Is the Space of bounded functions with the maximums norm a Banach space and even a Banach Algebra?

Solution 1:

Let $\|f\|:= \sup_{x\in X} |f(x)|$.

For proving that your space $(B(X),\|\cdot \|)$ is complete I would begin with : Let $(f_n)_{n=0}^\infty$ be a Cauchy sequence in $B(X)$. Let $\epsilon >0 $. Then we find $N \in \mathbb N$ such that for all $n,m \geq N: \|f_n - f_m\| < \epsilon$. This implies that each $(f_n(x))_{n=0}^\infty$ converges pointwise because $\mathbb R$ is complete. Then we may define $$ f: X \rightarrow \mathbb R: x\mapsto \lim_{n \rightarrow \infty} f_n(x) $$ We now show that $f_n \rightarrow f$ uniform on $X$. Let $\epsilon > 0$. Then we find $N \in \mathbb N$ such that for all $n,m \geq N$ : $$ |f_n(x)-f_m(x)| < \epsilon, \forall x \in X $$ Taking the limit in $m \rightarrow \infty$ (this must be verified as legitimate step, too) leads to $$ \forall n \geq N: |f_n(x)-f(x)| < \epsilon, \forall x \in X $$ which proves uniform convergence of the $f_n$ to $f$. Thus any Cauchy sequence in $B(x)$ converges which proves completeness of $(B(x),\|\cdot\|)$.

Further is boundedness preserved by uniform convergence.

What I mean by "thist must be verified": For a given sequence $(a_n)_{n=0}^\infty$ in $\mathbb R$ with limit $a$ we have for any metric d: $$ \lim_{n \rightarrow \infty} d(a_n,x) = d(a,x) $$ where $x$ is some point in your space.