Q:

Convexity definition when $\lambda \in \mathbb{R} \setminus (0,1)$

We are given the standard definition
$$f(\lambda x + (1 - \lambda) y) \leq \lambda f(x) + (1-\lambda) f(y)$$
for $\lambda \in (0,1)$. I am trying to prove that the opposite inequality is true when $\lambda \in \mathbb{R} \setminus (0,1)$, i.e.
$$f(\lambda x + (1 - \lambda)y) \geq \lambda f(x) + (1 - \lambda) f(y).$$
How can I derive this second inequality from the first? Or is it just a given? (My teacher uses this second inequality in a exercise solution, particularly Baby Rudin 4.23, when we are only given the first as a definition).

A:

Suppose that for some $\lambda_0\notin [0,1]$ we have that 
$$
f(x_0):= f(\lambda_0 x+ (1-\lambda_0)y) < \lambda_0 f(x) + (1-\lambda_0)f(y).
$$
Notice that $x_0,\ x, \ y$ are all collinear and either $y$ is between $x_0$ and $x$ (in case $\lambda_0<0$), or else $x$ is between $x_0$ and $y$ (in case $\lambda_0>1$). Assume it's the former. Then for $s\in [0,1]$ given by $s=-\lambda_0/(1-\lambda_0)$ we must have $y=sx+(1-s)x_0$, and by convexity
$$
f(y)=f(sx+(1-s)x_0) \leq sf(x)+(1-s)f(x_0).
$$
Using the bound for $f(x_0)$ that we got by assumption at the start we arrive at
$$
f(y)<\frac{-\lambda_0}{1-\lambda_0}f(x) + \frac{1}{1-\lambda_0} \left( \lambda_0 f(x) +(1-\lambda_0) f(y) \right).
$$
Simplifying we get $0<0$, a contradiction.

