Is any multiplicative linear functional $ \Phi $ on $ C([a,b]) $ bounded? Denote by $ X $ the function space
$$
  C([a,b])
= \{
  u: [a,b] \to \mathbb{C} \mid u ~ \text{is continuous at each} ~ t \in [a,b]
  \}
$$
over the scalar field $ \mathbb{C} $ of complex numbers, as usual, in which the norm of $ u \in X $ is given by
$$
\| u \| = \max_{t \in [a,b]} |u(t)|.
$$
Suppose that $ \Phi: X \to \mathbb{C} $ is a linear functional satisfying
$$
\forall x,y \in X: \qquad
\Phi(x y) = \Phi(x) \Phi(y).
$$
Can we conclude that $ \Phi $ is bounded?
At this stage, I guess that there exists $ t_{0} \in [a,b] $ such that $ \Phi(x) = x(t_{0}) $ for all $ x \in X $.

Here is a more direct proof of your original question without proving your guess, or using measure theory.


*

*$\Phi(1)=1$, where $1$ denotes the constant function with value $1$.

*For any function $x\in C[a,b]$ without zeroes, it is $\Phi(x)\neq 0$, because $\Phi(x)\Phi(x^{-1})=\Phi(xx^{-1})=\Phi(1)=1$, where $x^{-1}$ is the pointwise inverse of $x$.

*For any $x\in C[a,b]$, the function $(x-\Phi(x))$ has a zero because $\Phi(x-\Phi(x))=0$. This implies $|\Phi(x)|\le ||x||$, i.e. $||\Phi||\le 1$

