If I define on $[0,1]$, $f_n(x)=x^n$. Then the pointwise limit $f$ exists and is given by $$f(x)=\begin{cases} 0 &,\quad x \in [0,1) \\ 1 &,\quad x=1\end{cases}$$
Convergence is not uniform, since we have $||f_n-f||_{\infty}=1$ for all $n$.
So, I would just like to check whether my thinking is right in the process of calculating the $||f_n-f||_{\infty}= \sup |f_n-f|$
So, Sup $|f_n-f|=$ $\sup|x^n-1|$
For $n=1$ $|x^1-1|$ takes max value $|-1|=1$ for $x \in [0,1]$
For $n=2$ $|x^2-1|$ take max value $|-1|=1$ for $x \in [0,1]$
And so on for every $n$
So, this is the right way to deduce this? Is there a faster, more creative way? Or do I have to compute the supremum manually until at some point, I just do it in my head really quickly (ie just have enough experience that I do it very quick)?
You want to figure out how to compute $$ \|f_n-f\|_{\infty} $$ where $f$ takes two values, $0$ on $[0,1)$ and $1$ if $x=1$; moreover $f_{n}(x)=x^n$.
Now since $f_{n}(1)-f(1)=1-1=0$, the supremum has to be considered on $[0,1)$.
But on this interval $f$ is $0$, thus
$$ \|f_{n}-f\|_{\infty,[0,1]} =\|f_{n}-f\|_{\infty,[0,1)} =\|f_{n}\|_{\infty,[0,1)} =\sup_{x\in[0,1)}x^n $$ and the latter is clearly 1.