Extending a norm-like function defined on a cone Let $(X,||\cdot||)$ be a Banach space and let $\mathcal{C}$ be a cone in $X$ (that is, if $x\in\mathcal{C}$ then $tx\in\mathcal{C}$ for $t\geq 0$; also assume that if $x,y\in\mathcal{C}$ and $t,s\geq 0$ then $tx+sy\in\mathcal{C}$). Assume that there is a function $||\cdot||_{\mathcal{C}}$ defined on $\mathcal{C}$ that satisfies $||\lambda x||_{\mathcal{C}}=|\lambda|||x||_{\mathcal{C}}$ for real $\lambda$ and that $||x+y||_{\mathcal{C}}\leq||x||_{\mathcal{C}} + ||y||_{\mathcal{C}}$ for all $x,y\in\mathcal{C}$ and finally that $||x||_{\mathcal{C}}=0$ iff $x=0$. Additionally, for the specific reason I ask, we may assume that $||x||\leq||x||_{\mathcal{C}}$; however we can't assume any form of the opposite inequality (i.e. even with multiplicative constants). 
The question is this: can $||\cdot||_{\mathcal{C}}$ be extended to a norm $||\cdot||_{X}$ on $X$? There need not be any relation between $||\cdot||$ (the given norm on $X$) and this extension (in particular I don't need to have $||x||\leq ||x||_{X}$ as above). Also, we cannot assume that $||\cdot||_{\mathcal{C}}$ is ``monotone''; that is, if $x-y\in\mathcal{C}$ then we can't assume that $||y||_{\mathcal{C}}\leq||x||_{\mathcal{C}}$ (calling this monotone makes sense in the context I care about in which $\mathcal{C}$ is the cone of non-negative functions.)
As a simple example, let $(X,||\cdot||)$ be $L^1$ on the interval $[0,1]$ and let $\mathcal{C}$ be the cone of non-negative functions. Define $||f||_{\mathcal{C}}:=|\int_{0}^{1}f(t)dt|$. Then this satisfies the properties I listed above and the extension is just the $L^1$ norm. 
In the context I care about, $\mathcal{C}$ consists of the non-negative functions on a cube in $\mathbb{R}^n$. 
I have a feeling that this is either not true in general (and maybe I'm even missing an ``easy'' counter example) or that there is some way to put together Hahn-Banach type theorems that I can't see. I'd also appreciate any information you might have regarding my question. For example, does the thing I am asking about already have a name? 
 A: I don't know if it has a name, but it looks like a  straightforward construction. Please fill in the details in case I have overlooked something:
First step is to   extend the norm $\|\cdot\|_C$ on the subspace generated by $C$, that is on $Y=C-C$. We do so, by defining the norm of $x\in C-C$ to be 
$$\|x\|_Y=\inf\{\|y\|_C+\|z\|_C: x=y-z, \ \ y,z\in C\}.$$
Only the triangle inequality is a little tricky: Let $x, x'\in Y$ and $\varepsilon>0$. By the definition of $\|\cdot\|_Y$, there exist $y, y', z, z'\in C$ such that $x=y-z$, $x'=y'-z'$ and 
\begin{align}\|x\|_Y&>\|y\|_C \ +   \|z\|_C \ -\frac{\varepsilon}{2}\\
\|x'\|_Y&>\|y'\|_C+\|z'\|_C-\frac{\varepsilon}{2}\end{align}
Then \begin{eqnarray}\|x+x'\|_Y&\leq& \|y+y'\|_C+\|z+z'\|_C  
 \\ &\leq&\|y\|_C+\|z\|_C+\|y'\|_C+\|z'\|_C\\
 &<&\|x\|_Y+\|x'\|_Y+\varepsilon.\end{eqnarray}
Since $\varepsilon>0$ can be arbitrary small, we get the desired inequality.
So we extended the norm on $Y=C-C$. In your example (the non negative functions on the cube) the cone is generating, that is $X=C-C$, so you are done. In general the cone need not be generating, so we need one more step:
Second step is to extend the norm from $Y$ to $X$. 


Lemma: Let $Y$ be  a linear subspace of $X$ and $\|\cdot\|_Y$ be a norm on $Y$. Then there exists a norm on $X$ which extends $\|\cdot\|_Y$.
Proof:
    Let $B_Y$ be a  Hamel basis of $Y$ and we extend it to a Hamel basis $B_X\supset B_Y$ of $X$. Every $x\in X$ can be written uniquely in the form $x=y+\sum_{i \in F} \lambda_ib_i$, where $y\in Y$ and all of the $b_i$'s belong in $B_X\setminus B_Y$. We define 
    $$\|x\|=\|y+\sum_{i \in F}\lambda_ib_i \|=\|y\|_Y+\sum_{i\in F} |\lambda_i|.$$
Then $\|\cdot\|$ is a norm on $X$  that extends $\|\cdot\|_Y$. 


Notice also that if $Y$ has infinite codimension, then this norm can not be a Banach norm. Whether or not this is a useful norm extension is debatable. As you can see, since we didn't have any conditions that the extended norm should satisfy, it was easy to construct many such extensions using Zorn's lemma.  For example, if we demanded the norm extension to give a Banach space, or a vector lattice, then we wouldn't have so much freedom. The good thing is that this step is not needed in your context.
