# Every Multiplicative Linear Functional from a Banach Algebra is Bounded

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.

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

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