1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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
  • 1
    $\begingroup$ 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
3
$\begingroup$

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 $\lVert\cdot\rVert$.

$\endgroup$

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.