Does this sequence of function converges uniformly?

Question

I was practicing for my final and stumbled on this problem.

Let $f_n$:$[0,1]$ $\mapsto$ $R$ such that

$f(x)_n$ = \begin{cases} nx, & 0\leq \text{ $x$ } \lt1/n\\ 1, & 1/n\leq\text{ $x$ }\leq 1 \end{cases}

I could find its piecewise limit, which is

$f(x)$ = \begin{cases} 0, & \text{ $x$ =0} \\ 1, & \text{ $otherwise$ } \end{cases}

I was thinking about using the $sup$$|f(x)_n-f(x)|<\epsilon$

then I realized there were more than one interval of both $f_n$ and $f$.

So I used $x$ to indicate the supremum of each interval

for $x=0$

Clearly $sup$$|f(x)_n-f(x)|=0<\epsilon$

Since both function yield $0$

for $1/n\leq\text{ $x$ }\leq 1$

Clearly $sup$$|f(x)_n-f(x)|=0<\epsilon$

Since both function yield $1$

Now I am stuck here

for $0\lt \text{ $x$ } \lt1/n$

What should i do with this interval?

And does $f(x)_n$ converges uniformly?

Answer 1

The sequence doesn't converge uniformly, since the uniform limit of a sequence of continuous functions is again a continuous function.

Answer 2

If you have some theorems under your belt, you know that this sequence of functions cannot converge uniformly, since each $f_n$ is continuous but $f$ is not. On the interval you are considering, try and take a very small neighborhood of $0$, say $(0,\epsilon/n^2)$, and consider the difference between $f_n$ and $f$ on that subinterval.