Let $(X,d)$ be a complete separable metric space. ($d$ is complete.) For any bounded Lipschitz function, we define the norm $||\cdot||$. $$ ||f||_1=\sup\left\lbrace\frac{|f(x)-f(y)|}{d(x,y)}\middle|\,x,y\in X, x \ne y \right\rbrace \\ ||f|| = ||f||_\infty+||f||_1 \quad\text{where }||f||_\infty=\sup\{f(x)\mid x \in X\} $$ I wanted to show that the space of bounded Lipschitz function under the norm $||\cdot||$ is complete, but I don't know how to.

My attempt: Let $(x_n)_n$ be a sequence dense in $X$. Let $\epsilon>0$. Let $(f_n)_n$ be a Cauchy sequence with respect to $||\cdot||$. Since the sup norm is complete and $||\cdot||_\infty \le ||\cdot||$, we get a uniform convergence $f_n \rightrightarrows f$ under $||\cdot||_\infty$. There exists $N\in\Bbb N$ such that \begin{align} & \forall\,m,n\ge N, ||f_m-f_n||_\infty+||f_m-f_n||_1<\epsilon \\ & \forall\,n\ge N, ||f_n-f||_\infty<\epsilon \end{align}

Since $(f_n)_n$ is Cauchy for the sup norm and for $||\cdot||_1$, it's bounded for those two norms. As $f_n\rightrightarrows f$, $f$ is also bounded. There exists $M>0$ so that \begin{align} & \forall\,n\in\Bbb{N},\forall x \in X, |f_n(x)|\le M \\ & \forall\, x \in X, |f(x)|\le M \\ & \forall\,n\in\Bbb{N},\forall\, x,y \in X, |f_n(x)-f_n(y)| \le M\,d(x,y) \end{align}

As I write out the expression for $||f-f_n||_1$, I can do nothing to control the denominator $d(x,y)$. For any $x,y\in X$ with $x\ne y$, $$ \frac{|(f-f_n)(x)-(f-f_n)(y)|}{d(x,y)} < \frac{2\epsilon}{d(x,y)}. $$ I tried finding some $x_*$ and $y_*$ in the dense sequence $(x_n)_n$ so that $d(x,x_*),d(y,y_*)<\epsilon$, but I found this useless as the fraction still appears to be $O(\epsilon)/d(x,y)$. As $d(x,y)\to 0$, this fraction will explode. Any help is welcome.

  • $\begingroup$ The quantity $\|\cdot\|_1$ is not a norm. $\endgroup$ – copper.hat Dec 15 '17 at 20:43
  • $\begingroup$ @copper.hat Can you explain why? We have the triangular inequality. (Fix $x,y$ and it follows from $|\cdot|$ in $\Bbb R$.) If it's zero, then the function inside will map everything to zero. $\endgroup$ – GNUSupporter 8964民主女神 地下教會 Dec 15 '17 at 20:49
  • $\begingroup$ It's a seminorm, $\lVert f\rVert_1 = 0$ if $f$ is constant. $\endgroup$ – Daniel Fischer Dec 15 '17 at 20:50
  • $\begingroup$ @DanielFischer Thank you. I'll fix it. The second one $||\cdot||$ should be a norm, so the rest should be unaffected. $\endgroup$ – GNUSupporter 8964民主女神 地下教會 Dec 15 '17 at 20:52
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    $\begingroup$ Perhaps it would help to think of $\|f\|_1 = \inf \{ L \ge 0 | \forall x,y, \ |f(x)-f(y)| \le L d(x,y) \}$? $\endgroup$ – copper.hat Dec 15 '17 at 20:57

Fix $\varepsilon > 0$. Since $(f_k)$ is a Cauchy sequence, there is an $N$ such that $\lVert f_k - f_m\rVert \leqslant \varepsilon$ for $k,m \geqslant N$. Fixing $m$ and letting $k \to \infty$, it follows that $\lVert f - f_m\rVert_{\infty} \leqslant \varepsilon$ for all $m \geqslant N$. Again fixing $m$, but now also fixing arbitrary $x, y\in X$, and letting $k \to \infty$, it follows that

$$\lvert (f - f_m)(x) - (f - f_m)(y)\rvert \leqslant \varepsilon d(x,y).$$

Since this holds for all $x,y\in X$, it follows that $\lVert f - f_m\rVert_1 \leqslant \varepsilon$. This holds for all $m \geqslant N$. Hence $\lVert f - f_m\rVert \leqslant 2\varepsilon$ for all $m \geqslant N$, so $f_k \to f$ with respect to the norm $\lVert\cdot\rVert$.


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