# Extension of Hölder continuous function

Let $f:[0,c] \rightarrow \mathbb R$ be Hölder continuous with constant $M>0$ and power $p \in (0,1)$ and satisfies $f(0)=f(c)$. Let $g:[0,2c] \rightarrow \mathbb R$ be given by:

$$g(x)=f(x) \ \ \textrm{for} \ \ x \in [0,c]$$ and $$g(x)=f(x-c) \ \ \textrm{for} \ \ x \in [c,2c].$$

Then $g$ is Hölder continuous with the same power but I don't know if constant $M$ is the same.

I try in this way. Let $x,y \in [0,2c]$, $x<y$. It is obvious that if $x,y \in [0,c]$ or $x,y \in [c,2c]$ then $|g(x)-g(y)|\leq M |x-y|^p$. For $x\in [0,c]$, $y\in [c,2c]$ we have

$$|g(x)-g(y)| \leq |g(x)-g(c)|+|g(c)-g(y)| =|f(x)-f(c)|+|f(0)-f(y-c)|\leq M [(x-c)^p+(y-c)^p]\leq M 2^{1-p} (c-x+y-c)^p= M 2^{1-p} |y-x|^p.$$

(I have used inequality: $2^{p-1}(u^p+v^p)\leq (u+v)^p$ for $u,v \geq 0$, $p\in (0,1)$, which follows from concavity of $[0,\infty) \ni t\mapsto t^p$.)

Does constant $M 2^{1-p}$ can be improved to $M$ ?

Thanks

-

## 1 Answer

Suppose $f(x) = x^p$ for $0 \le x \le \epsilon$ and $f(x) = -(c-x)^p$ for $c-\epsilon \le x \le c$, which (at least if $c$ is large enough and you define $f$ suitably on $[\epsilon, c-\epsilon]$) has constant $M=1$. Then taking $x=c+\epsilon$, $y=c-\epsilon$, $$|g(x) - g(y)| = 2 \epsilon^p = 2^{1-p} |x-y|^p$$ and you can't get rid of the $2^{1-p}$.

-
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 31, "mathjax_display_tex": 4, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": false}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9628298282623291, "perplexity_flag": "head"}