# Proof for uniform convergence of sequence of functions

I was given this problem:

These are my calculations and I'm asking for verification:

Pointwise limit:

$$\lim_{n \to \infty} f_{n}(x) = \lim_{n \to \infty} \frac{x^{2n}}{1+x^{2n}} = \lim_{n \to \infty} \frac{x^{n}}{\frac{1}{x^n}+x^{n}} = 1$$

Uniform convergence:

$$\mid f_{n}(x)-f(x)\mid = \mid f_{n}(x) - 1\mid = \mid\frac{x^{2n}}{1+x^{2n}} -1 \mid= \mid \frac{x^{n}}{\frac{1}{x^n}+x^{n}} - \frac{\frac{1}{x^n}+x^{n}}{\frac{1}{x^n}+x^{n}}\mid = \mid -\frac{\frac{1}{x^n}}{\frac{1}{x^n}+x^{n}}\mid = \frac{\frac{1}{x^n}}{\frac{1}{x^n}+x^{n}} \leq \frac{1}{x^n}$$

Thus:

$$\lim_{n \to \infty} sup\{\mid f_{n}(x)-f(x)\mid : x \in [R, \infty)\} = \lim_{n \to \infty} \frac{1}{x^n} = 0.$$

From this follows that $$f_n(x)$$ is uniform convergent

• I think the last line should be $\lim_{n \to \infty} \sup$ and not $\lim \sup_{n \to \infty}$. – devianceee May 29 at 11:30
• Yeah it should. – Fo Young Areal Lo May 29 at 11:31

You have written "$$|f_n(x) - f(x)| = f_n(x) - 1$$". That is not correct. The RHS is negative.

To do it, an easy way is to note that for every $$n \in \Bbb N$$, the function $$f_n$$ is increasing on $$[R, \infty)$$. And also that $$f_n(x) < 1$$ for every $$n \in \Bbb N$$ and $$x \le R$$.

Now, you know that $$\lim_{n\to\infty} f_n(R) = 1.$$ So, given any $$\epsilon > 0$$, choose $$N \in \Bbb N$$ such that $$|f_n(R) - 1| < \epsilon$$ for all $$n \ge N$$.
Since $$f_n(R) \le f_n(x) < 1$$ for all $$n \in \Bbb N$$ and $$x > R$$, it follows that $$|f_n(x) - 1| < \epsilon,$$ for all $$x > R$$ and $$n \in \Bbb N$$, as desired.

EDIT: This is after the post was edited.
You write

$$\lim_{n \to \infty} \sup\{\mid f_{n}(x)-f(x)\mid : x \in [R, \infty)\} = \lim_{n \to \infty} \frac{1}{\color{#FF0000}x^n} = 0.$$

It should actually be

$$\lim_{n \to \infty} \sup\{\mid f_{n}(x)-f(x)\mid : x \in [R, \infty)\} = \lim_{n \to \infty} \frac{1}{\color{#FF0000}R^n} = 0.$$

• Thank you for your reply, if corrected my calculations. Does it work like this – Fo Young Areal Lo May 29 at 11:41
• Have updated my answer as well. – Aryaman Maithani May 29 at 11:45

For all $$x \in [R, +\infty)$$, we find that, since $$R > 1$$, therefore we have \begin{align} \lim_{n \to \infty} f_n(x) &= \lim_{n \to \infty} \frac{ x^{2n} }{ 1 + x^{2n} } \\ &= \lim_{n \to \infty} \frac{1}{ \frac{1}{x^{2n}} + 1} \\ &= \frac{1}{0+1} \qquad [\mbox{ as x \geq R > 1, so \lim_{n \to \infty} \frac{1}{x^{2n}} = 0 } ]\\ &= 1. \end{align} Now let $$f \colon [R, +\infty) \rightarrow \mathbb{R}$$ be defined by the formula $$f(x) \colon= 1 \qquad \mbox{for all } x \in [R, +\infty). \tag{0}$$ Then the sequence $$\left( f_n \right)_{n \in \mathbb{N}}$$ converges pointwise to the function $$f$$ on $$[R, +\infty)$$.

Let us now check if this convergence is uniform.

We note that, for all $$n \in \mathbb{N}$$ and for all $$x \in [R, +\infty)$$, we have \begin{align} \left\lvert f_n(x) - f(x) \right\rvert &= \left\lvert \frac{x^{2n}}{1+x^{2n}} - 1 \right\rvert \\ &= \left\lvert \frac{-1}{1+x^{2n}} \right\rvert \\ &= \frac{1}{\left\lvert 1+x^{2n} \right\rvert } \\ &= \frac{1}{ 1+x^{2n} } \\ &< \frac{1}{x^{2n} } \\ &\leq \frac{1}{R^{2n}}. \tag{1} \end{align}

Now as $$R > 1$$, so $$\lim_{n \to \infty} \frac{1}{R^{2n}} = 0.$$

Thus from (2) we can conclude that, given a real number $$\varepsilon > 0$$, we can find a natural number $$N = N(\varepsilon)$$ such that $$\left\lvert \frac{1}{R^{2n}} - 0 \right\rvert = \frac{1}{R^{2n}} < \varepsilon$$ for any natural number $$n > N$$. In fact, we can take $$N$$ to be any natural number greater than the quantity $$\begin{cases} \frac{ - \ln \varepsilon }{ \ln R} \ \mbox{ if } \varepsilon \neq 1, \\ 1 \ \mbox{ if } \varepsilon = 1. \end{cases}$$

So using (1) we can conclude that, given a real number $$\varepsilon > 0$$, we can find a natural number $$N$$ such that $$\left\lvert f_n(x) - f(x) \right\rvert < \varepsilon$$ for all $$x \in [R, +\infty)$$ and for any natural number $$n > N$$.

Hence the sequence $$\left( f_n \right)_{n \in \mathbb{N}}$$ indeed converges uniformly to the function $$f$$ defined by the formula (0) above.