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