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This is an extract from Douglas: "Banach Algebra Techniques in operator theory".

"For Banach algebras and, in particular, for $C(X)$ the importan idea is that of multiplicative linear functional. [...] Except for the zero functional, which is obviously both multiplicative and linear, every multiplicative linear functional $\varphi$ satisfies $\varphi(1) = 1$ since $\varphi \not= 0$ means there exists an $f$ in $C(X)$ with $\varphi(f) \not= 0$ and then the equation $\varphi(1)\varphi(f)=\varphi(f)$ implies $\varphi(1)=1$"

I don't understand the reason that he want to prove that $\varphi(1)=1$. Isn't that property in the definition itself of Banach Algebra?

And after that, I didn't understand the "proof" he provides:

$\varphi \not= 0$ means there exists an $f$ in $C(X)$ with $\varphi(f) \not= 0$ OK

and then $\varphi(1)\varphi(f)=\varphi(f)$ implies $\varphi(1)=1$" WHY?

Ok, that's because "multiplicative" means "homomorphic", but why can I say that I can take $\varphi(f 1)$? The identity element in C(X) should be something like 1(x) = x no?

Then $\varphi(f 1)=\varphi(f(x) 1(x))$ and I cannot conclude that $\varphi(1)\varphi(f)=\varphi(f)$.

Please, can you explain me the reason of this in detail? Thank you!

Benzio
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  • Multiplicative does not mean isomorphic. – Mariano Suárez-Álvarez Oct 28 '13 at 18:56
  • If a and b are real numbers, b is not zero and ab=b, what can you say about a? – Mariano Suárez-Álvarez Oct 28 '13 at 18:57
  • Ok then it means that $(\varphi_1\varphi_2)(f)=\varphi_1(f)\varphi_2(f)$.. right? This explain even less what I read in the book. – Benzio Oct 28 '13 at 18:57
  • Doesn't the text you are reading tell you? – Mariano Suárez-Álvarez Oct 28 '13 at 18:58
  • Yes... But only in a group. We are not in a group, right? – Benzio Oct 28 '13 at 18:59
  • Hm. No. Here multiplicative means that $\phi(fg)=\phi(f)\phi(g)$. – Mariano Suárez-Álvarez Oct 28 '13 at 19:00
  • Your questions make me wonder if you should be reading a book about Banach algebras :-/ – Mariano Suárez-Álvarez Oct 28 '13 at 19:01
  • I wonder if I should be reading it too, you can trust me... But I have to. :-p – Benzio Oct 28 '13 at 19:08
  • Oh! I wrote "isomorphic", I meant "homomorphic". That was a writing error! – Benzio Oct 28 '13 at 19:10
  • Have you been able to answer the question in my second comment above? – Mariano Suárez-Álvarez Oct 28 '13 at 21:54
  • Yes, now it's everything clear. I was exaust after a lot of study and I wasn't able to see anything. I didn't realized that C(X) obviously has the element 1, that is the element 1(x) = 1, and NOT 1(x) = x. Anyway you should not demoralize people like you did with me with your second-last comment. I was here because there were some easy questions that I did cannot answer because of misreading of the text and a lot of stress. You don't really know what my level is, and your comment was not much helpful. – Benzio Oct 28 '13 at 23:05
  • Well, there is nothing more demoralizing than trying to read something for which one does not have the required background — mathematical knowledge is inherently hierarchical. The gist of your question here, which is what I hinted at in my second comment, is a piece of very, very, VERY basic knowledge regarding the real numbers. I honestly do not think there is any point setting out to learn about Banach algebras without having an immensely solid grasp on the basic properties of the real numbers. I do not say such things lightly, believe me. – Mariano Suárez-Álvarez Oct 29 '13 at 02:54

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