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Consider the Banach space ($C[0,1], \vert\vert\cdot\vert\vert_{\infty}$) of all continuous complex-valued functions on $[0,1]$ w.r.t $\vert\vert\cdot\vert\vert_{\infty}$. A linear functional $Z: C([0,1]) \to \mathbb{C}$ is called multiplicative if $Z(fg) = Z(f)Z(g)$ for all $f,g \in C([0,1])$. How do I show now that the functionals $$ d_x: C([0,1]) \to \mathbb{C},\quad d_x(f) = f(x)$$ are the only multiplicative functionals on $C([0,1])$ apart from the zero functional? I am familiar with Riesz representation theorem. Since the underlying set $[0,1]$ is compact, it holds that $C([0,1]) = C_0([0,1]) = C_c([0,1])$: Therefore, according to the Riesz representation theorem every positive linear functional has the form $$Z(f) = \int_{[0,1]} f d\mu.$$ This doesn't help me though since we are not necessarily talking about positive functionals. Should I assume that there is another linear functional that is multiplicative and then, on grounds of Riesz, show that this would lead to a contradiction. What should I consider here?

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  • $\begingroup$ $C[0,1]$ is actually a Banach Algebra, do you know the form that maximal ideals have in this algebra? $\endgroup$ – s.harp Feb 5 '17 at 14:28
  • $\begingroup$ Also, all multiplicative linear functionals on a $C^*$-algebra are positive. $\endgroup$ – Teebro Prokash Mar 1 '19 at 18:22
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The first point is to show that $Z$ corresponds to a positive norm $1$ measure. The next ingredient is to use this to see that if $f,g$ are positive maps with disjoint support $Z(0)=Z(fg)=Z(f)Z(g)=0$ and either $Z(g)$ or $Z(f)$ is zero. But since $Z$ corresponds to a positive measure $\mu$ you must have zero mass either on $g^{-1}(\mathbb R_{>0})$ or on $f^{-1}(\mathbb R_{>0})$.

This allows you to see, for any two disjoint closed sets one of the two must have mass zero. This should however imply that the measure was a dirac measure.

To see the first point: mass one follows from $1=Z(1)=\int1d\mu$. Positivity first requires that the measure is real, which can be done by assuming it has an imaginary component "located" somewhere and integrating the square of bump functions near that place, positivity then follows from any positive function having a positive root and then $Z(f^2)=Z(f)^2≥0$.


There is a sort of standard story to this result, which is told in a Banach Algebra setting.

$C[0,1]$ is a Banach algebra, as the product of two continuous functions is again continuous and $\|f\cdot g\|≤\|f\|\cdot\|g\|$. If $Z$ is a multiplicative linear map $C[0,1]\to\Bbb C$ you have that $\ker(Z)$ is an ideal. Indeed it must be a maximal ideal:

$Z$ induces a multiplicative map $\psi:C[0,1]/\ker(Z)\to \Bbb C$ with kernel $\{0\}$. This implies that $C[0,1]/Z$ is $\Bbb C$, since $\psi( x-\psi(x)1\,)=0$ implies $x=\psi(x)1$, so proportional to identity (since $Z(1)\overset!=1$). So $Z$ has to be a maximal ideal.

On the other hand maximal ideals in $C[0,1]$ are always of the form $N_x=\{f(x)=0\mid f\in C[0,1]\}$ for some $x$. This follows from showing that for an ideal $I$ the set $N(I)=\{x\mid f(x)=0\ \forall f\in I\}$ is not empty whenever $I\neq C[0,1]$ and then any proper ideal is contained in some $N_x$.

To see $N(I)=\emptyset$ implies $I=C[0,1]$ let $f_x\in I$ be any function not-vanishing at $x$. It follows $|f_x|^2$ lies in $I$ and from continuity $|f_x|^2$ is strictly greater than $0$ in some open neighbourhood of $x$. One has this for every $x$ and since $[0,1]$ is compact one can thus find a finite amount of $x_1,..,x_n$ so that $\sum |f_{x_i}|^2>0$ everywhere. But this is in $I$ and invertible so $I=C[0,1]$.

$\ker(Z)=N_x$ means that $f-Z(f)\,1$ must be $0$ at $x$ and thus $\delta_x(f -Z(f)\,1)=f(x)-Z(f)=0$.

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  • $\begingroup$ Impressive. Though that is way beyond my background knowledge. I marked it as the selected answer. However, do you know a way to prove it using more or less elementary results concerning Riesz representation theorem and the spaces $C_c([0,1]$ or $C_0([0,1])$? Otherwise the answer would not quite fit to the advanced analysis course the question was posed at. $\endgroup$ – Taufi Feb 5 '17 at 15:42
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    $\begingroup$ Unfortunately I cannot see any argument that works like that. It probably exists though. Had I known one I would have written it and only mentioned this way as a curiosity. $\endgroup$ – s.harp Feb 5 '17 at 20:06
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    $\begingroup$ @Taufi actually it turned out to be doable, so I wrote it in the answer as well. $\endgroup$ – s.harp Feb 6 '17 at 9:32

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