= Solution
The upper bound follows immediately from <convexity>, since
$$
y_0+\lambda r=(1-\lambda)y_0+\lambda(y_0+r).
$$
For the lower bound, write
$$
y_0=\frac1{1+\lambda}(y_0+\lambda r)
+\frac\lambda{1+\lambda}(y_0-r).
$$
Applying convexity and rearranging gives
$$
\boxed{(1+\lambda)f(y_0)-\lambda f(y_0-r)
\leq f(y_0+\lambda r)
\leq(1-\lambda)f(y_0)+\lambda f(y_0+r)}.
$$
Choose a fixed $r_0>0$ with $[y_0-r_0,y_0+r_0]\subset(a,b)$. Applying these bounds with $r=r_0$ and $r=-r_0$ shows that the difference $f(y_0+h)-f(y_0)$ is trapped between quantities of order $|h|/r_0$. Both tend to zero, so every <convex function> on an open interval is continuous.
Now suppose $c$ is a local minimum but some $x$ satisfies $f(x)<f(c)$. For sufficiently small $\lambda>0$, the point $z=(1-\lambda)c+\lambda x$ lies in the neighbourhood on which $c$ is minimal, whereas convexity gives
$$
f(z)\leq(1-\lambda)f(c)+\lambda f(x)<f(c),
$$
a contradiction. Thus
$$
\boxed{f(x)\geq f(c)\text{ for every }x\in(a,b)}.
$$
Differentiability at the minimizer is unnecessary: $f(x)=|x-c|$ is convex, has its global minimum at $c$, and is not differentiable there.
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