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Let $f: (0,\infty) \to \mathbb{R}$ be some continuous function. We say that $f$ is subadditive if the bound \begin{align} f(x+y) \leq f(x) + f(y) \tag{1} \end{align} holds. I was attempting to understand a bit better what it means for $f$ to be subadditive by understanding what the condition says about the associated function $g:(0,\infty) \to \mathbb{R}$ given by $g(x) = \frac{f(x)}{x}$ (note that $f$ can be recovered from $g$). One finds that $f$ is subadditive if and only if $g$ satisfies \begin{align} \frac{g(x+y) - g(x)}{y} + \frac{g(y+x) - g(y)}{x} \leq 0 \tag{2} \end{align} i.e. the sum of the slopes of two particular secants of $g$ is always nonpositive. This gives a bit of intution for subadditivity. For instance, one sees immediately that a sufficient condition for $f$ to be subadditive is that $g$ be (weakly) decreasing (so that both of the above summands are nonpositive). But, not all subadditive $f$ have $g$ decreasing and, in general, this condition on $g$ seems no less confusing than subadditivity. My vague question is:

Can someone "interpret" this condition on $g$? Or somehow else shed some light on subadditivity? Roughly, should I be able to tell (as one can for linear functions) whether subadditivity holds by glancing the graph of $f$ or $g$?

To stimulate discussion, let me close by pointing out a method of generating interesting examples of subadditive functions. If $x \mapsto \Phi_x$ is a semigroup homomorphism (i.e. $\Phi_{x+y} = \Phi_x \circ \Phi_y$ holds) from $(0,\infty)$ to the isometries of some metric space $(M,d)$, and $p \in M$, then, $f(x) = d(p, \Phi_x(p))$ is a subadditive function. For example, if $\Phi_x = \begin{pmatrix} \cos x & - \sin x \\ \sin x & \cos x \\ \end{pmatrix} $ viewed as an isometry of Euclidean space $\mathbb{R}^2$ and $p=(1,0)$, then this construction yields the subadditive function $f(x) = \sqrt{ (\cos x - 1)^2 + \sin^2 x}$. The associated $g$ is not decreasing in this case and, just from looking at the graph, it seems far from obvious that (2) holds.

Mike F
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    One result is Fekete's lemma which states that so long as $f$ is measurable, $\lim_{x\rightarrow \infty} g(x)$ exists. – vadim123 Jun 06 '13 at 00:32
  • @vadim123: Thanks for pointing this out. This sort of result is precisely the reason I introduced $g$ as an auxillary function. I should have mentioned this as motivation. – Mike F Jun 06 '13 at 00:38

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