Proving that a point is a local minimum using $\varepsilon$-$\delta$ proofs

Question

I theoretically understand this question and we've covered the basics of $\delta$-$\epsilon$ proofs but I really have no clue where to start here so please don't ask what I've got so far because I've got nothing. This is for a question sheet handed out during my real analysis course - no answers are provided so no way to get hints or anything.

Suppose $f: \mathbb{R} \rightarrow \mathbb{R}$ is $C^2$ everywhere and suppose that $c \in \mathbb{R}$ satisfies $f' (c) = 0$ and $f'' (c) > 0$

a) Prove that there exists $\delta > 0$ such that $|x - c| < \delta \implies f'' (x) > 0$

b) Prove that $f'$ is increasing on $[c - \delta, c + \delta]$ and deduce that $f' (x) > 0$ for $x \in (c, c + \delta)$

c) Prove that $c$ is a local minimum for $f$

Answer 1

A useful result:

Suppose that $f : \mathbb R \to \mathbb R$ is continuous at $x_0 \in \mathbb R$ and that $f(x_0)>0$. Then $f$ is positive in a neighborhood of $x_0$, that is, there exists $\delta>0$ such that $f(x)>0$ for all $x$ in the interval $(x_0-\delta,x_0+\delta)$.

Proof : Since $f$ is continuous at $x_0$, for $\varepsilon = f(x_0)/2$ there is a $\delta>0$ such that $|f(x)-f(x_0)| < \varepsilon$ for all $x \in (x_0-\delta,x_0+\delta)$. But this means that, for all $x \in (x_0-\delta,x_0+\delta)$, we have $$f(x)-f(x_0)>-f(x_0)/2,$$ and then $f(x)>f(x_0)/2>0$.

Answer 2

a) is basic.

$f''(c)>0$ so use that as your epsilon. ' As $f$ is $C^2$, $f'$ is continuous at $c$ so for $\epsilon = f''(c)>0$ there is a $\delta$ so that $|x-c| < \delta \implies |f''(x) - f''(c)| < f''(c)\implies -f''(c) < f''(x)-f''(c) < f''(c)\implies 0 < f''(x) < 2f''(c)\implies f''(x) > 0$.

That's all there was to it.

b) As $f''(x)> 0$ on $[c-\delta, c+\delta]$ the $f'(x)$ is increasing on that interval. As $f'(c) = 0$ and $f'(x)$ is increasing on $(c-\delta,c+\delta)$ then $f'(x) > 0$ for $c < x < x+\delta$. (And FWIW $f'(x) < 0$ for $x \in (c-\delta, c)$.

c) well $f'(x)>0$ on $(c, c+\delta)$ then $f(x)$ is increasing, and as $f'(x)< 0$ on $(c-\delta, c)$ then $f(x)$ decreasing so $f(c)$ must be a local minimum.