Involution and Gelfand Transform Properties Let $\mathcal{B}$ be a commutative unital Banach algebra, and let for each $x\in\mathcal{B}$ $\hat{x}$ be the Gelfand transform. I assume that $\mathcal{B}$ has an involution *.
I want to show that:

1) $\|\hat{x}\|=\|\hat{x^*}\|$;
2) if $x=x^*$, then $\|x\|=\|\hat{x}\|$;
3) if $\|x^*x\|=\|x^*\|\|x\|$ for each $x\in\mathcal{B}$, then $\mathcal{B}$ is a $C^*$-algebra.

I observe that 
\begin{align}
\|\hat{x}\|&=\sup_{\varphi}|{\hat{x}(\varphi)}|\\
           &=\sup_{\varphi}|{\varphi(x)}|\\
           &=\sup_{\varphi}|{\overline{\varphi(x)}}|\\
           &=\sup_{\varphi}|{\varphi(x^*)}|\\
           &=\sup_{\varphi}|{\hat{x^*}(\varphi)}|=\|\hat{x^*}\|\\
\end{align}
Now, I assume that $x$ is self-adjoint, i.e. $x=x^*$. There are proofs where the C* property is used, i.e. $\|x^*x\|=\|x\|^2$ to show that Gelfand representations are isometric. However, I want to show that self-adjointness is enough. Unfortunately, I wasn't able to do this.
Next, we see that $\mathcal{B}$ is a commutative unitial Banach algebra. So, the only thing I need to show here is that 
$$
\|x^*x\|=\|x\|^2
$$
holds. In other words, I need to show that $\|x^*x\|=\|x^*\|\|x\|$ implies $\|x^*x\|=\|x\|^2$. Here, I considered $x=u^*u$, but it wasn't very helpful. 
Can anyone help me? Any help is appreciated.
 A: If all you assume about your involution is that it's an involution. that is, $(x+y)^*=x^*+y^*$, $(xy)^*=x^*y^*$, $x^{**}=x$ and $(cx)^*=\overline cx^*$, then most of what you expect doesn't follow. In particular you assume above that $\phi(x^*)=\overline{\phi(x)}$, and that doesn't follow:
Consider $C([-1,1])$. Define $$f^*(t)=\overline{f(-t).}$$ That's an involution with $||f^*||=||f||$ but $\phi(f^*)\ne\overline{\phi(f)}$.
It's easy to make things worse. Define a non-standard norm on $C([-1,1])$ by $$||f||=\max(\sup_{0\le t\le 1}|f(t)|,\sup_{-1\le t<0}2|f(t)|).$$That norm makes $C([-1,1])$ into a Banach algebra isomorphic to the usual $C([-1,1])$, but the involution above has $||f^*||\ne||f||$.

Bonus Not that you asked, but on the topic of proving the basic facts about $C^*$ algebras under weaker hypotheses, you might note that it's impossible to give purely algebraic proofs of various things that look like they should be nothing but algebra.


Example A gizmo that satisfies all the properties of a $C^*$ algebra except that the norm is not complete, but such that $\phi(x^*)\ne\overline{\phi(x)}$ for every complex homomorphism $\phi$, and $x^*=x$ does not imply $\phi(x)\in\Bbb R$


Let $A$ be the algebra of trigonometric polynomials $$f(t)=\sum_{n=-N}^Na_ne^{int},$$with  pointwise operations, $||f||=\sup_{t\in\Bbb R}|f(t)|$, and the usual involution $$f^*(t)=\overline{f(t)}=\sum_{n=-N}^N\overline{a_{-n}}e^{int}.$$That satisfies all the axioms except for completeness of the norm. But if $$\phi(f)=\sum_{n=-N}^Na_n2^n$$then $\phi$ is a complex homomorphism of $A$ with $\phi(f^*)\ne\overline{\phi(f)}$. And in fact $f^*=f$ does not imply $\phi(f)\in\Bbb R$.
(A few years ago I noticed that the proof that $x^*=x$ implies $\phi(x^*)\in\Bbb R$ in a $C^*$ algebra involved completeness, and this seemed "wrong". Spent a long time looking for a purely algebraic proof... When I finally noticed that example I decided that from then on when I was doing $C^*$ algebra things I'd assume I had a $C^*$ algebra, heh-heh.)

Edit It appears that all this started with an exercise.
Part (a) of the exercise is fine, and in fact we don't need to show that $\phi(x^*)=\overline{\phi(x)}$ to prove it. Instead we define an involution on the maximal ideal space by $\phi^*(x)=\overline{\phi(x^*)}$, and (a) follows easily.
But part (b)  is simply false (unless the definition of "involution" in the exercise includes something we haven't beem told about). 
Some authors require $||e||=1$ in the definition of Banach algebra with identity and some do not. If $||e||\ne1$ is allowed then it's clear that (b) is false: Given $A$ as in the exercise, define $$|||x|||=2||x||.$$Then $A$ is also a Banach algebra with the new norm, and (b) cannot be true of both norms (since $||\hat x||$ is independent of the choice of norm on $A$).
If $||e||=1$ is part of the definition, let $A$ be a commutative Banach algebra with an involution but without an identity. Append an identity in the usual way: Let $A'=A\times\Bbb C$, define a multiplication so that $e=(0,1)$ is an identity, and define $$||(x,\lambda)||=||x||+|\lambda|.$$Consider $$|||(x,\lambda)|||=2||x||+|\lambda|.$$
Part (c) is ok, in fact the hypothesis in (c) is what is usually taken as the definition. See Wikipedia for a conjecture regarding what the definition in the exercise might be. The equivalence of the two conditions in that Wikipedia article is a standard thing; however we prove it we'd better not use (b)...
