Sequence of Functions - Pointwise and Uniform Convergence

Question

I'm learning about sequences of functions and need some help with this problem:

Show that the sequence of function $f_n(x)$ where $$f_n(x) = \begin{cases} \frac{x}{n}, & \text{if $n$ is even} \\ \frac{1}{n}, & \text{if $ n $ is odd,} \end{cases}$$ converges pointwise but not uniformly on $\mathbb{R}$.

Since I'm having difficulties disproving uniform convergence I'll be explaining my work for showing pointwise convergence.

My work:

We note that $f_n(x)$ is independant of $x$ for $n$ odd and the sequence $\frac{1}{n}$ tends to $0$ as $n$ approaches infinity.

For $x = 0$: $\lim_{n \to \infty} f_n(0) = \lim_{n \to \infty} \frac{0}{n} = 0$.

In addition, for any fixed $x \in \mathbb{R}$ and $n$ even we have: $\lim_{n \to \infty} f_n(x) = \lim_{n \to \infty} \frac{x}{n} = 0$.

Therefore, we conclude that $f_n(x)$ converges pointwise to the zero function on $\mathbb{R}$.

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Is my work correct so far? How do I show that $f_n(x)$ does not converge uniformy on $\mathbb{R}$? Is it possible to find a subsequence of $f_n(x)$ that converges uniformly on $\mathbb{R}$?

Answer 1

Your work is ok for the pointwise convergence.

For uniform convergence just use

$$ \lim_{n->+\infty}\sup_{x\in\mathbb{R}} \left | f_n(x) \right |=\lim_{n->+\infty} 1 \ne 0 $$

We have chosen $x=n$ for $n$ odd. You can also disprove uniform convergence by Cauchy uniform convergence test. Just use odd indexes and again as here use $x=n$ and it will result in the same - function $f_n(x)$ is not uniformly convergent on $\mathbb{R}$