3

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.

Aweygan
  • 23,232
LMW
  • 445

2 Answers2

3

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) $\|\phi\| \le 1$.

Disintegrating By Parts
  • 87,459
  • 5
  • 65
  • 149
1

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.

Sarvesh Ravichandran Iyer
  • 76,105