Let $V$ be the space of real valued lipschitz functions over $[a,b]$,we define:

$M_f=sup_{x\neq y}\frac{|f(x)-f(y)|}{|x-y|}$

and lipschitz norm:


prove that $V$ with lipschitz norm is a complete normed vector space.

it is easy to see $V$ is a vector space,but what about the completeness?

is there any hint?

thank you very much.


You should start from a hypothetical Cauchy sequence of functions (Cauchy in the sense of the newly defined norm). Prove that the values of that function at the left end $a$ converge as real numbers, and then prove that the functions converge pointwise at any $x\in[a,b].$ This allows you to write $f(x)$ for the pointwise limit. Now establish that the sequence actually converges to $f$ in the sense of the new norm.

Explicitly, let $\epsilon>0$ be given. Pick $N$ such that for $m,n\geq N,$ $M_{f_m-f_n}<\epsilon/4.$

For every $x,y\in[a,b]$ choose $m_{x,y}\geq N$ such that $|f_{m_{x,y}}(x)-f(x)|<\epsilon|x-y|/4$ and $|f_{m_{x,y}}(y)-f(y)|<\epsilon|x-y|/4.$ Then

$$\eqalign{ \forall x,y\in[a,b]:&|(f(x)-f_n(x))-(f(y)-f_n(y))|\\ &\leq |f(x)-f_{m_{x,y}}(x)|+|f_{m_{x,y}}(x)-f_n(x)+f_n(y)-f_{m_{x,y}}(y)| +|f_{m_{x,y}}(y)-f(y)| \\ &\leq \epsilon|x-y|. }$$

  • $\begingroup$ I showed for every $x $ $f_n (x)$ is cauchy, so the limit is function $g $,now I must say g is the limit of $f_n$'s with lip norm.isn't it true? $\endgroup$ – user115608 Dec 13 '15 at 15:44
  • $\begingroup$ It doesn't! I only needed pointwise convergence to be able to legitimately write the expression $f(x).$ Convergence in the Lipschitz norm follows from the Cauchy nature of the sequence. $\endgroup$ – Justpassingby Dec 13 '15 at 15:44
  • $\begingroup$ sorry if I don't understand you,but how "Convergence in the Lipschitz norm follows from the Cauchy nature of the sequence."? $\endgroup$ – user115608 Dec 13 '15 at 15:47
  • $\begingroup$ would you please have a look at this? math.stackexchange.com/questions/1569194/… $\endgroup$ – user115608 Dec 13 '15 at 15:57

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