I have to prove that, given $f\in C^{1, \frac{\alpha}{2}}([a, b])$, such that $\|f\|_{\infty}<L$ for some $L$, $f\geq0$ and $\alpha\in(0, 1)$, there exists a constant $K$ such that $$ \|e^f-1\|_{1, \frac{\alpha}{2}}\leq K\|f\|_{1, \frac{\alpha}{2}}. $$ where $\|\cdot\|_{1, \frac{\alpha}{2}}$ is the usual $(1, \frac{\alpha}{2})$-Hölder norm.

This is not a real exercise. It is only a thing to prove to deduce the second inequality in Theorem 3.1, (i), by Proposition 3.1, in the paper https://arxiv.org/pdf/1402.2467. In this case $f=C_T$.

My attempt. We have $$ \|e^f-1\|_{1, \frac{\alpha}{2}}=\sup_{[a, b]}|e^f-1|+\sup_{[a, b]}|f'e^f|+\sup_{x\neq y}\frac{|f'(x)e^{f(x)}-f'(y)e^{f(y)}|}{|x-y|^{\frac{\alpha}{2}}}, $$ $$ \|f\|_{1, \frac{\alpha}{2}}=\sup_{[a, b]}|f|+\sup_{[a, b]}|f'|+\sup_{x\neq y}\frac{|f'(x)-f'(y)|}{|x-y|^{\frac{\alpha}{2}}}. $$ For simplicity, we call the $\sup$ in the $\|e^f-1\|_{1, \frac{\alpha}{2}}$ expression with $S_1$ and the $\sup$ in the $\|f\|_{1, \frac{\alpha}{2}}$ expression with $S_2$. Observe that $$ \|e^f-1\|_{1, \frac{\alpha}{2}}\leq \sup_{[a, b]}|e^f-1|+\sup_{[a, b]}|f'|\sup_{[a, b]}|e^f|+S_1\leq\sup_{[a, b]}|e^f-1|+\|f\|_{1, \frac{\alpha}{2}}\sup_{[a, b]}|e^f|+S_1. $$ Now, by the arguments in Another definition of a Holder norm, the norms $\|e^f-1\|_{1, \frac{\alpha}{2}}$ and $$ \|e^f-1\|^*_{1, \frac{\alpha}{2}}:=\sup_{[a, b]}|e^f-1|+S_1 $$ are equivalent, that is there exist two constants $c, C>0$ such that $$ c\|e^f-1\|_{1, \frac{\alpha}{2}}\leq\|e^f-1\|^*_{1, \frac{\alpha}{2}}\leq C\|e^f-1\|_{1, \frac{\alpha}{2}}. $$ In particular, for $c=C=\frac{1}{2}$, we have $$ \|e^f-1\|^*_{1, \frac{\alpha}{2}}\leq\frac{1}{2}\|e^f-1\|_{1, \frac{\alpha}{2}}. $$ Then $$ \|e^f-1\|_{1, \frac{\alpha}{2}}\leq\frac{1}{2}\|e^f-1\|_{1, \frac{\alpha}{2}}+\|f\|_{1, \frac{\alpha}{2}}\sup_{[a, b]}|e^f|, $$ that is $$ \|e^f-1\|_{1, \frac{\alpha}{2}}\leq2\sup_{[a, b]}|e^f|\|f\|_{1, \frac{\alpha}{2}}=K\|f\|_{1, \frac{\alpha}{2}} $$ where $K=2\sup_{[a, b]}|e^f|$. Is this proof right?

Thank You

  • $\begingroup$ There must be a missing assumption here, something like $\|f\|_\infty \le C$ for some $C$. Otherwise the sequence of functions $f_n(x) = n, \, n \ge 1$ is a counterexample. $\endgroup$ – Hans Engler Jan 19 at 15:57
  • $\begingroup$ @HansEngler Yes, sorry. There is also the assumption that $\|f\|_{\infty}\leq C$ for some $C$. Is it now correct? $\endgroup$ – Jeji Jan 19 at 17:28
  • $\begingroup$ please edit your post accordingly. As to your proof attempt, you cannot assume that $c = C = \frac{1}{2}$. $\endgroup$ – Hans Engler Jan 19 at 21:54
  • $\begingroup$ @HansEngler I modifed my post. Why can’t I assume $c=C=\frac{1}{2}$? Then, how can I prove it? $\endgroup$ – Jeji Jan 20 at 13:35
  • 1
    $\begingroup$ The constants $c$ and $C$ are only known to exist. You cannot assume that they have specific values. You also need to replace $e^{f'(x)}$ etc. with $e^{f(x)}$ in $S_1$. Finally, please explain where this problem is coming from. It appears to be too complicated for an exercise. $\endgroup$ – Hans Engler Jan 20 at 16:55

The estimate is correct for all $(1,\alpha)$ Holder norms.

We can start out as in the OP's attempt to solve the problem. Recall that $$ \|f\|_{1, \alpha}=\sup_{[a, b]}|f|+\sup_{[a, b]}|f'|+\sup_{x\neq y}\frac{|f'(x)-f'(y)|}{|x-y|^{\alpha}} = A_0(f) + A_1(f) + A_2(f) $$ Similarly $$ \|e^f-1\|_{1, \alpha}=\sup_{[a, b]}|e^f - 1|+\sup_{[a, b]}|f'e^f|+\sup_{x\neq y}\frac{|f'(x)e^{f(x)}-f'(y)e^{f(y)}|}{|x-y|^{\alpha}} =: B_0 + B_1 + B_2 $$

From the answer to the question quoted in your post, we know that there is a constant $C_0$ such that $$ \boxed{A_1(f) \le C_0 A_0(f)^\gamma A_2(f)^{1-\gamma}} $$ where $\gamma = \frac{\alpha}{1+\alpha}$.

Finding $K_0$ such that $B_0 \le K_0 A_0(f)$

Set $K_0 = e^L$. Since $0 \le f(x) \le L$ on $[a,b]$, it follows that $$ 0 \le e^{f(x)} - 1 = e^{f(x)} - e^0 = e^\xi f(x) \le K_0 f(x) \le K_0 A_0 $$ for all $x$. Take the supremum on the left to deduce $B_0 \le K_0 A_0$.

Note that also $A_0(f) \le K_0$.

Finding $K_1$ such that $B_1 \le K_1 A_1(f)$

Set $K_1 = e^L$. Then, since $0 \le f(x) \le L$, for $x \in [a,b]$ $$ |f'(x) e^{f(x)}| \le e^L |f'(x)| = K_1 |f'(x)| \le K_1 A_1\, . $$ This implies $B_1 \le K_1 A_1$,

Finding $K_2$ such that $B_2 \le K_2 A_2(f)$

Set $K_2 = K_0 + 2 C_0 K_0^{1 + \alpha}$ where $C_0$ is from the boxed inequality. Let $a \le x < y \le b$. Then $$ |f'(x)e^{f(x)} - f'(y)e^{f(y)}| \le |f'(x)e^{f(x)} - f'(x)e^{f(y)}| + |f'(x)e^{f(y)} - f'(y)e^{f(y)}| $$ The second term may be estimated by $$ \dots \le |f'(x) - f'(y)| e^{f(y)} \le A_2(f) |x-y|^\alpha K_0 \, . $$ The first term on the right may be estimated, using the mean value theorem, by $$ \dots \le |f'(x)||e^{f(x)} - e^{f(y)}| = |f'(x)||x-y||f'(\zeta) e^{f(\zeta)}| \le |x-y| A_1^2(f) K_0 $$ We can alternatively estimate the first term on the right hand side by $$ \dots \le 2 K_0 A_1(f) $$ Raise the first estimate to the power $\alpha$ and the second to the power $1 - \alpha$ and multiply them together. The result is $$ |f'(x)e^{f(x)} - f'(x)e^{f(y)}| \le 2|x-y|^\alpha A_1^{1 + \alpha}(f) K_0 \le 2 C_0 K_0 A_2(f)A_0(f)^\alpha |x-y|^\alpha \, . $$ We may replace $A_0(f)^\alpha$ by $K_0^\alpha$. Combining the estimates and dividing by $|x-y|^\alpha$, we obtain $$ |\frac{|f'(x)e^{f(x)} - f'(y)e^{f(y)}|}{|x-y|^\alpha} \le K_0A_2(f) + 2C_0 K_0^{1+\alpha} A_2(f) = K_2 A_2(f) \, . $$ Taking the supremum we see that $B_2 \le K_2 A_2(f)$.

Combining all three estimates implies $$ B_0 + B_1 + B_2 \le K(A_0(f) + A_1(f) + A_2(f)) $$
for some $K$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.