Unclear Question (MVT)

Question

The following is taken from question 1/II/11D (page 8):

Show that if $g: \mathbb{R} \to \mathbb{R}$ is twice continuously differentiable then, given $\varepsilon >0$, we can find a constant $L$ and $\delta(\varepsilon)>0$ such that $$|g(t)-g(\alpha)-g'(\alpha)(t-\alpha)| \le L|t-\alpha|^{2}$$ for all $|t-\alpha|<\delta(\epsilon)$

IMO this is a bit of a mess because it isn't clear where all the variables are quantified (and hence what they depend on), and I can't quite tell what the question is asking. It seems there are two possible approaches, a 'uniform' one and a 'non-uniform' one (I don't want to type them out in full but if it's unclear I will). Given the remainder of the question, I suspect the correct reading is this:

Let $\alpha \in \mathbb{R}$ and $g: \mathbb{R} \to \mathbb{R}$ be twice continuously differentiable at $\alpha$. Then, given $\varepsilon>0$, show that there exist $\delta$ and $L$ such that, for all $t$ such that $|t-\alpha|<\delta$, $$|g(t)-g(\alpha)-g'(\alpha)(t-\alpha)| \le L|t-\alpha|^{2}$$

My solution would then be to define $h(t) = g(t)-g(\alpha)-g'(\alpha)(t-\alpha)$, so that $h(\alpha) = h'(\alpha) = 0$, and $h''(t) = g''(t)$ for all $t$. By continuity of $h''(t)$ at $\alpha$, we can find a $\delta$ such that $|t-\alpha|<\delta \implies |h''(t)-g''(\alpha)| < 1$, say. Then $$g''(\alpha)-1<h''(t)<g''(\alpha)+1$$ so that $|h''(t)| < \max(|g''(\alpha)+1|,|g''(\alpha)-1|)$ (call the RHS $2L$). I claim then that for all $t$ such that $|t-\alpha|<\delta$, $$|h'(t)|<2L|t-\alpha|$$ by the mean value theorem: suppose not, then there exists a $t' \in (t-\alpha,t+\alpha)$ such that $|h'(t')|>2L|t'-\alpha|$. By MVT on $[\alpha,t']$ or $[t',\alpha]$, we get $t'' \in (t-\alpha,t+\alpha)$ with $$|h''(t'')| = \left|\frac{h'(t')-h'(\alpha)}{t'-\alpha}\right| > \frac{2L|t'-\alpha|}{|t'-\alpha|} \ge 2L$$ a contradiction. Similarly again we get $|h(t)| \le L|t-\alpha|^{2}$, as claimed.

A few questions:

1. Does my interpretation sound right?
2. Does my solution sound right?
3. Should I be concerned that my solution doesn't depend on $\varepsilon$ whatsoever?!
4. (bonus) Is it possible to wrap up this argument into a lemma? (i.e. roughly $|f'(t)|<k \implies |f(t)|<k|t-\alpha|$)

Answer 1

In what follows $\alpha $ is a fixed number. Also the letter $f$ has been used instead of the $g$ given in question.

We have via MVT $$f(t) - f(\alpha) - f' (\alpha) (t-\alpha) =\{f'(\beta)-f'(\alpha) \} (t-\alpha) \tag{1}$$ for some $\beta$ between $\alpha$ and $t$ so that $|\beta-\alpha|<|t-\alpha|$.

Next we have via definition of derivative $$f'(\beta) - f'(\alpha) =(\beta-\alpha) \{f''(\alpha) +\phi(\beta)\} $$ where $\phi(\beta) \to 0$ as $\beta\to\alpha $. Thus given any $\epsilon>0$ we can find a $\delta>0$ such that $|\phi(\beta) |<\epsilon $ whenever $0<|\beta-\alpha|<\delta$. Let $L=|f''(\alpha) |+\epsilon $ and then we can see that $$|f' (\beta) - f'(\alpha) |<L|\beta - \alpha|\tag{2}$$ Now our job is done. Starting with $\epsilon>0$ we have defined a constant $L>0$ and another number $\delta>0$ both dependent on $\epsilon$. If $|t-\alpha|<\delta$ then by $(1)$ and $(2)$ we have the equation $$|f(t) - f(\alpha) - f'(\alpha) (t-\alpha) |<L|t-\alpha|^{2}$$ You can see that only existence (and not continuity) of $f''$ at $\alpha$ is needed.