Proving an Inequality for Upper Right-Hand Dini Derivatives

Question

Suppose $f$ and $g$ are functions from $\Bbb{R} → \Bbb{R}$. I wish to show that for any point $x_0 ∈ \Bbb{R}$, we have $D^+(f + g)(x_0) ≤ D^+f(x_0) + D^+g(x_0)$ where $D^+$ denotes the upper right-hand Dini derivative. So I tried to start with

$$ \limsup_{h \to 0^+} \frac{(f+g)(x_0 + h) - (f+g)(x_0)}{h}≤\limsup_{h \to 0^+} \frac{f(x_0 + h) - f(x_0)}{h}+\limsup_{h \to 0^+} \frac{g(x_0 + h) - g(x_0)}{h} $$

but I am not sure how to proceed.

Answer 1

First use that $(f+g)(x)=f(x)+g(x)$ by definition.

$\limsup_{h \to 0^+} \frac{(f+g)(x_0 + h) - (f+g)(x_0)}{h}=\limsup_{h \to 0^+} \frac{f(x_0+h)+g(x_0 + h) - f(x_0)-g(x_0)}{h}=\limsup_{h \to 0^+}\frac{f(x_0+h)- f(x_0)+g(x_0 + h) -g(x_0)}{h}=\limsup_{h \to 0^+}\frac{f(x_0+h)- f(x_0)}{h}+\frac{g(x_0 + h) -g(x_0)}{h} (\ast)$.

Now, using subadditivity of $\limsup$

$(\ast) ≤\limsup_{h \to 0^+} \frac{f(x_0 + h) - f(x_0)}{h}+\limsup_{h \to 0^+} \frac{g(x_0 + h) - g(x_0)}{h}.$

And you're done.