1
$\begingroup$

I have come across the following proposition in the book "Complete Normed Algebras" by F. F. Bonsall and J. Duncan in section 16 on page. The section denotes $ A $ as a Banach algebra.

Definition: A multiplicative linear functional on $ A $ is a non-zero linear functional $ \phi $ on $ A $ such that

$$ \phi(xy) = \phi(x) \phi(y) $$

for all $ x, y \in A $.

Proposition 3: Let $ \phi $ be a multiplicative linear functional on $ A $. Then $ \phi $ is continuous and $ \| \phi \| \leq 1 $.

The proof in the text is as follows:

Proof: Suppose that there exists $ x \in A $ with $ \| x \| < 1 $ and $ \phi (x) = 1 $, and let $ y = \sum_{n=1}^{\infty} x^n $. Then $ x + xy = y $, and so

$$ 1 + \phi(y) = \phi(x) + \phi(x)\phi(y) = \phi(x + xy) = \phi(y) $$

which is absurd. $ \blacksquare $

I understand every step of the above proof - the only problem is that I don't see how this proves that $ \phi $ is bounded. Any help would be greatly appreciated.

$\endgroup$
  • $\begingroup$ The assumption $||x||< 1=\phi(x)=||\phi(x)||$ leads to a contradiction; doesn't this prove the bounded condition? $\endgroup$ – SystematicDisintegration Jun 16 '18 at 6:31
1
$\begingroup$

Well, we have found a contradiction, starting from the statement : "there exists $x$ such that $||x|| < 1$ and $\phi(x) = 1$".

Therefore, the negation of this statement is true : for all $x$ such that $||x|| < 1$, we have $\phi(x) \neq 1$. Call this statement $(*)$.

However, if there was a $y$ such that $||y|| < 1$ and $\phi(y) > 1$ then one may consider the vector $\frac{y}{\phi(y)}$, which satisfies $\left\|\frac{y}{\phi(y)}\right\| < 1$ but also has $\phi\left(\frac y{\phi(y)}\right) = 1$. Taking $x$ as this vector contradicts the statement $(*)$.

Consequently, for all $x$ such that $||x|| < 1$, we have $\phi(x) \leq 1$. This implies that $\phi$ is bounded : for any $x_0$ such that $||x_0|| \leq 1$, we have $\phi(\lambda x_0) < 1$ for all $\lambda < 1$, so $\phi(x_0) < \frac 1{\lambda}$ for all $\lambda < 1$. Hence $\phi(x_0) \leq 1$.

Consequently, $||\phi|| \leq 1$, since $\phi$ is bounded by $1$ on the unit sphere.

$\endgroup$
1
$\begingroup$

You want to show that $\phi$ is bounded with $\|\phi\| \le 1$. If $\phi$ were not bounded, or if it were bounded with $\|\phi\| > 1$, then there would exist $x$ with $\|x\|< 1$ such that $|\phi(x)| =1$. And that leads to a contradiction, which proves that (a) $\phi$ is bounded and (b) $\|\phi\| \le 1$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.