Why for $f$ here it must be for $x, y \in [0, + \infty)$ then $f(x + y) \le f(x) + f(y)$

Question

I am reading in" A. Aizpuru, Density by moduli and statistical convergence, Quaest. Math. 37(4) (2014), 525–530."

Its link

In page 529:

Lemma $3.4.$ If $A \subset \mathbb N$ is infinite then there exists an unbounded module $f$ such that $d_f (A) = 1$.

Proof. Let us define $g : N \to N$ by $g(1) = 1$, $g(2) = \min\{n : |A(n)| = 2\}$ and if $k \ge 2$ then $g(k + 1) = \max \{ \min \{n : |A(n)| = 1 + g(k)\}, 2g(k) − g(k − 1)\}$

We have by construction that g is increasing and $|A(g(k + 1))| \ge 1 + g(k).$ Now define $f : [0,\infty) \to [0,\infty)$ by $f(0) = 0$, for $n \in \mathbb N $ let $f(g(n)) = n$ and finally extend $f$ to be piecewise linear in the remaining intervals. We observe that $\frac{f(g(2))−f(g(1))}{g(2)−g(1)} ≤ 1$ and if $k \ge 2$ it is $g(k + 1) − g(k) \ge g(k) − g(k − 1) $ and that is why $\frac{(k+1)−k}{g(k+1)−g(k)} \le \frac{k-(k−1)}{g(k)−g(k−1)} $, this is, $\frac{f(g(k+1))−f(g(k))} {g(k+1)−g(k)} \le \frac{f(g(k))−f(g(k−1))}{g(k)−g(k−1)}$ , therefore the corresponding slopes of the segments that form the graph of $f$ are decreasing and thus if $x, y \in [0, + \infty)$ then $f(x + y) \le f(x) + f(y)$

I can't see why if these slopes of segments are decreasing then it must be if $x, y \in [0, + \infty)$ then $f(x + y) \le f(x) + f(y)$. I tried to proving that by using the information in this proof above or by assuming the converse but I can't find any thing. Any help ?

Answer 1

We define $n(x)$ as the integer such that $g(n) \leq x < g(n+1)$. Our goal is to prove the following statement by induction:

Statement $\mathcal{P}(k)$: For all $x, y \in \mathbb{R}$ with $x < y $, if $n(y) = n(x) + k$, then $$\frac{y-x}{g(n(y) + 1) - g(n(y))} \leq f(y) - f(x) \leq \frac{y-x}{g(n(x) + 1) - g(n(x))}$$

Base Case: $k = 0$

If $n(x) = n(y)$, then $x$ and $y$ lie within the same interval $[g(n), g(n+1))$. Since $f$ is defined such that $$ f(y) - f(x) = \frac{y-x}{g(n+1) - g(n)}, $$ we see that both inequalities hold with equality.

Inductive Step

We assume $\mathcal{P}(k)$ holds for some $k$, and we aim to prove $\mathcal{P}(k+1)$.

Suppose $x < y$ with $n(y) = n(x) + (k+1)$. We decompose the difference $ f(y) - f(x)$ as follows: $$ f(y) - f(x) = \big( f(y) - f(g(n(y))) \big) + \big( f(g(n(y))) - f(g(n(x) + 1)) \big) + \big( f(g(n(x) + 1)) - f(x) \big). $$

By applying the induction hypothesis to $g(n(y))$ and $g(n(x) + 1)$, and using the fact that $g(n+1) - g(n)$ is increasing, we can establish the desired inequalities for $k+1$. Thus, $\mathcal{P}(k)$ holds for all $k$.

Conclusion

We have proven that for any $x < y$, $$ \frac{y-x}{g(n(y) + 1) - g(n(y))} \leq f(y) - f(x) \leq \frac{y-x}{g(n(x) + 1) - g(n(x))}. $$

Proving Concavity of $f$

Now, let $x < y$ and consider a convex combination $z = (1-t)x + ty$ with $t \in (0,1)$.

\begin{align} f(z) - (1-t) f(x) - t f(y) &= (1-t)(f(z) - f(x)) + t(f(z) - f(y))\\ &\ge (1-t)\frac{z - x}{g(n(z) + 1) - g(n(z))} + t\frac{z - y}{g(n(z) + 1) - g(n(z))}\\ &= 0 \end{align}

Thus, $f$ is concave.