Boundedness of high order derivatives

Question

Let $f:\mathbb{R}\to\mathbb{R}$ a function differentiable $p$ times in $\mathbb{R}$, such that $f$ and $f^{(p)}$ bounded. Consequently, I would show that all intermediate derivatives $f^{(1)},...,f^{(p-1)}$ are also bounded on $\mathbb{R}$.

By myself, I succedeed in the case $p=2$ in this way:

$$f(x_0+h)=f(x_0)+f^{(1)}(x_0)h+f^{(2)}(\xi _1)\frac{h^2}{2}$$ $$f(x_0-h)=f(x_0)-f^{(1)}(x_0)h+f^{(2)}(\xi _2)\frac{h^2}{2}$$

that are true $\forall x_0\in\mathbb{R}$ and $\forall h>0$ with $\xi_1\in (x_0, x_0+h)$ and $\xi_2\in (x_0-h, x_0)$.

So subtracting the first equation and the second:

$$f(x_0+h)-f(x_0-h)=2hf^{(1)}(x_0)+\frac{h^2}{2}\left( f^{(2)}(\xi _1)-f^{(2)}(\xi _2) \right)$$

and so:

$$f^{(1)}(x_0)=\frac{f(x_0+h)-f(x_0-h)}{2h}-\frac{h}{4}\left( f^{(2)}(\xi _1)-f^{(2)}(\xi _2) \right)$$

taking the absolute value of both members and using the triangular inequality at right member:

$$|f^{(1)}(x_0)|\leq \frac{|f(x_0+h)|+|f(x_0-h)|}{2h}+\frac{h}{4}\left( |f^{(2)}(\xi _1)|+|f^{(2)}(\xi _2)| \right)$$

from which:

$$|f^{(1)}(x_0)|\leq \frac{M_{0}}{h}+\frac{h}{2}M_{2}$$

where $M_{0}:=\sup_{x\in\mathbb{R}}|f(x)|$ and $M_{2}:=\sup_{x\in\mathbb{R}}|f^{(2)}(x)|$.

Consequently:

$$\sup_{x\in\mathbb{R}}|f^{(1)}(x)|=:M_1\leq \frac{M_{0}}{h}+\frac{h}{2}M_{2}$$.

In a similar way I also proved the claim in the case $p=3$, but I can't start the induction chain.

Can you help me? Thanks.

Answer 1

I propose an abstract approach, based on the closed graph theorem.

Lemma. Suppose that $T_1, T_2, \ldots, T_n$ are closed linear operators on the Banach space $X$, with nested domains $$D(T_1)\subset D(T_2)\subset \ldots \subset D(T_n).$$ Then there is a constant $C>0$ such that $$\tag{1}\|T_j x\|\le C(\|T_1 x\|+\|x\|), $$ for every $j=1, 2, \ldots, n$.

Proof of Lemma. By assumption, the graph $$ G(T_1):=\{(x, T_1 x)\ :\ x\in D(T_1)\}, $$ is a Banach space with the norm $$ \|(x, T_1 x)\|:=\|x\|+\|T_1x\|.$$ We define a linear operator on $G(T_1)$ by $$ V(x, T_1 x):=T_jx. $$ This operator is closed because $T_j$ is. By the closed graph theorem, $V$ is bounded. We conclude that (1) holds. $\Box$

Remark. It would be sufficient that the operators $T_j$ with $j>1$ are pre-closed, instead of closed. (See Hörmander 1955, Theorem 1.1).

This lemma can be applied to the linear operators on $X=C([a, b])$ given by $$T_j f:= f^{(n-j+1)},\qquad D(T_j)=C^{n-j+1}([a, b]),$$ where, of course, $$\|f\|:=\sup_{x\in [a, b]} |f(x)|.$$ The result is that $$ \|f^{(h)}\|\le C(\|f\|+\|f^{(n)}\|),\qquad \forall h=1,2,\ldots, n.$$