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Daniele Tampieri
Daniele Tampieri
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Let $f: [0, 1] \to \mathbb R$ be a continuous function. Define the local Hölder exponent $H(f, x)$ of $f$ at $x \in [0, 1]$ by

$$H(f, x) := \sup\{0 \leq \alpha \leq 1\, | \,\lim_{\delta \to 0_+} \sup_{y, z \in B_\delta (x)} \frac{|f(y) - f(z)|}{|y-z|^\alpha}\ \, \text{is finite\}}.$$$$H(f, x) := \sup\left\{0 \leq \alpha \leq 1\mid\lim_{\delta \to 0_+} \sup_{y, z \in B_\delta (x)} \frac{|f(y) - f(z)|}{|y-z|^\alpha}\ \, \text{is finite}\right\}.$$

Let $F: \mathbb R \to \mathbb R$ be an another continuous function. We define the local Hölder exponents for $F$ in the same way. Is it true that for all $x \in [0, 1]$, we have

$$H(F \circ f, x) \leq 1 + H(f, x) - H(F, f(x))?$$

Let $f: [0, 1] \to \mathbb R$ be a continuous function. Define the local Hölder exponent $H(f, x)$ of $f$ at $x \in [0, 1]$ by

$$H(f, x) := \sup\{0 \leq \alpha \leq 1\, | \,\lim_{\delta \to 0_+} \sup_{y, z \in B_\delta (x)} \frac{|f(y) - f(z)|}{|y-z|^\alpha}\ \, \text{is finite\}}.$$

Let $F: \mathbb R \to \mathbb R$ be an another continuous function. We define the local Hölder exponents for $F$ in the same way. Is it true that for all $x \in [0, 1]$, we have

$$H(F \circ f, x) \leq 1 + H(f, x) - H(F, f(x))?$$

Let $f: [0, 1] \to \mathbb R$ be a continuous function. Define the local Hölder exponent $H(f, x)$ of $f$ at $x \in [0, 1]$ by

$$H(f, x) := \sup\left\{0 \leq \alpha \leq 1\mid\lim_{\delta \to 0_+} \sup_{y, z \in B_\delta (x)} \frac{|f(y) - f(z)|}{|y-z|^\alpha}\ \, \text{is finite}\right\}.$$

Let $F: \mathbb R \to \mathbb R$ be an another continuous function. We define the local Hölder exponents for $F$ in the same way. Is it true that for all $x \in [0, 1]$, we have

$$H(F \circ f, x) \leq 1 + H(f, x) - H(F, f(x))?$$

[Edit removed during grace period]; edited body
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Nate River
Nate River
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Let $f: [0, 1] \to \mathbb R$ be a continuous function. Define the local Hölder exponent $H(f, x)$ of $f$ at $x \in [0, 1]$ by

$$H(f, x) := \sup\{0 \leq \alpha \leq 1\, | \,\lim_{\delta \to 0_+} \sup_{y, z \in B_\delta (x)} \frac{|X(y) - X(z)|}{|y-z|^\alpha}\ \, \text{is finite\}}.$$$$H(f, x) := \sup\{0 \leq \alpha \leq 1\, | \,\lim_{\delta \to 0_+} \sup_{y, z \in B_\delta (x)} \frac{|f(y) - f(z)|}{|y-z|^\alpha}\ \, \text{is finite\}}.$$

Let $F: \mathbb R \to \mathbb R$ be an another continuous function. We define the local HolderHölder exponents for $F$ in the same way. Is it true that for all $x \in [0, 1]$, we have

$$H(F \circ f, x) \leq 1 + H(f, x) - H(F, f(x))?$$

Let $f: [0, 1] \to \mathbb R$ be a continuous function. Define the local Hölder exponent $H(f, x)$ of $f$ at $x \in [0, 1]$ by

$$H(f, x) := \sup\{0 \leq \alpha \leq 1\, | \,\lim_{\delta \to 0_+} \sup_{y, z \in B_\delta (x)} \frac{|X(y) - X(z)|}{|y-z|^\alpha}\ \, \text{is finite\}}.$$

Let $F: \mathbb R \to \mathbb R$ be an another continuous function. We define the local Holder exponents for $F$ in the same way. Is it true that for all $x \in [0, 1]$, we have

$$H(F \circ f, x) \leq 1 + H(f, x) - H(F, f(x))?$$

Let $f: [0, 1] \to \mathbb R$ be a continuous function. Define the local Hölder exponent $H(f, x)$ of $f$ at $x \in [0, 1]$ by

$$H(f, x) := \sup\{0 \leq \alpha \leq 1\, | \,\lim_{\delta \to 0_+} \sup_{y, z \in B_\delta (x)} \frac{|f(y) - f(z)|}{|y-z|^\alpha}\ \, \text{is finite\}}.$$

Let $F: \mathbb R \to \mathbb R$ be an another continuous function. We define the local Hölder exponents for $F$ in the same way. Is it true that for all $x \in [0, 1]$, we have

$$H(F \circ f, x) \leq 1 + H(f, x) - H(F, f(x))?$$

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Nate River
Nate River
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How much can you improve a Hölder function by composing it with another?

Let $f: [0, 1] \to \mathbb R$ be a continuous function. Define the local Hölder exponent $H(f, x)$ of $f$ at $x \in [0, 1]$ by

$$H(f, x) := \sup\{0 \leq \alpha \leq 1\, | \,\lim_{\delta \to 0_+} \sup_{y, z \in B_\delta (x)} \frac{|X(y) - X(z)|}{|y-z|^\alpha}\ \, \text{is finite\}}.$$

Let $F: \mathbb R \to \mathbb R$ be an another continuous function. We define the local Holder exponents for $F$ in the same way. Is it true that for all $x \in [0, 1]$, we have

$$H(F \circ f, x) \leq 1 + H(f, x) - H(F, f(x))?$$