Real Analysis - Convergence of a Function

Question

I am given that

For $n \in \mathbb{N}$ define $f_n:(0,1) \to \mathbb{R}$ by

$$f_n(x)=\frac{1}{nx}.$$

I must determine whether the sequence $(f_n)$ converges pointwise on $(0,1)$, whether $(f_n)$ converges uniformly on $(0,1)$ and whether $(f_n)$ converges uniformly on $[1/4,1/2]$.

I have shown that it converges pointwise on $(0,1)$ by saying

$$\lim_{n \to \infty}f_n(x)=\lim_{n \to \infty} \frac{1}{nx} = 0.$$

Which means that $f(x)=0 \: \: \forall x \in (0,1)$ is the pointwise limit of $(f_n)$.

Also, I think I have shown that it converges uniformly on $(0,1)$ by doing the following:

Let $\varepsilon >0$ and $N>n$, hence

$$\left| f_n(x)-f(x) \right|=\frac{1}{nx} < \varepsilon \implies n > \frac{1}{\varepsilon x},$$

so let $N>\frac{1}{\varepsilon x}$ and so

$$\left| f_n(x)-f(x) \right|=\frac{1}{nx}<\frac{1}{Nx}<\frac{1}{\left( \frac{1}{\varepsilon x} \right) x}=\varepsilon.$$

Any help would be great!

Answer 1

The convergence can't be uniform since the $f_n$ are even not uniformly bounded. It's easy to compute the $\sup_{(0,1)}|f_n-f|$ and see the it doesn't converge to $0$.

For $[1/4,1/2]$ the convergence is indeed uniform. It's easy to compute $\sup_{[1/4,1/2]}|f_n-f|$ and see that it converge to $0$.