Topologically equivalent metrics but not strongly equivalent in $\mathrm{Lip}_M(\mathbb{R})=\{f:[0,1]\rightarrow\mathbb{R}:|f(y)-f(x)|\leq M·|y-x|\}$

Question

Let's consider this set $\mathrm{Lip}_{M}(\mathbb{R})=\{f: [0,1] \rightarrow \mathbb{R} : |f(y)-f(x)|\leq M\cdot|y-x| \}$ (i.e Lipschitz functions in $[0,1]$). How can I prove that $(\mathrm{Lip}_{M}(\mathbb{R}), d_{\infty})$ and $(\mathrm{Lip}_{M}(\mathbb{R}), d_{1})$ are topologically equivalent (this means that if a sequence converges for $d_{\infty}$ then it converges for $d_{1}$ and viceversa or that a subset $A$ is open for $d_{\infty}$ if and only if it is open for $d_{1}$) but that they are not strongly equivalent (i.e there is no constant C such that $d_{\infty}(f,g)\leq C \cdot d_{1}(f,g)\ \ \forall f, g \in \mathrm{Lip}_{M}(\mathbb{R})$)?

Here is my proof for the latter. By definition,

$$d_{\infty}(f,g)=\sup_{x \in [0,1]} |f(x)-g(x)|$$

$$d_{1}(f,g)=\int_{0}^{1} |f(x)-g(x)|\,dx$$

Clearly,

$$ \begin{multline}d_{1}(f,g)=\int_{0}^{1} |f(x)-g(x)|\,dx \leq \int_{0}^{1} \sup_{x \in [0,1]}|f(x)-g(x)|\,dx \\ = \sup_{x \in [0,1]}|f(x)-g(x)| \cdot (1-0) = d_{\infty}(f,g). \end{multline} $$

To show that these metrics are not strongly equivalent, let's consider the sequence of functions $f_{n}(x) = (1/n) \cdot \cos(n \pi x/2)$, each of which is Lipschitz in $[0,1]$. So, $d_{\infty}(f_{n},0) = 1/n$, but $$d_{1}(f_{n},0) = \int_{0}^{1} \left|\frac{1}{n}\cdot\cos\left(nx \frac{\pi}{2}\right)\right|\,dx = \frac{1}{n} \cdot \sin\left(nx \frac{\pi}{2}\right)\Big|_{0}^{1} \cdot \frac{2}{n\pi}=(-1)^n\ \frac{2}{n^2\pi}$$ So if there is a constant $C$ such that $d_{\infty}(f,g)\leq C \cdot d_{1}(f,g)\ \ \forall f, g \in \mathrm{Lip}_{M}(\mathbb{R})$, we should have that, for all $n \in \mathbb{N}$, $$\frac{1}{n} \leq C \cdot (-1)^n\ \frac{2}{n^2\pi} \quad\text{and hence}\quad n\leq C\cdot \frac{2}{\pi}$$ which is absurd. Is this correct?

But I don't know how to show that this metrics are topologically equivalent.

Answer 1

That's a creative problem. I'll address the topological equivalence.

Suppose you have a sequence of functions $\{ f_{n} \}_{n=1}^{\infty}$ for which $$ |f_{n}(x)-f_{n}(y)| \le M|x-y|,\;\;\; x,y \in [0,1]. $$ In one case, you get to assume that $\{ f_{n}\}$ converges in the uniform norm to some $f$ satisfying the above, and you must show that $\{ f_{n}\}$ also converges to $f$ in the $L^{1}[0,1]$ norm. That's not so hard because the integral norm of $f-f_{n}$ is bounded by a constant times the uniform norm of $f-f_{n}$. In the other case, you get to assume that there is some $f$ satisfying the above Lipschitz inequality to which $\{ f_{n}\}$ converges in the $L^{1}[0,1]$ norm, and you must show that $\{ f_{n}\}$ converges to $f$ in the uniform norm. Because $\{ f_{n} \}$ converges in $L^{1}$ to $f$, then there is a subsequence $\{ f_{n_{k}}\}$ which converges pointwise a.e. to $f$. Say that it converges on the set $S$ with $[0,1]\setminus S$ of measure zero. Note that the set $S$ is dense in $[0,1]$ because no interval $(x-\epsilon,x+\epsilon)$ can be fully contained in $[0,1]\setminus S$.

Now you can use an $\epsilon/3$ argument to show that $\{ f_{n_{k}}\}$ converges uniformly to $f$. More explicitly, let $\epsilon > 0$ be given. Choose $N$ so that the norm of the partition $\{0,\frac{1}{N},\frac{2}{N},\ldots,\frac{N}{N}\}$ is bounded by $\frac{\epsilon}{3(M+1)}$ where $M$ is the uniform Lipschitz constant given above. Choose $s_{j} \in S\cap (\frac{j-1}{N},\frac{j}{N})$ for $1 \le j \le N$, and choose $K$ large enough that $$ |f_{n_{k}}(s_{j})-f(s_{j})| < \frac{\epsilon}{3}, \;\;\; k \ge K,\;\; 1 \le j \le N. $$ Every $x \in [0,1]$ is in some interval $[\frac{j-1}{N},\frac{j}{N}]$ and, hence, for $k \ge K$, $$ |f_{n_{k}}(x)-f(x)| \le |f_{n_{k}}(x)-f_{n_{k}}(s_{j})| + |f_{n_{k}}(s_{j})-f(s_{j})|+|f(s_{j})-f(x)| < \left[\frac{2M}{3(M+1)}+\frac{1}{3}\right]\epsilon. $$ It follows that $\|f_{n_{k}}-f\|_{\infty}< \epsilon$ for $k \ge K$. Because $\epsilon > 0$ was arbitrary, then $\{ f_{n_{k}}\}$ converges in the uniform norm to $f$. Finally note that every subsequence of $\{ f_{n}\}$ has a subsequence which converges to $f$ in the uniform norm. Therefore $\{ f_{n}\}$ converges in the uniform norm to $f$.

Answer 2

Very nice about the strong equivalence.

Now about topological equivalence.I'll give you some hints:

If $d_{\infty}(f_n,f)\to 0$ then $f_n\to f$ uniformly. Thus $\int f_n \to \int f$.

And if $\int_{0}^{1} |f_n(x)|dx\to 0$ find an area where $|f_n(x)|<ε$ and thus $sup|f_n(x)|<ε$ .

Answer 3

Your proof of the second statement ($d_1$ and $d_\infty$ are not strongly equivalent) is a solid attempt, but it's not quite there: some details slipped through.

For instance, $$\sin\left(nx \frac{\pi}{2}\right) \Big|_{x=0}^1 = \sin\left(n\cdot 1 \cdot \frac{\pi}{2}\right) - \sin\left(n\cdot 0 \cdot \frac{\pi}{2}\right) = \sin\left(n\frac{\pi}{2}\right)$$ which is not equal to $(-1)^n$—rather, it cycles between $1$, $0$, $-1$ and $0$ starting at $n = 1$. Also note that if that were true, then the inequality would yield $\frac{1}{n} \le -C \frac{2}{n^2\pi}$ for odd $n$, which is impossible since $n$ and $C$ are positive. (By the way, you seem to have made the $(-1)^n$ factor magically disappear in the following inequality ;-) .)

But more importantly, you can't ignore the absolute value sign in the integral: $$\int_{0}^{1} \left|\frac{1}{n}\cdot\cos\left(nx \frac{\pi}{2}\right)\right|\,dx$$ is not the same as $$\int_{0}^{1} \frac{1}{n}\cdot\cos\left(nx \frac{\pi}{2}\right)\,dx = \frac{2}{n^2\pi} \cdot \sin\left(nx \frac{\pi}{2}\right)\Big|_{0}^{1} $$ because $\cos(nx\frac{\pi}{2})$ changes sign $\lfloor\frac{n}{2}\rfloor$ times in between 0 and 1.

A way we can compute this integral is to exploit the periodicity and symmetry of $\cos$: the area under the curve of $|\cos(x \frac{\pi}{2})|$ between $0$ and $n$ is made up of a number of "bumps" of the same area, so we can simply compute the area of one of these half-bumps and count the number of half-bumps on the interval. For example, for $n=6$, there are $6$ copies of the area labelled $A$:

In fact, for any $n \in \mathbb{N}$, there are exactly $n$ half-bumps. This is because at every multiple of $\frac{\pi}{2}$, $\cos$ has either a maximum, a minimum, changes sign from positive to negative, or from negative to positive, and these are precisely the points that limit our half-bumps; for any $n$, these multiples are at $k = 0, 1, 2, \ldots, n$, so there are $n + 1$ of them, meaning there are $(n + 1) - 1 = n$ half-bumps.

Also notice that the number of half bumps in $|\cos(nx \frac{\pi}{2})|$ between $0$ and $1$ is the same as the number of half bumps in $|\cos(x \frac{\pi}{2})|$ between $0$ and $n$.

Finally, the area of one of these half-bumps (for $|\cos(nx\frac{\pi}{2})|$ now, not $|\cos(x\frac{\pi}{2})|$) is $$\int_{0}^{\frac{1}{n}} \cos\left(nx \frac{\pi}{2}\right)\,dx = \frac{1}{n} \int_{0}^{1} \cos\left(x \frac{\pi}{2}\right)\,dx = \frac{2}{n\pi} \cdot \sin\left(x \frac{\pi}{2}\right)\Big|_{0}^{1} = \frac{2}{n\pi}$$

Therefore, we have $$\int_{0}^{1} \left|\frac{1}{n}\cdot\cos\left(nx \frac{\pi}{2}\right)\right|\,dx = \frac{1}{n} \int_{0}^{1} \left|\cos\left(nx \frac{\pi}{2}\right)\right|\,dx = \frac{1}{n} \cdot n \cdot \frac{2}{n\pi} = \frac{2}{n\pi}$$

Now, if we apply the inequality, we don't get anything contradictory:

$$\frac{1}{n} \le C \frac{2}{n\pi} \quad\rightsquigarrow\quad C \ge \frac{\pi}{2}$$

Note that this doesn't mean that there is such a $C$, just that this is not sufficient proof to show that there isn't.