X is an arbitrary , non empty set, B(X) the set of bounded functions $f:X\rightarrow \mathbb{R}$ and $||f||_\infty = \sup_{x\in X }|f(x)|$.
Is $(B(X),||.||_\infty )$ a Banach Algebra?
My attempt at showing that this is true:
Definition of a Banach Algebra: A normed space E with elements f,g,… is called normed Algebra if it is an Algebra and the multiplication with the norm fulfills: $$||fg||\le ||f||\cdot||g||$$
A normed algebra is a Banach algebra , if it is complete as a space (if it is a Banach space).
* Defintion of an Algebra:* If K is a field , A a vector space equipped with multiplication operation in form of $A \times A \rightarrow A$, then A is an algebra if for $x,y,z \in A $ and $a,b \in K$ scalars it holds that: $$1. (x+y)\cdot z = xz+yz \\2: x\cdot(y+z)=xy+xz \\ 3: (ax)\cdot (by)=(ab)(x\cdot y)$$
In this case A is B(X) and x,y,z are bounded functions, $a,b\in \mathbb{R}$ and it fulfills (1-3) of the Algebra definition.
Now for the step from Algebra to normed Algebra one has to check the submultiplicativity : $$\sup_{x\in X}|f(x)g(x)| \le \sup _{x \in X} |f(x)|\sup_{x\in X}|g(x)|$$
How to show this ???
Best Answer
Completeness of $(B(X), \lVert \cdot \rVert_\infty)$ was discussed in your previous question.
The only remaining difficulty seems to be that the pointwise product of bounded functions is bounded and that the norm is submultiplicative with respect to the pointwise product. To see this, note that by definition $\lvert f(x) \rvert \leq \lVert f \rVert_\infty$ and $\lvert g(x)\rvert \leq \lVert g\rVert_\infty$ so that $$ \lvert f(x) g(x) \rvert = \lvert f(x) \rvert \lvert g(x) \rvert \leq \lVert f \rVert_\infty \lvert g(x) \rvert \leq \lVert f \rVert_\infty \lVert g\rVert_\infty $$ and taking the supremum over $x \in X$ yields $$\lVert f \cdot g \rVert_\infty = \sup_{x\in X}{\lvert f(x) g(x) \rvert} \leq \lVert f \rVert_\infty \lVert g \rVert_\infty. $$
I trust that you can now check that pointwise multiplication is an algebra structure on $B(X)$ so that $B(X)$ is indeed a Banach algebra.