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

X is a arbitrary non empty set , B(X) the set of bounded functions $f:X\rightarrow \mathbb{R}$ and $||f||_\infty = \sup_{x\in X} |f(x)|$

Completeness: Let $(f_n(x))_{n \in \mathbb{N}}$ be a cauchy sequence, then: $$||f_n-f_m||\le \frac{\epsilon}{2} \ \text{for n,m greater than some N}$$

the cauchy sequence $f_n$ will have a limit $f(x)$ for $x \in \mathbb{R}$, so there must be a $f_{n_k}$ with a $n_k > N$ such that : $|f_{n_k}(x)-f(x)|\le \frac{\epsilon}{2}$ so one can put:

$$|f_n(x)-f(x)| \le ||f_n(x)-f_{n_k}(x)||+ |f_{n_k}(x)-f(x)| \le \epsilon$$

And for every $x\in \mathbb{R}$: $$|f(x)|\le |f_{n_k}(x)|+|f_n(x)-f(x)|\le ||f_{n_k}(x)||+\epsilon < \infty$$

Is this sufficient to say that it was shown that $(B(X), ||.||_\infty )$ is a Banach space?

Is it also a Banach Algebra?

• assuming that it is complete : I don't understand. Jan 19, 2013 at 17:35
• It is just $\ell^{\infty}(X)$... May 11, 2019 at 5:31

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 $$\forall n \geq N: |f_n(x)-f(x)| \le \epsilon, \forall x \in X$$ 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.

• Thanks, it does answer the question in the title. Jan 19, 2013 at 20:46
• I am not familiar with "Banach Algebra" so sorry for that ^^
– user42761
Jan 20, 2013 at 9:37
• There's a small error; in the last step, the inequality should not be strict, i.e., it should be $\forall n \geq N: |f_n(x)-f(x)| \leq \epsilon, \forall x \in X$. Apr 18, 2023 at 18:33