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Do you know an explicit norm on $\mathcal{C}^0(\mathbb{R},\mathbb{R})$? Using the axiom of choice, every vector space admits a norm but have you an explicit formula on $\mathcal{C}^0(\mathbb{R},\mathbb{R})$?
A related question is: Can we proved that $\mathcal{C}^0(\mathbb{R},\mathbb{R})$ has a norm without the axiom of choice?
The answer is no. There is no explicit norm on $\mathcal{C}^0(\mathbb{R}, \mathbb{R})$; constructing any norm on this space requires the axiom of choice to be used in an essential way.
In my answer to the (newer) question Inner product on $C(\mathbb R)$, I show that it is consistent with ZF+DC that there does not exist a norm on the vector space $\mathcal{C}^0(\mathbb{R}, \mathbb{R})$ (called $C(\mathbb{R})$ in that question).
Refining @Mebat's answer: the seminorms on $C^o(\mathbb R)$ (here meant to be continuous, real-valued functions on $\mathbb R$, with no decay or boundedness restrictions) given by $\nu_K(f)=\sup_{x\in K} |f(x)|$ for compact subsets $K$ of $\mathbb R$, give a Frechet-space (complete, locally convex, metric) structure on $C^o(\mathbb R)$. As @Mebat notes, there is a countable subset, e.g., $[-n,n]$ of compacts which give that topology. Then the usual trick of writing $$ d(f,g)=\sum_n {1\over 2^n}\cdot {\nu_{[-n,n]}(f-g)\over 1+\nu_{[-n,n]}(f-g)} $$ gives a (non-canonical) metric.
Significantly, this makes $C^o(\mathbb R)$ complete. We almost always want to "give" TVS's topologies with the best completeness properties possible.
But this is not a norm, only a metric.
There is a reasonable criterion for normability of TVS's (once a topology is given), namely, that every neighborhood of $0$ is "absorbing", meaning that sufficiently large dilates contain a given bounded set. In the present example, the fact that continuous functions can blow up arbitrarily fast enables construction of counter-examples to a claim of normability, with the natural topology.
Here some ideas to find an explicit norm on $\mathcal{C}^0(\mathbb{R},\mathbb{R})$. If we define $\Vert\cdot\Vert$ like $$\Vert f\Vert=\sup_{x\in\mathbb{R}} \,\,\frac{\vert f(x)\vert}{1+\vert f(x)\vert}$$
for every $f\in\mathcal{C}_0(\mathbb{R},\mathbb{R})$, then:
(i) $\Vert f\Vert=0$ iff $f=0$
(ii)$\Vert f+g\Vert\leq\Vert f\Vert+\Vert g\Vert$
but (iii) $\Vert \alpha f\Vert=|\alpha|\Vert f\Vert$, is not satisfied.
To fix (iii) we can do the following. Consider on $\mathcal{C}^0(\mathbb{R},\mathbb{R})$ the relation $\sim$ defined as follows: for $f,g \in \mathcal{C}^0(\mathbb{R},\mathbb{R})$, $f\sim g $ iff $\exists c\in\mathbb{R}, c\neq0$ such that $f=cg$. This is an equivalence relation. Then we have a partition of $\mathcal{C}^0(\mathbb{R},\mathbb{R})$ in classes. Denote by $[f]$ the class of $f\in\mathcal{C}^0(\mathbb{R},\mathbb{R})$. The class of the constant zero function is the singleton $\{0\}$. Notice that if $a\in \mathbb{R}$ is such that $f(a)\neq0$, then $g(a)\neq0$ for every $g\in[f]$. Hence, for every class $F=[f]$ different from $[0]$, we can choose $\alpha_F\in\mathbb{R}$ such that $f(\alpha_F)\neq0$ for each $f\in F$. Since we have chosen the numbers $\alpha_F$, now we can choose a representative function for each class in a unique way. For every class $F=[f]\neq[0]$ there is a unique function $\hat{f}\in F$ such that $\hat{f}(\alpha_F)=1$. Put $\hat{0}=0$. Now we can define $\Vert\cdot\Vert$ on $\mathcal{C}_0(\mathbb{R},\mathbb{R})$ by setting
$$\Vert f\Vert=\sup_{x\in\mathbb{R}} \,\,\frac{\vert f(x)\vert}{1+\vert \hat{f}(x)\vert}$$ for every $f\in\mathcal{C}_0(\mathbb{R},\mathbb{R})$.
This satisfies (i) and (iii) but we have lost (ii). If we could choose each $\hat{f}$ in such a way that $\vert\widehat{(f+g)}(x)\vert\leq\vert\hat{f}(x)\vert+\vert\hat{g}(x)\vert$ for all $x\in\mathbb{R}$ and for all $f,g\in\mathcal{C}_0(\mathbb{R},\mathbb{R})$, then the function $\Vert\cdot\Vert$ resultant will be a norm.
