How to prove that whether it is a Banach space or not?

Question

We consider the Banach space of all continuous functions on $X$ such that for each $f$ in the space, \begin{equation*} ||f||=\sup_{x\neq y}\frac{\left\vert f(x)-f(y)\right\vert }{\left\vert x-y\right\vert }. \end{equation*}

How I can prove that it is a Banach space?

Answer 1

In general it is kind of hard to guess whether a space is complete or not. Usually you may try to describe the property defining your norm/functions in geometric terms and try to see if you can construct a sequence of functions in your space so that their pointwise limit breaks the property defining your space. If you cannot find such thing, you could try proving that your space is complete and see what you get. If you find an obstruction in your proof, then you can try to see if that obstruction can help constructing a counterexample. It is a back and forth process, a lot of experimentation needed.

In this particular case, I would advice to try proving that the pointwise limit is in your space. Arzela-Ascoli's theorem might come in handy.

Answer 2

Suppose $\{f_n\}$ is a Cauchy sequence in $(CL^0,\|\cdot\|_{CL^0})$. Then

• $\{f_n\}$ is bounded (in the $\|\cdot\|_\infty$ norm), and
• $\{f_n\}$ is equicontinuous.

To see the first point, let $M:=\sup_n\|f_n\|_n$, which is finite. For any $n$, one has $$|f_n(x)|=|f_n(x)-f_n(0)| \le m(f_n)|x-0| \le M|x|\le M,$$

so $\|f_n\|_\infty \le M$. To see the second point, fix $\epsilon>0$ and let $\delta:=\epsilon/M$. Then for all $n$ and all $x,y\in[0,1]$, if $|x-y|<\delta$ then $$|f_n(x)-f_n(y)| \le m(f_n)|x-y| \le M\delta = \epsilon.$$

It follows from the Arzela-Ascoli theorem that $\{f_n\}$ has a uniformly convergent subsequence, i.e. there exists $\{n_k\}$ and a continuous function $f$ such that $f_{n_k}\to f$ uniformly. Of course, we immediately have $f(0)=0$ and $m(f) \le M$, so $f\in CL^0$. It remains to show $f_{n_k} \to f$ in $\|\cdot\|_{CL^0}$.

Fix $\epsilon>0$. There exists $K$ such that $\|f_{n_j}-f_{n_k}\|_{CL^0} < \epsilon/3$ for all $j,k\ge K$. Now let $x,y\in[0,1]$ with $x\neq y$. Since $f_{n_k}\to f$ uniformly, there exists $K^{x,y} \ge K$ such that $\|f_{n_{K^{x,y}}}-f\|_\infty < \epsilon|x-y|/3$. It follows that, if $k\ge K$, \begin{align*} &\frac{|(f_{n_k}-f)(x) - (f_{n_k}-f)(y)|}{|x-y|} \\ &\qquad\qquad\qquad\le \frac{|(f_{n_k}-f_{n_{K^{x,y}}})(x) - (f_{n_k}-f{n_{K^{x,y}}})(y)|}{|x-y|} + \frac{|(f_{n_k}-f)(x) - (f_{n_k}-f)(y)|}{|x-y|} \\ &\qquad\qquad\qquad\le \|f_{n_k}-f_{n_{K^{x,y}}}\|_{CL^0} + \frac{2\|f_{n_k}-f\|_\infty}{|x-y|} \\ &\qquad\qquad\qquad\le \frac{\epsilon}3 + \frac{2\epsilon}3 = \epsilon. \end{align*} Crucially, since $K$ does not depend on $x$ or $y$, we can take supremum over $x,y\in[0,1]$ with $x\neq y$ and conclude that, if $k\ge K$, $$\|f_{n_k}-f\|_{CL^0} \le \epsilon.$$

Answer 3

$m$ is not a norm on the set of the continuous function on $[0, 1]$ with $f(0)=0$, because not all continuous functions on $[0, 1]$ are Lipschitz. For example, take $f(x)=\sqrt{x}$. It's continuous, $f(0)=0$, but $$\frac{|\sqrt{x}-\sqrt{y}|}{|x-y|}=\frac{1}{|\sqrt{x}+\sqrt{y}|}$$ And taking $x=0$, we get that $$m(f)\geqslant \lim_{y \to 0+0} \frac{1}{\sqrt{y}}=+\infty$$