A typical example: Let $\varphi \in (C([-1,1]),\vert\vert\cdot\vert\vert_{\infty})^{*}$ and then define $\varphi(f):=\int_{-1}^{0}f(x)dx-\int_{0}^{1}f(x)dx$
It is clear that $\vert \vert \varphi \vert \vert_{*}\leq 2$ and in order to show that $\vert \vert \varphi \vert \vert_{*}= 2$, I have chosen the function $f(x):=1\chi_{[-1,0]}(x)-1\chi_{]0,1]}(x)$ it is clear that $\vert \vert \varphi (f)\vert \vert=2$ and $\vert \vert f\vert \vert_{\infty}=1$ but of course $f \notin C([-1,1])$ so I defined a sequence $(f_{n})_{n}$ where
$f_{n}(x):=\begin{cases} -1, x\in [\frac{1}{n},1] \\ -nx, x \in ]-\frac{1}{n},\frac{1}{n}[ \\ 1, x \in [\frac{1}{n},1] \end{cases}$
Now it is clear that $f_{n} \in C([-1,1])$ for all $n \in \mathbb N$ and for all $x \in [-1,1]$ we have $f_{n}(x)\xrightarrow{n \to \infty} f(x)$. But it is clear that this convergence is not uniform, so can I really use it to prove that:
$\vert \vert \varphi \vert \vert_{*}\geq 2 $? My problem:
We would essentially use an argument, like:
$\vert \vert \varphi \vert \vert_{*}=\sup\limits_{g \in C([-1,1])\setminus \{0\} }\frac{\varphi (g)}{\vert\vert g\vert \vert_{\infty}}\geq \sup\limits_{n \in \mathbb N}\frac{\varphi (f_{n})}{\vert\vert f_{n}\vert \vert_{\infty}}\geq \frac{\varphi(f)}{\vert\vert f\vert \vert_{\infty}}=2$
But surely I can only use the last inequality, i.e. $\sup\limits_{n \in \mathbb N}\frac{\varphi (f_{n})}{\vert\vert f_{n}\vert \vert_{\infty}}\geq \frac{\varphi(f)}{\vert\vert f\vert \vert_{\infty}}$ only if the convergence is uniform.
I suppose my main question is as follows: In order to approximate a function to show an upper bound for an operator, does pointwise convergence suffice and if so, why?
No, for this kind of problem you don't need the fact that $f_n \to f$ at all. All you need to do is compute $\varphi(f_n)$ directly and hence deduce that $$ \sup\limits_{n \in \mathbb N}\frac{\varphi (f_{n})}{\vert\vert f_{n}\vert \vert_{\infty}}=2. $$
Alternatively, it is also possible to view the convergence $f_n\to f$ in the space $L^1([-1,1])$, which is somewhat a correct space to work with for this kind of functional (some other $L^p$ also works). However, I don't think you need to go that far.