once, twice, thrice differentiable So I am going through abbotts 2nd edition of Understanding analysis and I would like to know if this is correct.
Exercise:
(a) Let $g : [0, a]→R$ be differentiable, $g(0) = 0$, and $|g'(x)| ≤M$ for all $x∈[0, a]$. Show $|g(x)| ≤Mx$ for all $x∈[0, a]$.
(b) Let $h : [0, a]→R$ be twice differentiable, $h'(0) = h(0) = 0$ and $|h''(x)| ≤ M$ for all $x∈[0, a]$. Show $|h(x)| ≤Mx^{2}/2$ for all $x∈[0, a]$.
(c) Conjecture and prove an analogous result for a function that is differentiable three times on $[0, a]$.
My solution:
(a)  $g : [0, a]→R$ and $g(0) = 0$ and $|g'(x)| \leq M$ for all $x \in [0,a]$. 
For any $x \in (0,a]$ we have by the mean value theorem there exists $c \in (0,x)$ such that $g(x) - g(0) = (x-0) g'(c) $[By MVT on $[0,a]$].
$|g(x)| = |x g'(c)| $ [as $g(0) = 0$]
$|g(x)|= |x| \cdot |g'(c)|$. As $|g'(c)| \leq M$ and $x \in (0,a]$. Therefore $|x|=x$. Hence $|g(x) | \leq Mx$ for all $x \in (0,a]$ when $x=0$ the inequality holds trivially. Thus $|g(x)| \leq Mx$ for all $x \in [0,a]$.
(b) $h: [0,a] \rightarrow \mathbb{R}$ is twice differentiable, $h'(0)=h(0)=0$. And $|h''(a)| \leq M \forall x \in [0,a]$. For any $x \in [0,a]$ we have 
$h(x) = h(0) + xh'(0) + (x^{2}/2!) h''(c)$ where $c \in (0,x)$. As $h(0)=h'(0)$, so $h(x)= (x^2/2) h''(c)$
$|h(x)| = |(x^2/2) h''(c)| \leq Mx^2/2$ [as $|h''(c)| \leq M$]
(c) Let $ f:[0,a] \rightarrow \mathbb{R}$ is three times differentiable and $f''(0)=f'(0)= f(0) =c$. Also $|f'''(x)| \leq M$ for all $x \in [0,a]$. For any $x \in [0,a]$, we have by the taylor series expantion of the given function $f(x)$about the point '$0$' in the interval [0,x]
$f(x) = f(0) + x f'(0) + (x^2/2!)f''(0) + (x^3/3!) f'''(c)$ where $c \in (0,x)$ 
So $|f(x)|= (x^3/3!) |f'''(c)|$, as $x\geq 0$
$|f(x)| \leq Mx^3/(3!)$ as $|f'''(c)| \leq M$. So the given statement is true in this case.

I only read your proof for (a), and it is slightly confusing. I think it might have the general idea right, but it is very unclear because of the poor presentation. You need to fix these issues:


*

*Currently, you say that by MVT on $[0,a]$, you know there exists some $c\in(0,x)$ such that $g(x)-g(0) = (x-0)\cdot g'(c)$, but this is simply untrue. MVT on $[0,a]$ tells you there is some $c$ in $(0, a)$ such that $g(a)-g(x) - (a-x)\cdot g'(c)$. You need to correct your justification here.

*You say $|g(x)|= |n||g'(c)|$, but you did not define $n$. This statement is currently meaninless.

*You say "Therefore $|x|=|x|$". This is a tautology that is completely irrelevant.

