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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 ???

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bakabakabaka
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  • Yes, I believe so. Are you having trouble proving it? – user108903 Jan 20 '13 at 10:12
  • Hi, yes, I don't know how to start. – bakabakabaka Jan 20 '13 at 10:38
  • You should start by looking up the definition of a Banach algebra. The second step is checking each condition in the definition. Some of these are easier to check than others. If you get stuck checking a particular condition, post back here and someone will probably help you. – user108903 Jan 20 '13 at 10:40
  • Continuity of pointwise multiplication $(f\cdot g)(x) = f(x) g(x)$ is expressed by $\lVert f \cdot g \rVert_\infty \leq \lVert f\rVert_\infty \lVert g\rVert_\infty$ which follows from $|f(x) g(x)| \leq \lVert f\rVert_\infty \lVert g\rVert_\infty$ and the rest is discussed in your previous question http://math.stackexchange.com/q/282121/ – Martin Jan 20 '13 at 11:39
  • How did you get this inequality $|f(x)g(x)| \le ||f|| \infty || g ||\infty$ ? – bakabakabaka Jan 20 '13 at 11:53
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    By definition $\lvert f(x) \rvert \leq \lVert f\rVert_\infty$ and $\lvert g(x) \rvert \leq \lVert g\rVert_\infty$, so $\lvert f(x) g(x) \rvert = \lvert f(x) \rvert \lvert g(x) \rvert \leq \lVert f \rVert_\infty \lvert g(x) \rvert \leq \lVert f \rVert_\infty \lVert g\rVert_\infty$. – Martin Jan 20 '13 at 11:56

1 Answers1

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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.

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Martin
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  • Is this what you meant by checking the algebra structure on B(X): If f(x),g(x),z(x) are bounded functions , a,b, scalars in $\mathbb{R}$ Then : $1: (f(x)+g(x))\cdot z(x)= f(x)z(x)+g(x)z(x) ; 2: f(x)\cdot (g(x)+z(x)) = f(x)g(x)+f(x)z(x) ; 3: (af(x))\cdot (bg(x)) = af(x)\cdot bg(x) = (ab)(f(x)\cdot g(x) $ ? – bakabakabaka Jan 20 '13 at 12:25
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    Yes, almost. You also need to know that $f(x)\cdot g(x) \in B(X)$ which follows from the norm inequality I proved. Moreover, you can also deduce your point 2) from your point 1) when you combine it with the observation that $f \cdot g = g \cdot f$ (i.e. multiplication is commutative). It might also be worth mentioning that $B(X)$ has a unit element, the constant function $1_X(x) = 1$. – Martin Jan 20 '13 at 12:28