This question came up when I was doing some reading into convolution squares of singular measures. Recall a function $f$ on the torus $T = [-1/2,1/2]$ is said to be $\alpha$-Hölder (for $0 < \alpha < 1$) if $\sup_{t \in \mathbb{T}} \sup_{h \neq 0} |h|^{-\alpha}|f(t+h)-f(t)| < \infty$. In this case, define this value, $\omega_\alpha(f) = \sup_{t \in \mathbb{T}} \sup_{h \neq 0} |h|^{-\alpha}|f(t+h)-f(t)|$. This behaves much like a metric, except functions differing by a constant will not differ in $\omega_\alpha$. My primary question is this:

1) Is it true that the smooth functions are "dense" in the space of continuous $\alpha$-Hölder functions, i.e., for a given continuous $\alpha$-Hölder $f$ and $\varepsilon > 0$, does there exists a smooth function $g$ with $\omega_\alpha(f-g) < \varepsilon$?

To be precise, where this came up was worded somewhat differently. Suppose $K_n$ are positive, smooth functions supported on $[-1/n,1/n]$ with $\int K_n = 1$.

2) Given a fixed continuous function $f$ which is $\alpha$-Hölder and $\varepsilon > 0$, does there exist $N$ such that $n \geq N$ ensures $\omega_\alpha(f-f*K_n) < \varepsilon$?

This second formulation is stronger than the first, but is not needed for the final result, I believe.

To generalize, fix $0 < \alpha < 1$ and suppose $\psi$ is a function defined on $[0,1/2]$ that is strictly increasing, $\psi(0) = 0$, and $\psi(t) \geq t^{\alpha}$. Say that a function $f$ is $\psi$-Hölder if $\sup_{t \in \mathbb{T}} \sup_{h \neq 0} \psi(|h|)^{-1}|f(t+h)-f(t)| < \infty$. In this case, define this value, $\omega_\psi(f) = \sup_{t \in \mathbb{T}} \sup_{h \neq 0} \psi(|h|)^{-1}|f(t+h)-f(t)|$. Then we can ask 1) and 2) again with $\alpha$ replaced by $\psi$.

I suppose the motivation would be that the smooth functions are dense in the space of continuous functions under the usual metrics on function spaces, and this "Hölder metric" seems to be a natural way of defining a metric of the equivalence classes of functions (where $f$ and $g$ are equivalent if $f = g+c$ for a constant $c$). Any insight would be appreciated.

  • $\begingroup$ I thnk the answer to 2) should be "yes". Certainly the trigonometric polynomials are dense in our space of $\alpha$-Holder functions -- this is usually known as a $Lip_\alpha$-space, by the way, see Katznelson's Intro to HA for instance for the terminology and the proof. If you haven't looked things up there, and can get hold of a copy of the book from your library or similar, that would be my first suggestion. $\endgroup$
    – Yemon Choi
    Jun 29, 2010 at 7:28
  • $\begingroup$ I dn't have my copy of the book to hand, but some foraging on Google Books suggests that the kind of approximation results you want can be extracted from the general machinery in Chapter I, Section 2 of Katznelson's book. $\endgroup$
    – Yemon Choi
    Jun 29, 2010 at 7:51
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    $\begingroup$ Thank you for the response Yemon! I have a copy of Katznelson in fact, and I think you are referring to Exercise 6? I didn't notice this earlier, but I believe this is slightly different from what I have. The exercise gives that the trigonometric polynomials are dense in the space $lip_\alpha$, which is in general a subspace of $Lip_\alpha$, the difference being that we require in addition that $\lim_{h \to 0} \sup_t |h|^{-\alpha}|f(t+h)-f(t)| = 0$. $\endgroup$
    – Vince
    Jun 29, 2010 at 18:02
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    $\begingroup$ @Yemon: ${\rm Lip}_\alpha$ consists of the functions which are Lipschitz for the metric $|s-t|^\alpha$. For this metric, every smooth function is "locally flat": we have $\frac{|f(t + h) - f(t)|}{|h|^\alpha} \to 0$ as $h \to 0$. The family of locally flat Lipschitz functions is a norm closed proper subspace of ${\rm Lip}_\alpha$ denoted ${\rm lip}_\alpha$. In fact the former is the second dual of the latter. $\endgroup$
    – Nik Weaver
    Apr 6, 2013 at 17:58
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    $\begingroup$ @Nik: Indeed, I was too hasty back then, as wangtwo's calculation shows $\endgroup$
    – Yemon Choi
    Apr 6, 2013 at 22:41

3 Answers 3


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.

  • 3
    $\begingroup$ Right. When $\alpha=1$ take $f=|x|$ and write $$\omega_1(f-g) \ge |\frac{g(x)-g(0)}{|x|}-\frac{f(x)-f(0)}{|x|}|.$$ The second fraction is 1, the first one tends to $g'(0)$ if $x\to0^+$ and to $-g'(0)$ if $x\to0^-$. If $g'(0)\ge0$ taking the left limit gives $\omega_1(f-g)\ge 1 + |g'(0)|$, if $g'(0)<0$ use the right limit. $\endgroup$ Aug 12, 2010 at 21:03

Smooth functions are not dense in the space of Hölder continuous functions, but it is possible to characterize those functions that can be approximated. This is done below.

Definition. Let $(X,d)$ be a metric space and $0<\alpha\leq 1$. The space $C^{0,\alpha}(X)$ of Hölder continuous functions is a space of bounded real valued functions such that $$ [f]_{C^{0,\alpha}}:=\sup\left\{\frac{|f(x)-f(y)|}{d(x,y)^\alpha}:\, x\neq y\right\}<\infty. $$

$C^{0,\alpha}(X)$ is a Banach space with respect to the norm $$ \Vert f\Vert_{C^{0,\alpha}}=\Vert f\Vert_\infty+[f]_{C^{0,\alpha}}. $$

Definition. Let $(X,d)$ be a metric space. For $0<\alpha\leq 1$ we define the space $C^{0,\alpha+}(X)$ to be a subspace of $C^{0,\alpha}(X)$ (equipped with the same norm) that consists of functions $f\in C^{0,\alpha}(X)$ such that for every compact set $K$, $$ \lim_{t\to 0+}\ \sup\left\{\frac{|f(y)-f(x)|}{d(x,y)^\alpha}:\ x,y\in K,\ d(x,y)\leq t,\ x\neq y\right\}= 0. $$

In other words $C^{0,\alpha+}(X)$ is a subspace of $C^{0,\alpha}(X)$ consisting of functions that, on every compact set, have a slightly better modulus of continuity than that one in the definition of the $C^{0,\alpha}$ norm.

Theorem. Let $0<\alpha<1$. Then a function $f\in C^{0,\alpha}(\mathbb{R}^n)$ can be approximated by a sequence smooth functions $f_k\in C^\infty$ so that for every compact set $K\subset\mathbb{R}^n$, $\Vert f_k-f\Vert_{C^{0,\alpha}(K)}\to 0$ as $k\to\infty$ if and only if $f\in C^{0,\alpha+}(\mathbb{R}^n)$.

Remark. The result is not true for $\alpha=1$. Lipschitz functions that are not $C^1$ cannot be approximated by smooth functions in the Lipschitz norm, because the Lipschitz norm is the same as $C^1$ norm.

Example. $|x|^{1/2}\in C^{0,1/2}(-1,1)\setminus C^{0,1/2+}(-1,1)$ and hence it cannot be approximated by smooth functions in the $C^{0,1/2}$ norm (see also the accepted answer).

Proof of the theorem. Suppose that $f\in C^{0,\alpha}(\mathbb{R}^n)$ can be approximates by smooth functions. We need to show that $f\in C^{0,\alpha+}(\mathbb{R}^n)$. Let $B_R$ be a ball of any radius. Let $\varepsilon>0$ be given. Then for a sufficiently large $k$, $$ |(f_k-f)(y)-(f_k-f)(x)|\leq\frac{\varepsilon}{2}|x-y|^\alpha \quad \text{for all $x,y\in B_R$.} $$ Let $M=\sup_{B_R}|\nabla f_k|$. Hence the mean value theorem yields $$ |f(y)-f(x)|\leq\frac{\varepsilon}{2}|x-y|^\alpha +|f_k(y)-f_k(x)|\leq \left(\frac{\varepsilon}{2}+M|x-y|^{1-\alpha}\right)|x-y|^\alpha $$ so $$ |f(y)-f(x)|\leq\varepsilon |x-y|^\alpha \quad \text{for all $x,y\in B_r$ satisfying $|x-y|<(\varepsilon/2M)^{1/(1-\alpha)}$.} $$ This proves that $f\in C^{0,\alpha+}$.

Suppose now that $f\in C^{0,\alpha+}(\mathbb{R}^n)$. We will show that the approximation by mollification $f_t$ has the desired property i.e., for every ball $B_R$, $\Vert f_t-f\Vert_{C^{0,\alpha}(B_R)}\to 0$ as $t\to 0$. Since $f_t\to f$ uniformly, it remains to estimate the constant in the H\"older estimate of the difference $f_t-f$. Let $\varepsilon>0$ be given. It follows from the definition of $C^{0,\alpha+}$ that there is $R>\tau>0$ such that if $x,y\in B_{2R}$, $|x-y|<\tau$, then $|f(x)-f(y)|\leq \frac{1}{2}\varepsilon|x-y|^\alpha$. Hence for $0<t<R$, $|f_t(x)-f_t(y)|\leq\frac{1}{2}\varepsilon|x-y|^\alpha$ for $x,y\in B_R$ satisfying $|x-y|<\tau$. This easily follows from the definition of $f_t$, because $f_t(x)$ is a weighted average of $f$ on the ball $B(x,t)\subset B_{2R}$. Therefore $$ |(f_t-f)(x)-(f_t-f)(y)|\leq \varepsilon |x-y|^\alpha \quad \text{for all $x,y\in B_R$ satisfying $|x-y|<\tau$.} $$ Let $0<\delta<R$ be such that $\Vert f_t-f\Vert_\infty<\varepsilon\tau^\alpha/2$ for $0<t<\delta$.

If $x,y\in B_R$, $|x-y|\geq\tau$, then $$ |(f_t-f)(x)-(f_t-f)(y)|\leq 2\Vert f_t-f\Vert_\infty<\varepsilon\tau^\alpha\leq\varepsilon |x-y|^\alpha. $$ We proved that if $0<t<\delta$, then $$ |(f_t-f)(x)-(f_t-f)(y)|\leq \varepsilon |x-y|^\alpha \quad \text{for all $x,y\in B_R$} $$ as desired. This completes the proof. $\Box$


The answer is yes (Edit: my "proof" below seems incomplete; read the comment below). I will prove it in $\mathbb{R}^N$; the proof is easily adapted to the torus. Let $\phi\colon\mathbb{R}^N\to\mathbb{R}^N$ be a smooth function with compact support, non-negative, and such that $\int_{\mathbb{R}^N}\phi(x)dx=1$. Define $\phi_\varepsilon(x)=\varepsilon^{-N}\phi(\varepsilon^{-1}x)$. Given an $\alpha$-Hölder function $f\colon\mathbb{R}^N\to\mathbb{R}^N$, $\phi_\varepsilon*f$ is smooth and $\lim_{\varepsilon\to0}w_\alpha(f-\phi_\varepsilon*f)=0$. To prove it, let $g_\varepsilon=f-\phi_\varepsilon*f$. Then, for any $x,y\in\mathbb{R}^N$, $$ |g_\varepsilon(x)-g_\varepsilon(y)|\le\int_{\mathbb{R}^N}\phi(t)|(f(x-\varepsilon t)-f(x))-(f(y-\varepsilon t)-f(y))|dt. $$ Let $$ G_\varepsilon(t)=\sup_{x\ne y}\frac{|(f(x-\varepsilon t)-f(x))-(f(y-\varepsilon t)-f(y))|}{|x-y|^\alpha} $$ $$ \qquad\qquad\qquad\le\sup_{x\ne y}\frac{|f(x-\varepsilon t)-f(y-\varepsilon t)|+|f(x)-f(y)|}{|x-y|^\alpha}\le2w_\alpha(f). $$ Then, for any $x,y\in\mathbb{R}^N$, $x\ne y$ $$ \frac{|g_\varepsilon(x)-g_\varepsilon(y)|}{|x-y|^\alpha} \le\int_{\mathbb{R}^N}\phi(t)\frac{|(f(x-\varepsilon t)-f(x))-(f(y-\varepsilon t)-f(y))|}{|x-y|^\alpha}dt\le\int_{\mathbb{R}^N}\phi(t)G_\varepsilon(t)dt. $$ Taking the supremum over $x,y\in\mathbb{R}^N$ we get $$ w_\alpha(g_\varepsilon)\le\int_{\mathbb{R}^N}\phi(t)G_\varepsilon(t)dt. $$ As we have seen, $G_\varepsilon(t)$ is bounded and, since $f$ is continuous, $\lim_{\varepsilon\to0}G_\varepsilon(t)=0$ for all $t\in\mathbb{R}^N$. The Lebesgue dominated convergence theorem implies the result.

  • $\begingroup$ Hi Julian, thanks for replying. I followed most of what you wrote, except I'm unclear about the last step. I am guessing you meant $g_\varepsilon$ in place of $g$? Could you please provide some insight as to how the inequality $\omega_\alpha(g_\varepsilon) \leq \int \phi(t)G_\varepsilon(t) dt$ holds? I ask because I thought I had a counterexample in mind in the boundary case of $\alpha = 1$, and it didn't seem like there was a dependence on the choice of $\alpha$, unless it appears in that particular inequality. Of course, I could be mistaken. $\endgroup$
    – Vince
    Jul 2, 2010 at 5:23
  • $\begingroup$ I have included some more detail, but I have also realizaed that it is not clear at all that $\lim_{\varepsilon\to0}G_\varepsilon(t)=0$. $\endgroup$ Jul 3, 2010 at 22:14
  • $\begingroup$ You want $\phi_{\varepsilon}(x)=\varepsilon^{-n}\phi(x/\varepsilon)$ to get a mollifier. $\endgroup$
    – Ben McKay
    Apr 6, 2013 at 18:14

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