Question: Let $\mathcal{L}$ be the normed space of all Lipschitz functions on a Banach space $X$ that are equal to $0$ at the origin, under the norm $$\|f\| = \sup\left\{ \frac{|f(x)-f(y)|}{\|x-y\|}:x,y\in X \right\}.$$ Show that $\mathcal{L}$ is a Banach space.

My attempt:

Let $(f_n)_{n=1}^\infty$ be a Cauchy sequence in $(\mathcal{L},\|\cdot\|).$ Define $f:X\to \mathbb{R}$ by $$ f(x) = \lim_{n\to\infty} f_n(x). $$ We claim that $f$ is well-defined, that is, the limit exists. Fix $x\in X\setminus \{0\}$ and $\epsilon>0.$ Since $(f_n)_{n=1}^\infty$ is Cauchy, there exists $N\in \mathbb{N}$ such that for any $m,n\geq N,$ $$ \|f_m-f_n\| <\frac{\epsilon}{\|x\|}. $$ Note that $$ \|f_m-f_n\| = \sup_{x,y\in X} \frac{|(f_m-f_n)(x) - (f_m-f_n)(y) |}{\|x-y\|} \geq \frac{|f_m(x) - f_n(x)|}{\|x\|}. $$ So we have $|f_m(x) - f_n(x)| < \epsilon.$ Therefore, $(f_n(x))_{n=1}^\infty$ is Cauchy in $\mathbb{R}.$ Since $\mathbb{R}$ is complete, so $\lim_{n\to\infty} f_n(x)$ exists, that is, $f$ is well-defined.

We claim that $f\in \mathcal{L}.$ By construction, $f(0) = 0.$ Since $(f_n)_{n=1}^\infty$ is Cauchy in $(\mathcal{L},\|\cdot\|),$ therefore it is bounded, that is, there exists $B>0$ such that $\|f_n\|\leq B$ for all $n\geq 1.$ Since $$f = f_N + (f-f_N),$$ so for any $x,y\in X,$ \begin{align*} |f(x)-f(y)| & \leq |f_N(x)-f_N(y)| + |(f-f_N)(x) - (f-f_N)(y)| \\ & \leq \|f_N\|\| x-y\| + \|f-f_N\| \|x-y\| \\ & \leq \|f_N\|\| x-y\| + \|x-y\| \\ & = (\|f_N\|+1)\|x-y\|. \end{align*} Therefore, $\|f\|$ is finite and hence $f\in \mathcal{L}.$

Lastly, we wish to prove that $(f_n)$ converges to $f$ in $(\mathcal{L},\|\cdot\|).$ Indeed, for every $\epsilon>0,$ by Cauchyness of $(f_n),$ there exists $N\in\mathbb{N}$ such that for any $m,n\geq N,$ $$ \|f_n-f_m\|<\frac{\epsilon}{2}. $$ Then $$ \|f-f_n\| = \|\lim_{m\to\infty} f_m - f_n\| = \lim_{m\to\infty} \|f_m-f_n\| < \epsilon. $$ We conclude that $\mathcal{L}$ is complete.

Is my proof above correct?

  • $\begingroup$ Why define $f(0) = 0$ instead of $f(0) = \lim_{n \to \infty} f_n(0)$? This means, if your Cauchy sequence is constantly, say, the constant function $1$, then $f(x)$ will be the function that is constantly $1$ except at $0$, where it's $0$. Such a function is not continuous, let alone Lipschitz. $\endgroup$ Feb 3, 2019 at 5:50
  • $\begingroup$ @TheoBendit Yes, you are right. Edited my proof. $\endgroup$
    – Idonknow
    Feb 3, 2019 at 5:52
  • $\begingroup$ Wait, I'm an idiot, I just noticed the requirement that $f(0) = 0$ for membership in the space. The constant function $1$ does not belong to the space in the first place! $\endgroup$ Feb 3, 2019 at 5:54
  • $\begingroup$ @TheoBendit Yes. But I think your suggestion still works and it is neater than my suggestion. $\endgroup$
    – Idonknow
    Feb 3, 2019 at 5:59

1 Answer 1


Your proof has all of the right ideas but suffers from a couple of defects since you end up implicitly doing things like assuming $f \in \mathcal{L}$ before you've proved it and confusing the pointwise limit with the limit in $\mathcal{L}$. To fix this, you just need to write everything down in terms of the definition of the norm and make sure you're working with pointwise limits. I include details below.

Firstly, when trying to prove that $f \in \mathcal{L}$, on the second line you write down $\|f-f_N\|$ and then bound this by $1$. Doing this already requires you to assume $f-f_N \in \mathcal{L}$ and hence $f \in \mathcal{L}$ and also something like $f_n \to f$ in $\mathcal{L}$.

Instead, write \begin{align*} |f(x) - f(y)| &= |\lim_{n \to \infty} (f_n(x) - f_n(y))| \\&= \lim_{n \to \infty} |f_n(x) - f_n(y)| \\& \leq \lim_{n \to \infty} \|f_n\| |x-y| \\& \leq C|x-y| \end{align*} where in the last line we used that $\|f_n\|$ is a bounded sequence. Note that in the second line, I only use that $f_n \to f$ pointwise and not in $\mathcal{L}$.

This is important because when trying to prove that $f_n \to f$, you take a pointwise limit and take it outside of the $\mathcal{L}$-norm. This doesn't work since you can only do that once you know that the limit converges in $\mathcal{L}$ and not just pointwise. Instead, work again with a pointwise limit by writing \begin{align*} |f(x) - f(y) - f_n(x) + f_n(y)| &= \lim_{m \to \infty} |f_m(x) - f_m(y) - f_n(x) + f_n(y)|\\ & \leq \lim_{m \to \infty} \|f_m - f_n\| |x-y| \end{align*} Now since $(f_n)$ is Cauchy in $\mathcal{L}$, there is $N$ independent of $x,y$ such that for $n,m \geq N$ $\|f_n - f_m\| \leq \varepsilon$. So we get $$|f(x) - f(y) - f_n(x) + f_n(y)| \leq \varepsilon |x-y|$$ which implies for $n \geq N$, $\|f-f_n\| \leq \varepsilon$ so $f_n \to f$ in $\mathcal{L}$.

  • $\begingroup$ Maybe some limits have to be replaced by limsup? For example $\lim\|f_n(x)-f_n(y)\|\leq\limsup\|f_n\||x-y|$ since we don't know whether the limit of the norms exists... $\endgroup$ Jan 27, 2021 at 19:35
  • 1
    $\begingroup$ @LorenzoCecchi $f_n$ is a Cauchy sequence for $\|\cdot\|$ and so by the reverse triangle inequality $\|f_n\|$ is Cauchy in $\mathbb{R}$ and hence converges. $\endgroup$ Jan 27, 2021 at 20:19

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