Problems of MATH202 # 17 of the last week

Since I am busy doing my work in another course, this weekly post is delayed til today :(.
Of the problems, only 5, 6, 7 are of our interest.

Problem 5. Let $  f:\mathbb{R}\to\mathbb{R}$, be a three times differentiable function. If $  f(x)$ and $  f'''(x)$ are bounded functions on $  \mathbb{R}$, show that $  f'$ and $  f''$ are also bounded functions on $  \mathbb{R}$.

My way.
Just make good use of the expansion $  f(x+h)=f(x)+f'(x)h+\frac{1}{2}f''(x)h^2+\frac{1}{3!}f'''(x+\theta h)h^3$, for some $  \theta \in (0,1)$.

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Problem 6. If $  f(x)$ and $  g(x)$ are $  n$ times differentiable and $  f^{(n-1)}(x),g^{(n-1)}(x)$ are both continuous in $  [a,b$. Then there exists a number $  c \in(a,b)$ such that \[ \frac{\displaystyle f(b)-f(a)-\sum_{k=1}^{n-1}\frac{(b-a)^k}{k!}f^{(k)}(a)}{\displaystyle g(b)-g(a)-\sum_{k=1}^{m-1}\frac{(b-a)^k}{k!}g^{(k)}(a)}=\frac{(m-1)!}{(n-1)!}(b-c)^{n-m}\left(\frac{f^{(n)}(c)}{g^{(m)}(c)}\right).\]

NO IDEA, I am just able to prove the case when $  m=n$.

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Problem 7. Let $  f$ be $  p$ times differentiable on $  \mathbb{R}$ and let $  M_k=\sup\{|f^{(k)}(x)|:x\in\mathbb{R}\}<\infty$, $  k=0,1,2,\dots,p$ and $  p\ge 2$. Prove that $  M_1\leq \sqrt{2M_0M_2}$ and \[M_k\leq 2^{\frac{k(p-k)}{2}}M_0^{1-\frac{k}{p}}M_p^{\frac{k}{p}}$, for $  k=1,2,\dots,p-1.\]
My way.
(1) Same as the case of problem 5, replace $  h$ by $  -h$ to construct another equation (remember to choose different $  \theta$), subtract two equation, observe that $  0\leq 2M_0 +2M_1h+M_2h^2$ for any $  h$, while discriminant $  \leq 0$, we are done.
(2) In exactly the same manner as (1), we conclude that $  M_{j+1}\leq \sqrt{2M_jM_{j+2}}$ for all $  j\leq p-2$. Now we take product $  \prod_{j=m}^{n}$ on both sides, having \[ \sqrt{M_{n+1}M_{m+1}}\leq (\sqrt{2})^{n-m+1}\sqrt{M_{n+2}M_m}.
\] Before we proceed, we first consider two cases. If $  M_k=0$, then the inequality we are asked to prove obviously holds since right hand side is always non-negative. In case if $  M_k>0$, then we take the product $  \prod_{m=0}^{k-1}$ on both sides of $  \sqrt{M_{n+1}M_{m+1}}\leq (\sqrt{2})^{n-m+1}\sqrt{M_{n+2}M_m}$, it results in \[ M_n^k\leq \left(\frac{M_0}{M_k}2^{k(2n-k+1)/2}\right)M_{n+1}^k.\]
(I remember I have replaced $  n$ by $  n-1$ to make the inequality seem better), we are interested in this because it is a beautiful (in the sense of solving the problem) reccurence relation, we have a direct consequence \begin{align*}
 M_n^k&\leq \left(\frac{M_0}{M_k}\right)^{p-k}2^{\sum_{j=n}^{n+p-k-1}k(2j-k+1)/2}M_{n+p-k}^k\\
&=\left(\frac{M_0}{M_k}\right)^{p-k}2^{k(p-k)(2n-2k+p)/2}M_{n+p-k}^k.
\end{align*} Finally, we take $  n=k$, $  M_k^p\leq M_0^{p-k}2^{\frac{k(p-k)p}{2}}M_p^{k}$.