# Proving an inequality between $(1, \frac{\alpha}{2})$-Hölder norms of two functions

I have to prove that, given $$f\in C^{1, \frac{\alpha}{2}}([a, b])$$, such that $$\|f\|_{\infty} for some $$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

• 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. – Hans Engler Jan 19 at 15:57
• @HansEngler Yes, sorry. There is also the assumption that $\|f\|_{\infty}\leq C$ for some $C$. Is it now correct? – Jeji Jan 19 at 17:28
• please edit your post accordingly. As to your proof attempt, you cannot assume that $c = C = \frac{1}{2}$. – Hans Engler Jan 19 at 21:54
• @HansEngler I modifed my post. Why can’t I assume $c=C=\frac{1}{2}$? Then, how can I prove it? – Jeji Jan 20 at 13:35
• 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. – 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 $$K$$.