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I have some questions about the problem I'm working on:

Find all intervals on which the sequence $f_n(x)=\frac{x^{2n}}{n+x^{2n}}$, $n \geq 1$, converges uniformly.

I am not sure if what I have is correct. I am studying for a real analysis final so this is practice to try and fully understand the concepts. The definition of uniform convergence I am using is

Given $S \subset \mathbb{R}^n$ and a sequence of functions $(f_k)$ in $C(S,\mathbb{R}^m)$, $(f_k)$ converges uniformly to $f$ if and only if $f_k - f \in C_b (S,\mathbb{R}^m)$ for all $k$ sufficiently large and

$$\lim_{k \to \infty} ||f_k - f||_{\infty} = 0.$$

Taking the limit as $n \to \infty$ yields

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

for $-1 \leq x \leq 1$ and

$$\lim_{n \to \infty}f_n(x) = 1$$

for $|x|>1$. Now, what are my next steps? Do I use my definition over these intervals? Also, how would I go about finding $\sup_{x \in [a,b]}|f_n (x)|$? That is, how would I find

$$||f_n(x)||_{\infty}$$

I am really looking for intuition on solving these types of problems. Any help in that regard would be greatly appreciated. Sometimes I can find $|| f_k ||_{\infty}$ without a problem because I can maximize the function in terms of $x$ by first derivative test, and then plug this value into the function and let $k \to \infty$ to get

$$\lim_{k \to \infty} ||f_k||_{\infty}.$$

It seems that the steps involved in checking for the uniform convergence of a sequence of functions $f_n (x)$ are:

  1. Find $f$ by taking $\lim_{n \to \infty} f_n (x)$
  2. Find the supremum of $f_n - f$, which in many cases ends up being $f_n - 0$, so find the supremum of $f_n$. Do this by finding the value for $x$ that maximizes the function.
  3. Plug this value into the function.
  4. Take the limit.

It seems to be finding the supremum that stumps me. Thanks for any help.

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greg_mankob
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1 Answers1

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We have the uniform convergence in $[0,1]$ since: $$\frac{x^{2n}}{n+x^{2n}}\leq \frac{1}{n}\quad\forall x\in[0,1]$$

In the interval $(1,+\infty)$ we have not the uniform convergence (and the same result in $(-\infty,-1)$ since the function is even), in fact:

Let $x_n=1+\frac{1}{n}$ then $$|\frac{x_n^{2n}}{n+x_n^{2n}}-1|\sim_\infty |\frac{e^2}{n+e^2}-1|\not \to 0$$

Added By a simple derivation we can see that the function $x\mapsto 1-\frac{x^{2n}}{n+x^{2n}}$ is decreasing on every interval $[a,+\infty)$ where $a>1$ then we have $$|1-\frac{x^{2n}}{n+x^{2n}}|\leq 1-\frac{a^{2n}}{n+a^{2n}}=\frac{n}{n+a^{2n}}\to0$$ hence we have the uniform convergence on every interval on the form $[a,+\infty)$ where $a>1$.