# If a sequence of unif. cont. Lipschitz functions $\{f_n\}$ converges unif. to an unif. cont. function f, then f is Lipschitz.

Let $f: (X,d) \to (\Omega,\rho)$ be a uniformly continuous function such that $\{f_n\}$ is a sequence of uniformly continuous functions that converges uniformly to $f$, where each $f_n$ is Lipschitz with constant $M_n$.

Show that if $\sup(M_n) < \infty$, then the function $f$ is Lipschitz . Show that $f$ may fail to be Lipschitz when $\sup(M_n) = \infty$.

My approach:

We must show that f is Lipschitz, then

$\exists M > 0, \forall x,y \in X, \rho(f(x),f(y)) \leq M d(x,y)$

We have that each $f_n$ is Lipschitz, then $\rho(f_n(x),f_n(y)) \leq M_n d(x,y), \forall x,y \in X$

Also by the triangle inequality,

$\rho(f(x),f(y)) \leq \rho(f(x),f_n(x)) + \rho(f_n(x),f(y))$

Using the triangle inequality again

$\rho(f(x),f(y)) \leq \rho(f(x),f_n(x)) + \rho(f_n(x),f_n(y)) + \rho(f_n(y),f(y))$

Since $f_n \to f$ uniformly, we have that

$\rho(f(x),f_n(x)) \to 0$ and $\rho(f(x),f_n(x)) \to 0$ as $n \to \infty$.

then

$\displaystyle\lim_{n\to\infty}\rho(f(x),f(y)) \leq \displaystyle\lim_{n\to\infty}\rho(f(x),f_n(x)) + \displaystyle\lim_{n\to\infty}M_n d(x,y) + \displaystyle\lim_{n\to\infty}\rho(f_n(y),f(y))$

Hence,

$\rho(f(x),f(y)) \leq 0 + \displaystyle\lim_{n\to\infty}M_n d(x,y) + 0 \leq \sup M_n d(x,y)$

Taking $M = \sup M_n$, since $0 < M < \infty$, we have that f is Lipschitz.

Now if $\sup M_n = \infty$, how can I show that $f$ "may fail to be Lipschitz"?. I know that my constant $M$ will not satisfy the definition of a Lipschitz function because it must be a real number, but I do not see how to prove that it "may fail to be Lipschitz". Any help would be very appreciated. Thanks.

• Saying that the $f_n$s are uniformly continuous is redundant, if you require them to be Lipschitz.
– user228113
Jun 16, 2017 at 15:41
• They are Lipschitz by hypothesis and also uniformly continuous. I need to proof that $f$ not $f_n$ are Lipschitz. The first part of this exercise was this one: math.stackexchange.com/questions/2322749/… Jun 16, 2017 at 15:42
• And my point is that a Lispchitz continuous function is a fortiori uniformly continuous.
– user228113
Jun 16, 2017 at 15:43
• Ok I get your point. Jun 16, 2017 at 15:45

If $M := \sup_n M_n < +\infty$, then $$\rho(f_n(x), f_n(y)) \leq M\, d(x,y) \qquad \forall x,y\in X, \quad \forall n\in\mathbb{N}.$$ Now, if the sequence $(f_n)$ converges pointwise to $f$, passing to the limit in the above relation gives $$\rho(f(x), f(y)) \leq M\, d(x,y) \qquad \forall x,y\in X,$$ i.e. $f$ is Lipschitz continuous.
For the last part: if you consider the sequence $$f_n(x) := \sqrt{x+\frac{1}{n}}, \qquad x\in [0,1],$$ then you see in a moment that each $f_n$ is of class $C^1([0,1])$, hence it is Lipschitz continuous in $[0,1]$ (with constant $M_n = \sqrt{n}/2$). On the other hand, the sequence $(f_n)$ converges uniformly in $[0,1]$ to the function $f(x) = \sqrt{x}$, which is not Lipschitz continuous.
• This is similar to my analysis but shorter because it does not use the triangle inequality. Thank you. However my problem is that I do not know how to verify that f may fail to be Lipschitz if the sup is $\infty$ Jun 16, 2017 at 15:48
• So basically you give me a sequence of uniformly continuous functions in [0,1] such that $\sup M_n = \infty$ and $f_n \to f$ uniformly, but f is not Lipschitz. I get it. Thank you very much. I just have to show that $M_n = \frac{\sqrt{n}}{2}$ and that $\sqrt{x}$ is not Lipschitz. Jun 16, 2017 at 15:55
• Both facts can be proved considering the behavior of the functions near the origin, and observing that $\sup f_n' = \lim_{x\to 0} f_n'(x) = M_n$, and $\sup f' = \lim_{x\to 0} f'(x) = +\infty$ (but $f$ is not differentiable at $x=0$). Jun 16, 2017 at 15:59
• @RichardClare Of course the triangle inequality is still there - hidden in (lower semi) continuity of $\rho(\cdot, \cdot)$. The answer uses the fact that $\rho(x_n,y_n) \to \rho(x,y)$ when $x_n \to x$, $y_n \to y$. Jun 16, 2017 at 18:48