# Lipschitz functions form a Banach space, and Analyzing its Dual

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.

• I have notified somebody to your question. I run out of idea. – user284331 Apr 2 '18 at 2:36
• 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 2 '18 at 7:17
• Ah... I have misunderstood that. But for the third part, how do we show the equality? I found it is tricky. – user284331 Apr 2 '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 2 '18 at 14:35
• @PhoemueX, do you know any clue for the third one? – user284331 Apr 5 '18 at 3:19

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 $f(x_{0})=\lim_{n\rightarrow\infty}f_{n}(x_{0})=0$.
• 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 2 '18 at 3:01
• 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 2 '18 at 3:05