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I'm having trouble with the following problem:

Let $(X, \rho)$ be a metric space containing the point $x_0$. Define $\text{Lip}_0(X)$ to be the set of real-valued Lipschitz functions $f$ on $X$ that vanish at $x_0$. The norm is given by: $$\|f\|=\sup_{x\neq y}\frac{|f(x)-f(y)|}{\rho(x,y)}$$

  1. Show that $\text{Lip}_0(X)$ is a Banach space.

  2. For each $x\in X$, define a linear functional $F_x(f) = f(x)$. Show that $F_x$ belongs to $L(\text{Lip}_0(X),\mathbb{R})$

  3. For all $x,y \in X$ show $\|F_x-F_y\| = \rho(x,y)$

  4. Use the preceding facts to show that every normed linear space is a dense subspace of a Banach space.

My attempts:

  1. I verified this was a normed linear space by checking axioms of a vector space and axioms of a norm. I'm having trouble showing it's complete. I start with a Cauchy sequence $(f_n)$. Then I need to produce a candidate limit and show it's in my space. I'm not sure how to argue here.

  2. Done (just added for #4)

  3. Edit: See comments of first answer. I have an argument for $\|F_x=F_y\| \leq \rho(x,y)$ but not the other inequality.

  4. By the above, if $(X,\rho)$ is a normed linear space (so certainly a metric space), then certainly $X \subset L(\text{Lip}_0(X))$ (#3 shows its an isometric subset, in fact). I'm getting mixed up arguing that it should be dense.

Giuseppe Negro
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yoshi
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    I have notified somebody to your question. I run out of idea. – user284331 Apr 02 '18 at 02:36
  • For part 4, you do not need to show density. Just note that the closure $B := \overline{X}\subset L(\mathrm{Lip}_0(X))$ is a Banach spaces in which $X$ is dense. I don't think that $X$ will be dense in $L(\mathrm{Lip}_0(X))$ I'm general. – PhoemueX Apr 02 '18 at 07:17
  • Ah... I have misunderstood that. But for the third part, how do we show the equality? I found it is tricky. – user284331 Apr 02 '18 at 14:33
  • That $|F_{x}-F_{y}|\leq\rho(x,y)$ is easy, but the other direction seems to be tricky. – user284331 Apr 02 '18 at 14:35
  • @PhoemueX, do you know any clue for the third one? – user284331 Apr 05 '18 at 03:19
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    @user284331: Sure: For $x \neq y$, consider $f : X \to \Bbb{R}, z \mapsto d(z,x) - d(x_0,x)$. Then $f \in Lip_0 (X)$ with $|f| = 1$ (check this!), and $F_x(f) - F_y(f) = d(x,x) - d(y,x) = - d(y,x)$, whence $|F_x - F_y| \geq |-d(y,x)| = d(x,y)$. Here, $d = \varrho$ in your notation. How did I come up with this? Essentially all that we can work with is the metric $\varrho$, so I tried to use it to build a suitable function. – PhoemueX Apr 05 '18 at 21:02

1 Answers1

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Given a Cauchy sequence $(f_{n})\subseteq\text{Lip}_{0}(X)$. In particular, we have for every $\epsilon>0$, an $N$ is such that $\|f_{n}-f_{m}\|=\sup_{x\ne y}\dfrac{|(f_{n}-f_{m})(x)-(f_{n}-f_{m})(y)|}{\rho(x,y)}<\epsilon$ for all $n,m\geq N$.

Realizing to $y=x_{0}$ and using the assumption that $f_{n}(x_{0})=0$ for each $n$, we have $|f_{n}(x)-f_{m}(x)|\leq\epsilon\rho(x,x_{0})$ for all $n,m\geq N$ and each fixed $x\in X$. So the sequence of real numbers $(f_{n}(x))$ is Cauchy and hence convergent, say, $\lim_{n\rightarrow\infty}f_{n}(x)=f(x)$.

Back to the $\|f_{n}-f_{m}\|$, we have \begin{align*} |f_{n}(x)-f_{m}(x)-(f_{n}(y)-f_{m}(y))|\leq\epsilon\rho(x,y),~~~~x,y\in X~~~~n,m\geq N. \end{align*} Taking $m\rightarrow\infty$ yields that $|f_{n}(x)-f(x)-(f_{n}(y)-f(y))|\leq\epsilon\rho(x,y)$, so $\|f_{n}-f\|\leq\epsilon$ for all such $n$, this shows that $f_{n}\rightarrow f$ in $\text{Lip}_{0}(X)$. Note that $f(x_{0})=\lim_{n\rightarrow\infty}f_{n}(x_{0})=0$.

user284331
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  • Thanks! Do you have ideas on the part 4? – yoshi Apr 02 '18 at 01:26
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    By the way, how do you show the third part? – user284331 Apr 02 '18 at 01:53
  • I think so: $| F_x - F_y | = \sup_{f \neq 0} \frac{|(F_x-F_y)(f)|}{|f|} = \sup_{f \neq 0} \frac{|f(x)-f(y)|}{\frac{|f(x)-f(y)|}{|\rho(x,y|}}$. Canceling gives the result. – yoshi Apr 02 '18 at 03:01
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    No, that is not correct. The norm of $f$ is defined as the supremum of all such $x,y$, now $x,y$ are fixed, there is a problem here. Of course, that inequality $\leq\rho(x,y)$ is easy. For the other direction, it seems that there is some delicate construction for that. – user284331 Apr 02 '18 at 03:05
  • ah yes, i was sloppy about this. I still have the other inequality to show. – yoshi Apr 02 '18 at 03:08
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    How do you know that $f$ is Lipschitz? – user0 Jul 17 '23 at 18:39