Let $f:\mathbb{R}^n\to\mathbb{R}$ be a function satisfying $f(x+y)\le f(x)+f(y)$ and $f(tx)=tf(x)$ for all $t\ge 0$, all $x,y\in\mathbb{R}^n$. Prove that the function $g(t):=\frac{1}{t}(f(x+ty)-f(x))$ is monotone for all $t\neq 0$.
I manage to show that $f$ is a convex function, but I struggle to proceed. I considered the difference: $$ \begin{align*} g(a)-g(b) = \frac{1}{ab} (bf(x+ay)-af(x+by)-(b-a)f(x)) \end{align*} $$ but due to the contraints $t\ge 0$, I couldn't make use of the convexity of $f$. Is there any promising way I can proceed?
For $\lambda>1,$ compute $$\begin{align} g(\lambda a)-g(a)=\frac{f(x+\lambda a y)-f(x)-\lambda f(x+ay)+\lambda f(x)}{\lambda a}\tag1\end{align}$$
So you need $$f(x+\lambda ay)-f(\lambda x+a\lambda y)+f((\lambda-1)x)\geq 0,$$ or possibly in the other direction. (But all basic examples of $f,$ norms of $\mathbb R^n,$ have $g(t)$ increasing and converging to $f(y),$ so we will check that first.)
$(1)$ is equivalent to:
$$f(\lambda x+a\lambda y)\leq f(x+\lambda ay)+f((\lambda-1)x).$$
But $$\lambda x+a\lambda y =(x+\lambda ay)+(\lambda-1)x.$$
This only solves for $b>a>0.$
We get it is increasing for $t<0$ by considering the same function for $(x_0,y_0)=(x,-y).$ Then $g_0(t)=-g(-t),$ and we know $g_0(t)$ is increasing for $t>0,$ so $g(t)$ is increasing for $t<0.$
Finally, we need $a<0<b.$ It seems we might want to show:
$$\lim_{t\to0^+} g(t)\geq \lim_{t\to 0^+} g(-t).$$
But this can be written as:
$$\lim_{t\to0^+}\frac{f(x+ty)-f(x-ty)}{t}\geq 0.$$
But $$f(x+ty)\leq f(x-ty)+tf(2y).$$
So we need to know that $f(2y)\geq 0?$ Since $y$ is arbitrary, this would mean that $f$ has to be non-negative.
Not sure where to go from there.
Using $\lambda a>a$ rather than $b>a$ makes the common denominator simpler, making the formula $(1)$ cleaner. We can always get back to $b>a$ by taking $\lambda=\frac ba.$