Let $a, b \in \mathbb{R}$ with $a<b$ and let $f:(a, b) \rightarrow \mathbb{R}$.

(a) Define what it means for $f$ to be continuous at $y_{0} \in(a, b)$.

$f$ is said to have a local minimum at $c \in(a, b)$ if there is some $\varepsilon>0$ such that $f(c) \leqslant f(x)$ whenever $x \in(a, b)$ and $|x-c|<\varepsilon$.

If $f$ has a local minimum at $c \in(a, b)$ and $f$ is differentiable at $c$, show that $f^{\prime}(c)=0$.

(b) $f$ is said to be convex if

$$f(\lambda x+(1-\lambda) y) \leqslant \lambda f(x)+(1-\lambda) f(y)$$

for every $x, y \in(a, b)$ and $\lambda \in[0,1]$. If $f$ is convex, $r \in \mathbb{R}$ and $\left[y_{0}-|r|, y_{0}+|r|\right] \subset(a, b)$, prove that

$$(1+\lambda) f\left(y_{0}\right)-\lambda f\left(y_{0}-r\right) \leqslant f\left(y_{0}+\lambda r\right) \leqslant(1-\lambda) f\left(y_{0}\right)+\lambda f\left(y_{0}+r\right)$$

for every $\lambda \in[0,1]$.

Deduce that if $f$ is convex then $f$ is continuous.

If $f$ is convex and has a local minimum at $c \in(a, b)$, prove that $f$ has a global minimum at $c$, i.e., that $f(x) \geqslant f(c)$ for every $x \in(a, b)$. [Hint: argue by contradiction.] Must $f$ be differentiable at $c$ ? Justify your answer.