Convex and bounded function is constant [duplicate]

Solution 1:

Suppose f is not constant, i.e., $\exists x,y\in\mathbb{R}:f(x)>f(y)$. Since f is convex, we have: $f(x)\leq\lambda f(\frac{x-(1-\lambda)y}{\lambda})+(1-\lambda)f(y)\;\;\;\forall\lambda\in(0,1).$

(This is just the definition of convexity, $f(\lambda x'+(1-\lambda)y')\leq\lambda f(x')+(1-\lambda)f(y')\;\;\;\forall\lambda\in(0,1)$, with $x=\lambda x'+(1-\lambda)y'$ and $y=y'$.)

Hence $\frac{f(x)-(1-\lambda)f(y)}{\lambda}\leq f(\frac{x-(1-\lambda)y}{\lambda}).$

Now, since $f(x)>f(y)$, $\frac{f(x)-(1-\lambda)f(y)}{\lambda}=\frac{f(x)-f(y)}{\lambda}+f(y)\rightarrow \infty$ as $\lambda\rightarrow0^+.$

Hence f is not bounded above.