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 $.
 A: 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. $||\Phi||\le 1$

A: Based on the proposition we have proved, we can show that 

If $T:X\to X$ is a linear operator with$$\forall x,y\in X:\ T(xy)=(Tx)(Ty).$$Then $T$ is bounded.

For any $x\in X$ with $\|x\|=1,$ let us verify that $\|Tx\|_X\leq1.$ It suffices to show that $|(Tx)(t_0)|\leq1$ holds for $t_0\in[0,1]$ fixed. To this end, we define a linear functional $f:X\to\mathbb C$ by$$f(u)=(Tu)(t_0),\forall u\in X.$$Then,$$\forall u,v\in X:\ f(uv)=(T(uv))(t_0)=(Tu)(t_0)\cdot(Tv)(t_0)=f(u)f(v),$$ then we obtain from $\|f\|_{X^*}\leq1$ that $|(Tx)(t_0)|\leq1,$ where $X^*$ is the dual space of $X$. Thus we have $|(Tx)(t)|\leq1$ for any $t\in[0,1],$ this yields that $\|Tx\|_X\leq1$, hence $T$ is bounded.
