How to prove that for all $k\in\mathbb N$, $h(kx)=kh(x)$ and $h(x+y)\le h(x)+h(y)$? Suppose $X$ is a commutative monoid and $f:X\to\mathbb R\cup\{\infty\}$ a function
and
$$g(x)=\inf\left\{\sum_{i=1}^nf(x_i)~\middle\vert~\sum_{i=1}^nx_i=x,n\in\mathbb N\right\}$$
$$h(x)=\inf\left\{\frac{g(mx)}m ~\middle\vert~ m\in\mathbb N\right\}$$
then how to prove that for all $k\in\mathbb N$, $h(kx)=kh(x)$ and $h(x+y)\le h(x)+h(y)$?
I proved that $g(x+y)\le g(x)+g(y)$, so I can write
$$h(kx)=\inf\left\{\frac{g(mkx)}m ~\middle\vert~ m\in\mathbb N\right\}=k\inf\left\{\frac{g(mkx)}{mk} ~\middle\vert~ m\in\mathbb N\right\}$$
since $g(x+y)\le g(x)+g(y)$, we have $\frac{g(mkx)}{mk}\le\frac{kg(mx)}{mk}$
So,
$$h(kx)\le k\frac{g(mx)}m$$
therefore
$$h(kx)\le kh(x)$$
on the other hand
$$kh(x)=\inf\left\{\frac{kg(mx)}m ~\middle\vert~ m\in\mathbb N\right\}$$
and since $\frac{kg(mx)}m\le\frac{km}mg(x)$ we have
$$kh(x)\le kg(x)$$
Any hint to continue?
 A: Let $k \in \mathbf N$. Note that 
\begin{align*} h(kx) &= k\inf \left\{ \frac{g(mkx)}{mk} : m \in \mathbf N\right\} \\&= k\inf\left\{\frac{g(m'x)}{m'} : m' \in k\mathbf N\right\} \\&\le k\inf \left\{ \frac{g(m'x)}{m'} : m' \in \mathbf N\right\} \\&= kh(x). \end{align*}
For the other direction, given $\epsilon > 0$, there is $m \in \mathbf N$ such that $g(mkx) -m\epsilon \le mh(kx)$. As, by definition of $h$, $h(x) \le g(mkx)/mk$, we have
\begin{align*}
  kh(x) &\le k\frac{g(kmx)}{km}\\
        &\le \frac{mh(kx) + m\epsilon}{m}\\
       &= h(kx) + \epsilon
\end{align*}As $\epsilon>0$ was arbitrary, $kh(x) \le h(kx)$. 
For the triangle inequality, given $\epsilon > 0$, choose $m_x, m_y \in \mathbf N$, such that $\frac{g(m_xx)}{m_x} \le h(x) + \epsilon$, and $\frac{g(m_yy)}{m_y} \le h(y) + \epsilon$. Let $m = m_xm_y$, then by what you have shown for $g$ 
\begin{align*}
  \frac{g(mx)}{m} &= \frac{g(m_y m_x x)}{m_x m_y}\\
                  &\le \frac{m_y g(m_x x)}{m_x m_y}\\
                  &\le h(x) + \epsilon
\end{align*}
Along the same lines $\frac{g(my)}m \le h(y) + \epsilon$. Now we have
\begin{align*}
  h(x+y) &\le \frac{g(mx + my)}{m}\\
         &\le \frac{g(mx) + g(my)}{m}\\
         &\le h(x) + h(y) + 2\epsilon
\end{align*}
And the result follows. 
