Determining if the following sequence of functions is uniformly convergent

Question

Hi Guys was attempting this question and was wondering if I was doing the question correctly?

Determine whether or not the sequence of functions is uniformly convergent:-

$$f_n:[0,2]\to \mathbb{R}$$ $$f_n(x) = 2^{-n}x^ne^2$$

Checking point wise convergence first

$$\lim_{n\to \infty}f_n(x) = \lim_{n\to \infty}2^{-n}x^ne^2$$

Taking the Limit as n $\to \infty$ gives the following

$$\lim_{n\to \infty}f_n(x) = \lim_{n\to \infty}2^{-n}x^ne^2$$

$$\implies e^2 \lim_{n\to \infty}2^{-n}x^n$$

substituting $n \to \infty$ gives

$$e^22^{-\infty}x^\infty = 0$$

Therefore $$\lim_{n\to \infty}f_n(x) = 0 = f(x)$$

$$M_n = sup|f_n(x)-f(x)|,x\in \mathbb{R}$$ $$M_n = sup|2^{-n}x^ne^2-0| \leq 2^{-n}2^ne^2 ,x\in \mathbb{R}$$

$$(2)^{-n}(2)^ne^2\to e^2 = 7.389$$

Therefore since $f_n(x) \to0 ,n \to \infty$ $$2^{-n}2^ne^2 \to e^2 = 7.389$$

This function does not converge uniformly.

Can anyone tell me if i have done this correctly as it relates tot the evaluation of the question and the procedure for justifying if the sequences of function is uniformly convergent.

Answer 1

The sequence converges pointwise to $$ f(x) = \begin{cases}0,& 0\leqslant x < 2\\ e^2,& x=2, \end{cases} $$ so it does not converge uniformly. Remember that if a sequence of continuous functions converges uniformly, then the limit function is continuous. This limit function is discontinuous, so the sequence does not converge uniformly.

Answer 2

A fact: given $(f_n\colon [a,b] \to \Bbb R)_1^\infty \subseteq \mathcal C [a,b]$, if $f_n \rightrightarrows f [n \to \infty, x \in [a,b]]$ [i.e., converges to $f$ uniformly on $[a,b]$], then $f$ shall be continuous [Proof omitted, hint: uniform continuity]. Compute the pointwise limit: $$ f(x) = \lim_{n \to \infty} f_n(x) = \begin{cases} 0, & x \in [0, 2), \\ \mathrm e^2, & x = 2, \end{cases} $$ which is discontinuous at $x = 2$, hence the convergence cannot be uniform.

To use the definition, note that $x = 2$ would be problematic, so for $f_n (x)= (x/2)^n \mathrm e^2$, consider $$ \sup _{x \in [0,2]} \vert f_n - f \vert \geqslant \vert f_n (2 - 2n^{-1}) - f(2-2n^{-1})\vert = (1 - 1/n)^n \mathrm e^2 \xrightarrow {n \to +\infty} \mathrm e^1 \neq 0, $$ hence the convergence is not uniform.