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Let $X$ be a compact space and let $|.|$ be an algebra norm on $ C(X)$ Show that $|f|_X leq |f|$ for every $fin C(X)$

Could anyone please suggest me how to deal with these questions

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 $xin mathbf{R}^n$. For $kin mathbf{N}$, 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 $krightarrow infty$ ; $|x|_X leq ||x||$ 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 $D_n$, $nin mathbf{N}$. 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 $C(D_n)$ :
$$|s|_{D_n} = sup_{xin D_n} |f(x)|$$

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 $fin C(X)$, 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 $|.|_{D_n}$) :

$$||s||_{D_n} = inf_{fin C(X), f|_{D_n} = s} ||f|| $$
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 $iin I$ then there is an algebra norm on $prod_{iin I} E_i$ 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 :
$||f|| = max_{x,y,z,tin X} ||| begin{pmatrix}
f(x) & f(y) \
f(z) & f(t)
end{pmatrix} |||$ 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).