Prove that $|f|_X\leq \|f\|$ for every $f\in C(X)$ Let $X$ be a compact space and let $\|.\|$ be an algebra norm on $ C(X)$  Show that $|f|_X \leq \|f\|$ for every $f\in C(X)$
Could anyone please suggest me how to deal with these questions
 A: You should note properly your question. You have a compact set $X$. You have a norm on $C(X)$ : $|f|_X = \sup_{x \in X} |f(x)|$ Let an another algebra norm $||.||$ on $C(X)$ and, the goal is to show that $||.||\geq |.|_X$.
Remark : you can check that $|.|_X$ is also a algebra norm.
A nice analogue of compacts is finite sets (with discrete topology). This avoids topological complications : the max and min occuring are obvious ; it also enable you to have plan to solve the problem in the general case. So let us solve the problem when $X$ is finite.
Then, $C(X)$ is just $\mathbf{R}^n$, the product is the product component by component, $|(x_1,..x_n)|_X$ is the max of components, and $||.||$ an algebra norm.
We are in a space of finite dimension, so the norms are equivalent : there exists $\lambda>0$ such that : 
$\frac{1}{\lambda}|.|_X \leq ||.|| \leq \lambda |.|_X$
Let $x\in \mathbf{R}^n$. For $k\in \mathbf{N}$, consider $\|| x^k ||^\frac{1}{k}$. We have : $\|| x^k ||^\frac{1}{k} \leq ||x|| $ ($||.||$ is a algebra norm) and $\|| x^k ||^\frac{1}{k} \geq \lambda^{-\frac{1}{k}} |x|_X $ (it is quite clear that $|.|$ commutes with the elevation to a certain power). So you obtain, by making $k\rightarrow \infty$ ; $|x|_X \leq ||x||$ and you have the result.
In the general case, you have two possibilities then to continue this kind of problem: 
The first way : find a way to generalize the min and max occuring in the finite case.
Here, it would be nice to show the following conjecture : 
Conjecture (?) : 
There exist $\lambda >0$ and $\mu>0$ such that : 
$\frac{1}{\mu} |.|_X \leq ||.|| \leq \lambda |.|_X  $
But... Theese inequalities are false (at least when ||.|| is not supposed to be an algebra norm : take $X = [0,1]$, and $||f|| = |f|_X + |f'|_X$) for the majoration, $ ||f|| = \int |f|$ for the minoration).
So this way does not look to give results.
The second way, use the finite case to try to make a demonstration. 
I suppose here that $K$ is metric.
Then $K$ is separable. So write $K$ as the closure of increasing union of finite set $D_n$, $n\in \mathbf{N}$. I will note $C(D_n)$ the space of (continuous, where $D_n$ has the discrete topology) functions $D_n \rightarrow \mathbf{R}$.
You have the induced norm on $C(D_n)$ :
$$|s|_{D_n} = sup_{x\in D_n} |f(x)|$$
Take $f_1, .. f_n$ in $C(X)$, and let $F$ the algebra generated by the $f_i$. We suppose that there exists an isomorphism of algebra (this is necessary to use the finite case) $i_F :  C(D_n) \rightarrow F$. 
So you have a norm induced by $||.||$ on $D_n$ :
$$||s||_{D_n, F} = ||i_F(s)||$$
It is by definition an algebra norm (it would not be the case if $i_F$ was just an isomorphism of vector spaces).
By the finite case we have $||s||_{D_n,F} \geq |s|_{D_n}$. The idea would be, for any $f\in C(X)$, to build $F$ big enough so that it can contain $f$.
But there is a huge problem : when you look at the images of the canonical basis $e_i$ by $i_F$, you see that $i_F(e_i)^2 = i_F (e_i)$, so $i_F (e_i)$ has values in $\{0,1\}$... So this kind of reasoning might work if $X$ is totally disconnected (nay if $X$ is a discrete space ! ... ), but will be a failure else.
I also tried to study the quantity, which is a priori only pseudo-norm (but, if the result is true, it must be a norm minorated by $|.|_{D_n}$) :  
$$||s||_{D_n} = inf_{f\in C(X), f|_{D_n} = s} ||f|| $$
This quantity is not infinite, using the fact that a function on a finit set admits an extension on $X$, because $X$ is compact, hence normal, and then it is possible to use Tietze Urysohn's extension theorem.
It is a pseudo-norm as an inf of (non generally algebra)-norms. To see it, consider all the sections of the application restriction $C(X)\rightarrow C(D_n)$ which induces linear isomorphisms $i_F : D_n \rightarrow F$ where $F$ is a subspace of $C(X)$, and show that the quantity I defined is the infimum of the $||i_F(s)||$.
But it is not obvious at all that it is an algebra (pseudo-)norm... Even if it was so, if is not obvious at all it is a norm. So I did not obtain result.
Conclusion : your question seems not obvious. It is quite hard to have documentation about algebra norms.
I've found a way to build some algebra norms with this lemma : 
Lemma : Let $I$ a set. If ||.|| is an algebra norm on $\mathbf{R}^I$ and $|.|_i$ an algebra norm on $E_i$ for $i\in I$ then there is an algebra norm on $\prod_{i\in I} E_i$ defined by : $(e_i) \rightarrow || (|e_i|_i) ||$ 
So as algebra norms, you have the max, the subordinate norms, and their compositions... But your result is quite trivial in theese cases (your result is compatible with composition).
For instance, if I try $X = [0,1]$ and :
$||f|| = max_{x,y,z,t\in X} ||| \begin{pmatrix}
f(x) & f(y) \\
f(z) & f(t)
\end{pmatrix} |||$ where |||.||| is some subordinate matricial norm, you can just use that the subordinate norm is greater than the max to show the result.
Maybe I am missing something obvious. But I don't see any proof nor counterexample (even if $X$ is not supposed compact).
