无处可导的连续函数

hbghlyj

这份讲义的第46页
2.1.3 Van der Vaerden's example
The following example of a continuous function on $\mathbb{R}$ which is nowhere differentiable was constructed by B. L. Van der Waerden [For your reading - I don't think I'll have time to work through this example].
Let us begin with a simple continuous function
$$
h(x)=\left\{\begin{array}{cc}
x &\text { if } 0 \leq x \leq 1 \\
2-x &\text { if } 1 \leq x \leq 2
\end{array}\right.
$$
and extend $h$ to be a periodic function with period 2, i.e. $h(x+2)=h(x)$ for $x \in \mathbb{R}$. Then $h$ is continuous on $\mathbb{R}$. Consider the series
$$
f(x)=\sum_{n=0}^{\infty}\left(\frac{3}{4}\right)^{n} h\left(4^{n} x\right)
$$
By the Weierstrass M-test, $\sum_{n=0}^{\infty}\left(\frac{3}{4}\right)^{n} h\left(4^{n} x\right)$ converges uniformly in $\mathbb{R}$, thus $f$ is continuous on $\mathbb{R}$ [Theorem 1.4.11] and
$$
|f(x)| \leq \sum_{n=0}^{\infty}\left(\frac{3}{4}\right)^{n}=4 \quad \text { for every } x \in \mathbb{R} .
$$
Let $x \in \mathbb{R}, m \in \mathbb{N}$ and set $k=\left[4^{m} x\right]$ the integer part of $4^{m} x: k$ is the unique integer such that
$$
k \leq 4^{m} x<k+1
$$
Let $\alpha_{m}=4^{-m} k$ and $\beta_{m}=4^{-m}(k+1)$. Obviously
$$
\alpha_{m} \leq x<\beta_{m}
$$
and
$$
\beta_{m}-\alpha_{m}=\frac{1}{4^{m}} \rightarrow 0 \text { as } m \rightarrow \infty
$$
In particular, $\lim _{m \rightarrow \infty} \alpha_{m}=\lim _{m \rightarrow \infty} \beta_{m}=x$. We are going to show that
$$
\lim _{m \rightarrow \infty} \frac{f\left(\beta_{m}\right)-f\left(\alpha_{m}\right)}{\beta_{m}-\alpha_{m}}
$$
does not exist, so that $f$ is not differentiable at $x$. Since $x$ is arbitrary, $f$ is nowhere differentiable.
If $n>m$, then $4^{n} \beta_{m}-4^{n} \alpha_{m}$ is an even number, and if $n \leq m$ then there is no integer between $4^{n} \beta_{m}$ and $4^{n} \alpha_{m}$. Therefore
$$
\left|h\left(4^{n} \beta_{m}\right)-h\left(4^{n} \alpha_{m}\right)\right|=\left\{\begin{array}{c}
0, \quad \text { if } n>m \\
4^{n-m}, \text { if } n \leq m
\end{array}\right.
$$
Hence
$$
\begin{aligned}
f\left(\beta_{m}\right)-f\left(\alpha_{m}\right) &=\sum_{n=0}^{\infty}\left(\frac{3}{4}\right)^{n}\left(h\left(4^{n} \beta_{m}\right)-h\left(4^{n} \alpha_{m}\right)\right) \\
&=\sum_{n=0}^{m}\left(\frac{3}{4}\right)^{n}\left(h\left(4^{n} \beta_{m}\right)-h\left(4^{n} \alpha_{m}\right)\right)
\end{aligned}
$$
so that
$$
\begin{aligned}
\left|f\left(\beta_{m}\right)-f\left(\alpha_{m}\right)\right| & \geq\left(\frac{3}{4}\right)^{m}-\sum_{n=0}^{m-1}\left(\frac{3}{4}\right)^{n}\left|h\left(4^{n} \beta_{m}\right)-h\left(4^{n} \alpha_{m}\right)\right| \\
&=\left(\frac{3}{4}\right)^{m}-\sum_{n=0}^{m-1} 4^{n-m}\left(\frac{3}{4}\right)^{n} \\
&=\left(\frac{3}{4}\right)^{m}-\frac{1}{4^{m}} \frac{3^{m}-1}{2} \\
&=\frac{1}{2}\left(\frac{3}{4}\right)^{m}+\frac{1}{2} \frac{1}{4^{m}} .
\end{aligned}
$$
Therefore
$$
\frac{\left|f\left(\beta_{m}\right)-f\left(\alpha_{m}\right)\right|}{\beta_{m}-\alpha_{m}} \geq \frac{3^{m}+1}{2} \rightarrow \infty \text { as } m \rightarrow \infty
$$
and it follows that $\lim \frac{f\left(\beta_{m}\right)-f\left(\alpha_{m}\right)}{\beta_{m}-\alpha_{m}}$ does not exist. Hence $f$ is not differentiable at any point $x$.

abababa

回复 1# hbghlyj

在《实分析中的反例》里也有这个例子，紧跟在魏尔斯特拉斯的例子的后面。

hbghlyj

An Introduction to Harmonic Analysis, page 105-106

Lacunarity can be used technically in a number of ways. Our first example is "local"; it illustrates how a Fourier coefficient that
stands apart from the others is affected by the behavior of the function in a neighborhood of a point.
Lemma: Let $f \in L^{1}(\mathrm{~T})$ and assume that $\hat{f}(j)=0$ for all $j$ satisfying $1 \leqq\left|n_{0}-j\right| \leqq 2 N$. Assume that $f(t)=O(t)$ as $t \rightarrow 0$. Then
$$
\left|\hat{f}\left(n_{0}\right)\right| \leqq 2 \pi^{4}\left(N^{-1} \sup _{|t|<N^{-1 / 4}}\left|t^{-1} f(t)\right|+N^{-2}\|f\|_{L^{1}}\right) .
$$
Corollary: Let $\left\{\lambda_{n}\right\}$ be a lacunary sequence and $f \sim \sum a_{n} \cos \lambda_{n} t$ be in $L^{1}(\mathbf{T})$. Assume that $f$ is differentiable at one point. Then $a_{n}=o\left(\lambda_{n}^{-1}\right)$.

Proof: Assume that $f$ is differentiable at $t=0$. Replacing it, if necessary, by $f-f(0) \cos t-f^{\prime}(0) \sin t$ we can assume $f(0)=f^{\prime}(0)=0$. It follows that $f(t)=o(t)$ as $t \rightarrow 0$. The lacunarity condition on $\left\{\lambda_{n}\right\}$ is equivalent to saying that there exists a positive constant $c$ such that, for all $n$, none of the numbers $j$ satisfying $1 \leqq\left|\lambda_{n}-j\right| \leqq c \lambda_{n}$ is in the sequence; hence $\hat{f}(j)=0$ for all such $j$. Applying the lemma with $n_{0}=\lambda_{n}$ and $2 N=c \lambda_{n}$ we obtain $a_{n}=2 \hat{f}\left(\lambda_{n}\right)=o\left(\lambda_{n}^{-1}\right)$.

Corollary: The Weierstrass function $\sum 2^{-n} \cos 2^{n} t$ is nowhere differentiable.

hbghlyj

回复 1# hbghlyj

这段文字和《Principles of Mathematical Analysis》141页7.18节，几乎完全一样。

PS、新论坛没有回复1#的按钮。