Let $f : \mathbb{R}_+ \to \mathbb{R}$ be a (differentiable) function that satisfies the following
- $f(x) = 0\iff x =0$
- $\forall x > 0, f(x) \geq 0$
- $f$ is stricly concave
Show that for any $x, y \in \mathbb{R}_+$ we have $f(x+y) \leq f(x) + f(y)$
My lecturer did the following proof in class
Lecturers proof: For any $x, y \in \mathbb{R}_+$ we have $$f(x + y) - f(y) = \int_o^x f'(s+y)ds \leq \int_o^x f'(s) ds = f(x) - f(0) = f(x)$$ which implies $f(x+y) - f(y) \leq f(x)$ which implies $f(x+y) \leq f(x) + f(y)$. $\square$
Now the only part in the proof I can't understand is how we can conclude $$\int_o^x f'(s+y)ds \leq \int_o^x f'(s) ds$$
Why does this inequality hold? Is it because $f$ is concave so it's derivative $f'$ is monotonically decreasing, hence $f'(s+y) \leq f'(s)$ for any $y \geq 0$, thus by the properties of integrals we arrive at the following inequality?
The strict concavity is not necessary, concavity is enough. As I said in my comment above, the differentiability assumption is also redundant.
Let $x,y\geqslant 0.$ If $x+y=0$ then $x=y=0$ and it is nothing to prove: $f(0+0)\leqslant f(0)+f(0)$, both sides are zero.
Let $x+y>0$. Then $$x=\frac{y}{x+y}\cdot 0+\frac{x}{x+y}\cdot(x+y).$$ The (nonnegative) scalars $\dfrac{x}{x+y}$ and $\dfrac{y}{x+y}$ sum up to $1$. By concavity and $f(0)=0$ we get $$f(x)\geqslant\frac{x}{x+y}\cdot f(x+y).$$ Similarly (by symmetry) $$f(y)\geqslant\frac{y}{x+y}\cdot f(x+y).$$ Summing up these inequalities we arrive at $$f(x)+f(y)\geqslant f(x+y)$$ and the proof is finished.