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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

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  • $\begingroup$ 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
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    $\begingroup$ 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
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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 $K$.

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