Lipschitz functions form a Banach space, and Analyzing its Dual

Question

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.

Answer 1

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 $f(x_{0})=\lim_{n\rightarrow\infty}f_{n}(x_{0})=0$.