Does there exist a norm on continuous real-valued function space? I know the space of continuous real-valued function on closed set can be given a norm by integral. How about the continuous funcion on the real line? It may be non-integrable, like f(x)=x^2. So, does there exist a norm on it, and how to construct it? 
 A: Assuming the axiom of choice, yes, it has many norms.  Use Zorn's lemma to choose a Hamel basis $B \subset C(\mathbb{R})$.  I claim $|B| = \mathfrak{c}$, i.e. $C(\mathbb{R})$ has Hamel dimension $\mathfrak{c}$.  Since evaluating at rationals gives an injection of $C(\mathbb{R})$ into $\mathbb{R}^{\mathbb{Q}}$, we have $|B| \le |C(\mathbb{R})| = |\mathbb{R}^{\mathbb{Q}}| = \mathfrak{c}$.  On the other hand, $C_0(\mathbb{R})$ (the functions vanishing at infinity) is a separable Banach space, hence has Hamel dimension $\mathfrak{c}$, and it embeds linearly into $C(\mathbb{R})$.
So if you pick any normed space $(X, \|\cdot\|_X)$ of Hamel dimension at least $\mathfrak{c}$, you can find an injection from $B$ to a Hamel basis of $X$, and extend this to a linear injection $T : C(\mathbb{R}) \to X$.  Now it's easy to verify that the pullback $\|f\|_T := \|Tf\|_X$ defines a norm on $C(\mathbb{R})$.
If you choose $X$ to have Hamel dimension exactly $\mathfrak{c}$, you can choose $T$ to be a linear isomorphism, and in that case the normed space $(C(\mathbb{R}), \|\cdot\|_T)$ is isometrically isomorphic to $(X, \|\cdot\|_X)$.  For instance, you could choose $X$ to be any separable Banach space,  in which case your norm $\|\cdot\|_T$ will be complete.  You could even choose $X = \ell^2$ and find a norm which makes $C(\mathbb{R})$ a separable Hilbert space.
This is of course rather non-constructive.  You might ask if there is an "explicit" norm for which we can write down a formula, or whose existence can be shown without the axiom of choice.  The answer to that is no.  As I showed in this Math.SE answer, it is consistent with ZF plus the axiom of dependent choice (DC, a weakened version of the axiom of choice which is still sufficient for most of real analysis) that there does not exist any norm on the vector space $C(\mathbb{R})$.  
The proof shows in ZF+DC that such a norm would let you construct a subset of a Polish space ($C(\mathbb{R})$ with the topology of local uniform convergence) which does not have the property of Baire, and then invokes the Solovay / Shelah result that it is consistent with ZF+DC that every subset of a Polish space has the property of Baire.
