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Joonas Ilmavirta
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In our PDE seminar, we met the same kinds of questions, and we think the answer is "WRONG". The smooth functions is NOT dense in H{"o}lderHölder spaces.

An example is, $$f(x) = |x|^{1/2} \quad x \in (-1,1)$$ it is easy to check that $f$ is $1/2$-H{"o}lderHölder continuous.

For details, for any $g \in C^{1}((-1,1))$, then the derivative of $g$ is continuous at $0$, so we have $$ \lim_{x \to 0} \frac{|g(x)-g(0)|}{|x|^{1/2}} = \lim_{x \to 0} |x|^{1/2}\frac{|g(x)-g(0)|}{|x|} = 0 $$
and $$ \omega_{1/2}(g-f) \ge \frac{|(g(x)-f(x))-(g(0)-f(0))|}{|x|^{1/2}} \ge |\frac{|(g(x)-g(0)|}{|x|^{1/2}}-\frac{|f(x)-f(0)|}{|x|^{1/2}}| $$ but
$$ \frac{|f(x)-f(0)|}{|x|^{1/2}}=1 \quad x \in (-1,1) \quad x \neq 0 $$ let $x \to 0$, we obtain $\omega_{1/2}(g-f) \ge 1$.

Thus, for any $g \in C^{1}((-1,1))$, we have $\omega_{1/2}(g-f)\ge 1$.

For $0< \alpha <1$ we can make similar examples, but when $\alpha = 1$, the proof of the counter-example may be different.

In our PDE seminar, we met the same kinds of questions, and we think the answer is "WRONG". The smooth functions is NOT dense in H{"o}lder spaces.

An example is, $$f(x) = |x|^{1/2} \quad x \in (-1,1)$$ it is easy to check that $f$ is $1/2$-H{"o}lder continuous.

For details, for any $g \in C^{1}((-1,1))$, then the derivative of $g$ is continuous at $0$, so we have $$ \lim_{x \to 0} \frac{|g(x)-g(0)|}{|x|^{1/2}} = \lim_{x \to 0} |x|^{1/2}\frac{|g(x)-g(0)|}{|x|} = 0 $$
and $$ \omega_{1/2}(g-f) \ge \frac{|(g(x)-f(x))-(g(0)-f(0))|}{|x|^{1/2}} \ge |\frac{|(g(x)-g(0)|}{|x|^{1/2}}-\frac{|f(x)-f(0)|}{|x|^{1/2}}| $$ but
$$ \frac{|f(x)-f(0)|}{|x|^{1/2}}=1 \quad x \in (-1,1) \quad x \neq 0 $$ let $x \to 0$, we obtain $\omega_{1/2}(g-f) \ge 1$.

Thus, for any $g \in C^{1}((-1,1))$, we have $\omega_{1/2}(g-f)\ge 1$.

For $0< \alpha <1$ we can make similar examples, but when $\alpha = 1$, the proof of the counter-example may be different.

In our PDE seminar, we met the same kinds of questions, and we think the answer is "WRONG". The smooth functions is NOT dense in Hölder spaces.

An example is, $$f(x) = |x|^{1/2} \quad x \in (-1,1)$$ it is easy to check that $f$ is $1/2$-Hölder continuous.

For details, for any $g \in C^{1}((-1,1))$, then the derivative of $g$ is continuous at $0$, so we have $$ \lim_{x \to 0} \frac{|g(x)-g(0)|}{|x|^{1/2}} = \lim_{x \to 0} |x|^{1/2}\frac{|g(x)-g(0)|}{|x|} = 0 $$
and $$ \omega_{1/2}(g-f) \ge \frac{|(g(x)-f(x))-(g(0)-f(0))|}{|x|^{1/2}} \ge |\frac{|(g(x)-g(0)|}{|x|^{1/2}}-\frac{|f(x)-f(0)|}{|x|^{1/2}}| $$ but
$$ \frac{|f(x)-f(0)|}{|x|^{1/2}}=1 \quad x \in (-1,1) \quad x \neq 0 $$ let $x \to 0$, we obtain $\omega_{1/2}(g-f) \ge 1$.

Thus, for any $g \in C^{1}((-1,1))$, we have $\omega_{1/2}(g-f)\ge 1$.

For $0< \alpha <1$ we can make similar examples, but when $\alpha = 1$, the proof of the counter-example may be different.

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wangtwo
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In our PDE seminar, we met the same kinds of questions, and we think the answer is "WRONG". The smooth functions is NOT dense in H{"o}lder spaces.

An example is, $$f(x) = |x|^{1/2} \quad x \in (-1,1)$$ it is easy to check that $f$ is $1/2$-H{"o}lder continuous.

For details, for any $g \in C^{1}((-1,1))$, then the derivative of $g$ is continuous at $0$, so we have $$ \lim_{x \to 0} \frac{|g(x)-g(0)|}{|x|^{1/2}} = \lim_{x \to 0} |x|^{1/2}\frac{|g(x)-g(0)|}{|x|} = 0 $$
and $$ \omega_{1/2}(g-f) \ge \frac{|(g(x)-f(x))-(g(0)-f(0))|}{|x|^{1/2}} \ge |\frac{|(g(x)-g(0)|}{|x|^{1/2}}-\frac{|f(x)-f(0)|}{|x|^{1/2}}| $$ but
$$ \frac{|f(x)-f(0)|}{|x|^{1/2}}=1 \quad x \in (-1,1) \quad x \neq 0 $$ let $x \to 0$, we obtain $\omega_{1/2}(g-f) \ge 1$.

Thus, for any $g \in C^{1}((-1,1))$, we have $\omega_{1/2}(g-f)\ge 1$.

For $0< \alpha <1$ we can make similar examples, but when $\alpha = 1$, the proof of the counter-example may be different.