Exercise 5.3.6

Question

\begin{enumerate}[label=(\alph*)] 
  \item Let $g:[0, a] \rightarrow \mathbf{R}$ be differentiable, $g(0)=0$, and $\left|g^{\prime}(x)\right| \leq M$ for all $x \in[0, a] .$ Show $|g(x)| \leq M x$ for all $x \in[0, a] .$
  \item Let $h:[0, a] \rightarrow \mathbf{R}$ be twice differentiable, $h^{\prime}(0)=h(0)=0$ and $\left|h^{\prime \prime}(x)\right| \leq$ $M$ for all $x \in[0, a] .$ Show $|h(x)| \leq M x^{2} / 2$ for all $x \in[0, a] .$
  \item Conjecture and prove an analogous result for a function that is differentiable three times on $[0, a]$.
   \end{enumerate}

Ulisse Mini · Accepted Answer

\begin{enumerate}[label=(\alph*)] 
  \item For $x \in [0,a]$, apply MVT to find a $c \in [0,x]$ with
    $$g'(c) = \frac{g(x)}{x} \implies g(x) = g'(c)x \implies |g(x)| \le Mx$$
  \item This is a special case of the theorem that if $f(0) = g(0) = 0$ and $f'(x) \le g'(x)$ for all $x \in [0,a]$ then $f(x) \le g(x)$. To prove this note how letting $h(x) = g(x) - f(x)$ changes the statement into $h'(x) \ge 0$ implying $h(x) \ge 0$. Which is true since MVT to get $c \in [0,x]$ implies $h'(c) = h(x)/x \ge 0$ thus $h(x) \ge 0$.

Now returning to $|h''(x)| \le Mx^2/2$ apply the above result to both cases in the inequality
    $$
    \begin{aligned}
      -Mx^2/2 \le h''(x) \le Mx^2/2 &\implies -Mx \le h'(x) \le Mx \
                                    &\implies -M \le h(x) \le M \
                                    &\implies |h(x)| \le M
    \end{aligned}
    $$
    Which proves $|h(x)| \le M$.
  \item I conjecture $|f(x)| \le x^3/6$ when $f(0) = f'(0) = f''(0) = 0$. The proof is the same as (b), except we differentiate one more time.
   \end{enumerate}

Exercise 5.3.6

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Exercise 5.3.6

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