Pointwise/Uniform Convergence of a function

Question

Let $f_n(x)$=$nx ,\space \space x\in[0,\dfrac{1}{n}]\brace 0, \space\space x\in(\dfrac{1}{n},1]$. Find the pointwise limit of $(f_n)$ and check if $(f_n)$ is uniformly convergent.

Answer 1

The sequence converges to 0 everywhere, but the convergence is only pointwise. To prove pointwise convergence, fix $x$. If $x$ is zero, all $f_n$ vanish on $x$, otherwise there exists an index $n$ from which all $f_n$ are zero on $x$, meaning that $f_n(x)=0$ for all $x$, if $n$ is big enough (how big depends on $x$)\

The convergence is not uniform though: uniform convergence means that for given $\epsilon$ there exists an index $n$ from which $f_n(x)<\epsilon$, for all $x$ in the domain. However, we see that when $\epsilon<1$ this does not happen, as $f_n(1/n)$ is always one.

Answer 2

Key Ideas

1. The pointwise limit is clear.

2. To consider uniform convergence, consider the value of $f_n \left( \frac{1}{n} \right)$ for each $n$.