Polynomials not dense in holder spaces

Question

How to prove that the polynomials are not dense in Holder space with exponent, say, $\frac{1}{2}$?

Answer 1

By exhibiting a function that cannot be approximated by polynomials in the norm of $C^{1/2}$, such as $f(x)=\sqrt{x}$ on the interval $[0,1]$. The proof is divided into steps below; you might not need to read all of them.

Let $p$ be a polynomial.

$(p(x)-p(0))/x\to p'(0)$ as $x\to 0^+$

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$|p(x)-p(0)|/x^{1/2}\to 0$ as $x\to 0^+$

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$|(f(x)-p(x))-(f(0)-p(0))|/x^{1/2}\to 1$ as $x\to 0^+$

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$\|f-p\|_{C^{1/2}}\ge 1$

Answer 2

I should have reported this long ago. We prove, in fact, that even the larger class $C^1[0,\ 1]$ of continuously differentiable functions on $[0,\ 1]$ is not dense in $H_{\alpha}^0$.

We record that $C^1 \subset H_{\alpha}^0$. For, if $g \in C^1$ and if $\Arrowvert g' \Arrowvert_C = \Lambda$, then $\delta_{c,\ \alpha}(g) \le \Lambda c^{1 - \alpha} \rightarrow 0$.

For $a \in (0,\ 1)$, let $x(t) = |t - a |^{\frac{1}{2}} $. Let $0 < \alpha < \frac{1}{2}$. We have, if $s < t$, $\frac{\big| |t - a|^{\frac{1}{2}} - |s - a|^{\frac{1}{2} } \big|}{|t - s|^{\alpha}}$ $= \frac{\big| |t - a|^{\frac{1}{2}} - |s - a|^{\frac{1}{2} } \big|}{|t - a - (s - a)|^{\frac{1}{2}}}\frac{|t - s|^{\frac{1}{2}}}{|t - s|^{\alpha}}$ $\le \frac{\big| |t - a|^{\frac{1}{2}} - |s - a|^{\frac{1}{2} } \big|}{|t - a - (s - a)|^{\frac{1}{2}}}$ $\le \frac{\big| |t - a|^{\frac{1}{2}} - |s - a|^{\frac{1}{2} } \big|}{\big||t - a| - |s - a|\big|^{\frac{1}{2}}}\le 1$. Hence $x \in H_{\alpha}$.

$\delta_{c,\ \alpha}(x) = \sup\limits_{|s - t| < c}\frac{\big| |t - a|^{\frac{1}{2}} - |s - a|^{\frac{1}{2} } \big|}{\big||t - a| - |s - a|\big|^{\frac{1}{2}}}|t - s|^{\frac{1}{2} - \alpha} \le c^{\frac{1}{2} - \alpha} \rightarrow 0$\ as $c \rightarrow 0$. Hence $x \in H_{\alpha}^0$. And then $g - x \in H_{\alpha}^0$

Now, $\Arrowvert g - x \Arrowvert_{\alpha}$ $ = \sup\limits_{0\le t,s\le 1,\ s \neq t} \frac{| g(t) - x(t) - \{g(s) - x(s)\} |}{|t - s|^{\alpha}}$ $ \ge \sup\limits_{0\le t\le 1,\ t \neq a} \frac{| g(t) - x(t) - \{g(a) - x(a)\}|}{|t - a|^{\alpha}}$ $= \sup\limits_{0\le t\le1,\ t \neq a} \big| \frac{| g(t) - g(a)|}{|t - a|^{\alpha}} - \frac{|x(t) - x(a)|}{|t - a|^{\alpha}} \big|$ $= \sup\limits_{0\le t\le1,\ t \neq a} \big| \frac{| g(t) - g(a)|}{|t - a|^{\alpha}} - |t - a|^{\frac{1}{2} - \alpha}\big|$ $\ge \sup\limits_{\substack{0\le t\le1,\ t \neq a\\|t - a| < (\frac{\varepsilon}{\Lambda})^{\frac{1}{1 - \alpha}}}} \big||g'(t^*)|{|t - a|^{1 - \alpha}} - \max\big( (1 - a)^{\frac{1}{2} - \alpha},\ a^{\frac{1}{2} - \alpha} \big)\big|$ $\ge \max\big( (1 - a)^{\frac{1}{2} - \alpha},\ a^{\frac{1}{2} - \alpha} \big) - \varepsilon$.

Hence the sub set of all functions which are continuously differentiable is not dense in $H_{\alpha}^0$.