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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.

  • @GiuseppeNegro your prove is interesting and beautiful but I think that can not be used to the case $p>2$ – Federico Fallucca Feb 26 '19 at 15:11
  • @GiuseppeNegro your link is only about the case $p=2$. I would like to generalize this result. – Nameless Feb 26 '19 at 18:09
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    @Nameless If you're still interested in an elementary proof of this, check out pp. 240–41 of Rădulescu, Rădulescu & Andreescu, Problems in Real Analysis: Advanced Calculus on the Real Axis (Springer, 2009). They give a complete proof of the Landau-Kolmogorov inequalities on $\mathbb{R}$ in all cases. – Patrick Li Feb 25 '25 at 18:28

1 Answers1

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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 $$ \|f^{(h)}\|\le C(\|f\|+\|f^{(n)}\|),\qquad \forall h=1,2,\ldots, n.$$

  • Thank you @Giuseppe Negro. I found this problem in Zorich, Mathematical Analysis I, so I think that it can be solved without advanced math but only knowing limits, continuous and differentiable functions. Can your proof be 'reduced' using simpler concepts? – Nameless Feb 26 '19 at 20:12
  • Woooow, it is beautiful. Perfect – Federico Fallucca Feb 26 '19 at 20:39
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    Now that I re-read it I am not totally sure it works, because I didn't prove that the operators $T_j$ are closed. I opened a new question on that. – Giuseppe Negro Feb 27 '19 at 09:48
  • @Nameless: this approach cannot be translated into simpler concepts because it relies on the closed graph theorem. This approach is not the best, either. It gives no indication on the value of the constant. So it would be better to find a "calculus-only" proof, instead of trying to adapt this one. Still, I thought that this proof could be interesting as an exercise in abstraction, that's why I posted it. – Giuseppe Negro Feb 27 '19 at 09:50
  • Okay, thank you for now. When I will have acquired more advanced concepts like these I probably will understand your proof. Now let's wait someone that succeds in a pure calculus proof. :) – Nameless Feb 27 '19 at 11:28