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$\def\R{\mathbb{R}}$

Consider a space $C^{1,\beta}$ of all continuous differentiable functions $f\colon\R\to\R$ such that their derivative $f'$ is Holder continuous with exponent $\beta$.

A standard way to define a norm on this space is to put

$$ \|f\|_{1,\beta}:=\sup_{x\in\R} |f(x)|+\sup_{x\in\R} |f'(x)|+\sup_{x\neq y}\frac{|f'(x)-f'(y)|}{|x-y|^\beta}. $$

My question is why do we need here the middle term $\sup_{x\in\R} |f'(x)|$? If we just put $$ \|f\|:=\sup_{x\in\R} |f(x)|+\sup_{x\neq y}\frac{|f'(x)-f'(y)|}{|x-y|^\beta}. $$ would this be a norm? Would the space equipped with this norm be a Banach space? Would this norm be equivalent to the standard norm $\|\cdot\|_{1,\beta}$?

Thanks!

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    $\begingroup$ Are you sure that the term with $\beta$ is not applied to $f'$ instead of $f$? $\endgroup$
    – Disintegrating By Parts
    Aug 9, 2016 at 13:35
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    $\begingroup$ thanks for your correction! That was a typo. $\endgroup$
    – Oleg
    Aug 9, 2016 at 13:43
  • $\begingroup$ Have you tried to check the three properties of a norm? If so, have you found any trouble? $\endgroup$
    – Siminore
    Aug 9, 2016 at 13:50
  • $\begingroup$ @Siminore with three properties of a norm I have no problem - it is indeed a norm. What is much less obvious (at least to me) is whether this new norm is equivalent to the standard norm or not. If these norms are equivalent, then it is not clear why people usually work with the first norm rather than with the second. $\endgroup$
    – Oleg
    Aug 9, 2016 at 14:53
  • $\begingroup$ Did you notice someone posted an answer? $\endgroup$
    – David C. Ullrich
    Aug 9, 2016 at 14:56

1 Answer 1

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Yes, if you omit the middle term you get an equivalent norm. (The reason the middle term is typically included is just to make various things simpler.)

This is a generalization of the classical Landau inequality, which says that a bound on $f$ and a bound on $f''$ imply a bound a $f'$. One can give a simple proof like so: Assume $f'$ is "large" at a point. The hypothesis on continuity of $f'$ shows that $f'$ is large on an interval of a certain length, which implies that $f$ must be large somewhere. In fact:

Theorem If $|f|\le a$ and $|f'(t)-f'(s)|\le b|s-t|^\beta$ then $$|f'|\le c_\beta a^{\frac{\beta}{\beta+1}}b^{\frac{1}{\beta+1}}.$$

Proof: Say $f'(0)=m>0$. Then $$f'(t)\ge f'(0)-bt^\beta\ge m/2, \quad(0<t<t_0=(m/2b)^{1/\beta}).$$Hence $$2a\ge f(t_0)-f(0)\ge\frac{m}{2}t_0,$$and the claimed inequality follows on inserting the definition of $t_0$ and unravelling things.

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    $\begingroup$ thanks again for your reply. Would the same trick work in general, when we are dealing with $C^{k,\beta}$ functions? Can we also omit "the middle term" (i.e. bounds on all the derivatives apart from the last) there? $\endgroup$
    – Oleg
    Aug 9, 2016 at 15:12
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    $\begingroup$ Yes. You could even use more or less the same argument, although there are probably more elegant ways. If you have a Holder condition on $f'''$ and $f'''$ is large at a point then $f'''$ is large on an interval of a certain length. This implies that $f''$ is large on a shorter interval, which implies that $f'$ is large on a yet shorter interval, so $f$ is large somewhere. $\endgroup$
    – David C. Ullrich
    Aug 9, 2016 at 15:19

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