Good morning.
So I was able to prove that $||f(x)||:=\sup|f(x)|$ spans a Banach space, but the main question is if $||f(x)||:= \max|f(x)|$ spans a Banach space. Since $||f(x)||:=\sup|f(x)|$ is bounded, doesn't it mean that proving that $||f(x)||:=\sup|f(x)|$ is a Banach space equivalent of proving that $||f(x)||:= \max|f(x)|$ is a Banach space?
Best regards.
I am not entirely certain if I understand your question, but I hope the following will provide some clarity:
Being able to find a supremum does not mean you can find a maximum. In the case of a finite dimensional vector space you can always find a maximum, i.e. for all $\mathbb{R}^{n}\ni f:\{1,...,n\}\rightarrow\mathbb{R}$ we have $$\|f\|_{\infty}=sup_{1\leq i\leq n}|f(i)|=\max_{1\leq i\leq n}|f(i)|.$$ This does not hold in infinite dimensions. To give two examples, the sequence space $\ell_{\infty}$ contains the sequence $f=(1-\frac{1}{n})_{n\in\mathbb{N}}$. Clearly $\|f\|_{\infty}=\sup_{n\in\mathbb{N}}|1-\frac{1}{n}|=1$, but there is no $n\in\mathbb{N}$ such that for all $m$ we have $f(n)\geq f(m)$ as $f(n)<f(n+1)$, so the maximum of the sequence does not exist.
The same holds for sequence space. Consider $L_{\infty}((0,1))$, the set of essentially bounded measurable functions, or, if you don't know measure theory, the set of bounded functions on $(0,1)$. The function $f(x)=x$ has supremum $1$, but again the maximum of the set $\{f(x):x\in (0,1)\}=(0,1)$ does not exist.