Let $f:[0,1]\rightarrow\mathbb{R}$ be a concave function with $f(1)=\sup_{t\in[0,1]} f(t)$. Then $f$ is non-decreasing in $[0,1]$. Does someone know how to prove this?
[Math] Increasing concave function
calculusconvex-analysisreal-analysis
calculusconvex-analysisreal-analysis
Let $f:[0,1]\rightarrow\mathbb{R}$ be a concave function with $f(1)=\sup_{t\in[0,1]} f(t)$. Then $f$ is non-decreasing in $[0,1]$. Does someone know how to prove this?
Best Answer
Assume there exist $x,y\in [0,1]$ with $x<y$ s.t. $f(x)>f(y)$. Then there exists $t\in (0,1)$ with $y = (1-t)x+t\cdot 1$ (because $y$ is in between $x$ and $1$). It follows that
$f(y) = f((1-t)x+t) \geq (1-t)f(x)+tf(1) > (1-t)f(y)+tf(1)$.
Hence $f(y) > f(1)$ (by subtracting $(1-t)f(y)$ and then dividing by $t$).
Contradiction.