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 $f=e^{-x} \forall x\in[0,\infty).$

  • $\begingroup$ You are right that $f_n\to e^{-x}$ pointwise (to prove this fix $x$ and choose $N > x$ - what can you say about $f_n(x)$ for $n>N$). To show that the convergence is uniform you need to find the maximum deviation from the limit function: $\max_{x\in[0,\infty)} |f_n(x) - f(x)|$ (for this I get $e^{-n-1}$) and show that this goes to $0$ when $n\to \infty$. $\endgroup$ – Winther Mar 12 '15 at 23:13

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

  • If $x\leq n$, then

$$|f_n(x)-f(x)|=|e^{-x}-e^{-x}|=0,\quad x\leq n$$

  • If $n<x\leq n+e^n$, then $$|f_n(x)-f(x)|=|e^{-2n}(e^n+n-x)-e^{-x}|\leq e^{-2n}(e^n+n-x)+e^{-x}\leq e^{-2n}(e^n+n)+e^{-n}=2e^{-n}+ne^{-2n}$$
  • If $x>n+e^n$, then $$|f_n(x)-f(x)|=e^{-x}\leq e^{-n-e^n}\leq e^{-n}$$

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 $n\to\infty$


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