Find $f(x)$ from $f_n(x)$.

Question

So I am learning about uniform convergence, pointwise convergence. In order to show uniform convergence, you must show $|f_n(x) - f(x)| = 0$, but I am really confused on how to find f(x) especially from the above function.

Thanks in advance.

Answer 1

For any $x \ge 0$, you have $f_n(x) = e^{-x}$ for all $n \ge x$. Thus $f_n(x) \to e^{-x}$ by definition.

You can estimate the difference between $f_n(x)$ and $e^{-x}$ in a piecewise manner:

• if $0 \le x \le n$ then $|f_n(x) - e^{-x}| = 0$
• if $n \le x \le n + e^n$ then $ 0 \le e^n + n - x \le e^n$ so that $0 \le e^{-2n}(e^n + n - x) \le e^{-n}$. It follows that $|f_n(x) - e^{-x}| \le e^{-n}.$
• If $x > n + e^n$ then $|f_n(x) - e^{-x}| = e^{-x} \le e^{-n}$.

Consequently $\sup_{x \ge 0} |f_n(x) - e^{-x}| \le e^{-n}$ so that $f_n(x) \to e^{-x}$ uniformly.