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Let $A$ be a Banach algebra and let $$A_1=\{(x,\alpha)\;;\;:x∈A, \alpha\in\mathbb{C}\}$$
with the following operations: $$ (x_1,\alpha_1 )+(x_2,\alpha_2 )=(x_1+x_2 ,\alpha_1+\alpha_2 )\qquad \lambda(x,\alpha)=(\lambda x,\lambda\alpha)$$ and $$ (x,\alpha)\cdot(y,\beta)=(xy+x\beta+y\alpha,\alpha\beta) $$ and with norm $$ \|(x,\alpha)\|=\|x\|+|\alpha|. $$ Show that $A_1$ is a Banach algebra with unit and that the mapping $x\longmapsto(x,0)$ is an isometric isomorphism from $A$ into $A_1$.

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Why do you repeat it with different variables (..or.. what is $F$)? Where did you stuck? –  Berci Mar 29 '13 at 12:39
    
SORRY - THE PROBLEM IS: Let A be a Banach Algebra and let A_1={(x,α):x∈A,α∈C} with folowing operations: (x_1,α_1 )+(x_2,α_2 )=(x_1+x_2 ,α_1+α_2 ) (x,α)*(y,β)=(xy+xb+yα,αβ) λ(x,α)=(λx,λα) With norm∶ ‖(x,α) ‖=‖x‖+|α| Show that A1 is a Banach algebra with unit and that mapping x→(x,0) is an isometric isomorphism on A into A1. –  elizabeta gigova Mar 29 '13 at 12:53
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@elizabetagigova: Welcome to MSE! It really helps readability to use MathJax for formatting (see FAQ). Regards –  Amzoti Mar 29 '13 at 12:54
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2 Answers 2

This is just (one of) the officional way to talk about the construction in which a unit is added freely to the Banach algebra $A$.

Informally, (even if $A$ already had a unit element), we adjoin a new element to $A$, let's denote it $\bf 1$ which would behave as unit w.r.t. product, i.e. $x\cdot{\bf 1}=x={\bf 1}\cdot x$ for all elements $x$. Since we want to arrive at a Banach algebra (that extends $A$ with a unit), we have to form also elements of the form $a+\lambda\cdot{\bf 1}$, fortunately, these elements are enough to form a vector space. Now, in your set, the formal $(a,\lambda)\in A_1$ stands for our element $a+\lambda\cdot{\bf 1}$, and the product rule is also defined accordingly.

Formally, you have to recall the definition of Banach algebra, and verify that the given $A_1$ satisfies those axioms. The fact that $\|(x,0)\|=\|x\|+|0|=\|x\|$ means exactly that the embedding $A\hookrightarrow A_1:\ x\mapsto (x,0)$ preserves the norm, i.e., as is linear, it is isometric isomorphism (onto its image $\subsetneq A_1$).

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Thank You Very Much –  elizabeta gigova Mar 29 '13 at 18:07
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This is called: adjoining a unit to $A$. Note that if $A$ already has a unit, it still works, but it's useless.

Banach space: as a normed vector space, $A_1=A\times\mathbb{C}$ is the $\ell^1$ sum of $(A,\|\cdot\|)$ and $(\mathbb{C},|\cdot|)$. Since both summands are Banach spaces, this makes $A_1$ a Banach space. Now here are some details.

The fact that it is a complex vector space follows from a routine verification: that's the canonical vector space structure on the cartesian product of two vector spaces.

Then check that the formula $\|(x,\alpha)\|=\|x\|+|\alpha|$ defines a norm on this product. That is: it is nonnegative and satisfies 1) separation, 2) positive homogeneity, 3) triangular inequality. This is really straightforward, using the norm properties of $\|\cdot\|$ and $|\cdot|$.

Since both summands are complete, $A_1$ is complete. Indeed, if $(x_n,\alpha_n)$ is a Cauchy sequence, it is easily seen that $(x_n)$ is a Cauchy in $A$ and $(\alpha_n)$ is Cauchy in $\mathbb{C}$. Thus $(x_n)$ converges to $x$ in $A$ and $\alpha_n$ converges to $\alpha$ in $\mathbb{C}$. Now $$ \|(x_n,\alpha_n)-(x,\alpha)\|=\|x_n-x\|+|\alpha_n-\alpha|\longrightarrow 0. $$ So $(x_n,\alpha_n)$ converges to $(x,\alpha)$ in $A_1$. So $A_1$ is complete, hence a Banach space.

Isometric embedding: It is immediate to verify that $\phi:x\longmapsto (x,0)$ is a linear bijection from $A$ onto its range. Also $\phi(xy)=(xy,0)=(x,0)(y,0)=\phi(x)\phi(y)$, so this is an algebra homomorphism. Finally, $\|\phi(x)\|=\|(x,0)\|=\|x\|+|0|=\|x\|$, so $\phi$ is an algebra isometric isomorphism from $A$ onto its range in $A_1$. We can call it an isometric embedding of $A$ into $A_1$. Note that it identifies $A$ isometrically with a closed subalgebra of $A_1$.

Banach algebra: it remains to check that the addition and the multiplication make $A_1$ a unital associative algebra such that $\|(x,\alpha)(y,\beta)\|\leq \|(x,\alpha)\|\|(y,\beta)\|$.

One easy thing is that $1:=(0,1)$ is the unit.

For the submultiplicativity of the norm, observe: $$ \|(x,\alpha)(y,\beta)\|= \|xy+\beta x+\alpha y\|+|\alpha\beta|\leq \|x\|\|y\|+|\beta|\|x\|+|\alpha|\|y\|+|\alpha||\beta| = \|(x,\alpha)\|\|(y,\beta)\| $$

Now observe that both mappings $((x,\alpha),(y,\beta))\longmapsto \alpha\beta$ and $((x,\alpha),(y,\beta))\longmapsto xy+\beta x+\alpha y$ are bilinear, so the product on $A_1$ is bilinear.

Here comes the really tedious part of the whole exercise (there is maybe a matrix formulation which would make that easier, but I couln't find one): associativity. We take three elements in $A_1$, $(x,\alpha),(y,\beta)$ and $(z,\gamma)$. We get $$((x,\alpha)(y,\beta))\cdot(z,\gamma)=(xy+\beta x+\alpha y,\alpha\beta)\cdot(z,\gamma)$$ $$ =(xyz+\alpha yz+\beta xz+\alpha\beta z+\gamma xy+\gamma\alpha y+\gamma\beta x+\alpha\beta z,\alpha\beta\gamma) $$ while $$ (x,\alpha)\cdot((y,\beta))(z,\gamma))=(x,\alpha)(yz+\beta z+\gamma y,\beta\gamma) $$ $$ =(xyz+\beta xz+\gamma xy+\beta\gamma x+\alpha yz+\alpha\beta z+\alpha\gamma y,\alpha\beta\gamma). $$ These are equal...which completes the check: $(A_1,+,\cdot)$ is an associative algebra.

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Thank You Very Much –  elizabeta gigova Mar 29 '13 at 18:08
    
@elizabetagigova You're welcome. Since you are new, note that you can accept an answer that satisfies you, if you want, by clicking on the accept button on the left. –  1015 Mar 29 '13 at 18:50
    
Thank you very much, you help me a lot! –  elizabeta gigova Mar 30 '13 at 15:48
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