thus we can see that
$$\sup_{x\geq 0}|f_n(x)-f(x)|\leq\max\{2e^{-n}+ne^{-2n},e^{-n}\}\to 0$$ as $n\to\infty$