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.

  • 1
    $\begingroup$ I have notified somebody to your question. I run out of idea. $\endgroup$ – user284331 Apr 2 '18 at 2:36
  • $\begingroup$ 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. $\endgroup$ – PhoemueX Apr 2 '18 at 7:17
  • $\begingroup$ Ah... I have misunderstood that. But for the third part, how do we show the equality? I found it is tricky. $\endgroup$ – user284331 Apr 2 '18 at 14:33
  • $\begingroup$ That $\|F_{x}-F_{y}\|\leq\rho(x,y)$ is easy, but the other direction seems to be tricky. $\endgroup$ – user284331 Apr 2 '18 at 14:35
  • $\begingroup$ @PhoemueX, do you know any clue for the third one? $\endgroup$ – 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 $f(x_{0})=\lim_{n\rightarrow\infty}f_{n}(x_{0})=0$.

| cite | improve this answer | |
  • $\begingroup$ Thanks! Do you have ideas on the part 4? $\endgroup$ – yoshi Apr 2 '18 at 1:26
  • 1
    $\begingroup$ By the way, how do you show the third part? $\endgroup$ – user284331 Apr 2 '18 at 1:53
  • $\begingroup$ 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. $\endgroup$ – yoshi Apr 2 '18 at 3:01
  • 1
    $\begingroup$ 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. $\endgroup$ – user284331 Apr 2 '18 at 3:05
  • $\begingroup$ ah yes, i was sloppy about this. I still have the other inequality to show. $\endgroup$ – yoshi Apr 2 '18 at 3:08

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.