Convex function with weights not summing to 1 Given a convex function, $x_i$ elements of some vector space (let's call them real numbers if we want), $\lambda_i \geq 0$, $\sum_i \lambda_i = 1$,  we have $f\left(  \sum_i \lambda_i x_i\right) \leq \sum_i \lambda_i f(x_i)$.
What are the most useful modifications of this when $\sum \lambda_i \neq 1$ (though $\lambda_i $ still non-negative)?
 A: Suppose $f:V\to\mathbb R$ is convex, where $V$ is a real vector space. Consider $g(x)=f(x)-f(0)$, which is also convex. For $0<\lambda<1$ and $x\in V$ we have $$g(\lambda x)\le \lambda g(x)+(1-\lambda) g(0) = \lambda g(x) \tag{1}$$
Given $\lambda_i\ge 0$ we can define $\lambda=\sum \lambda_i$ and apply convexity (1); then use convexity for $\lambda_i/\lambda$. The result is 
$$
g(\sum \lambda_i x)\le \lambda g\left(\sum (\lambda_i/\lambda)x_i\right) 
\le \lambda \sum (\lambda_i/\lambda)g(x_i) = \sum \lambda_i g(x_i)
$$
Returning to $f$, we conclude with 
$$
f(\sum \lambda_i x)-f(0) \le \sum \lambda_i (f(x_i)-f(0))
$$
which simplifies to 
$$
f(\sum \lambda_i x) \le f(0) \left(1-\sum \lambda_i\right)+\sum \lambda_i f(x_i) \tag{2}
$$
I don't know if (2) is actually useful when $f(0)\ne 0$ but I can imagine using it when $f(0)=0$.

old version
Dropping the requirement $\sum_i \lambda_{i}=1$, we arrive at the concept of a seminorm. Indeed, since both $$f(\lambda x)\le \lambda f(x)$$ and $$f(\lambda^{-1}(\lambda x))\le \lambda^{-1} f(\lambda x)$$  must hold for every $\lambda>0$, it follows that $$f(\lambda x) = \lambda f(x)$$ 
That is, $f$ is homogeneous (of degree $1$). Homogeneity and convexity are precisely what makes a function a seminorm. (It is a norm if it's also positive for nonzero $x$). 
