Establishing properties of set of Lipschitz functions

Question

I am trying to solve the following exercise in Kolmogorov's analysis textbook.

Let $M_K$ be the set of all functions $f$ in $C_{[a,b]}$ satisfying a Lipschitz condition, i.e., the set of all $f$ such that \begin{align*} |f(t_1) - f(t_2)| \leq K |t_1 - t_2| \end{align*} for all $t_1, t_2 \in [a,b]$, where $K$ is a fixed positive number.

a) $M_K$ is closed and in fact is the closure of the set of all differentiable functions on $[a,b]$ such that $|f'(t)| \leq K$.

b) The set $M = \bigcup_K M_K$ of all functions satisfying a Lipschitz condition for some $K$ is not closed.

c) The closure of $M$ is the whole space $C_{[a,b]}$.

I know that for part (a) I need to show that $M_k$ contains all of its limit points, though there are alternate definitions of closed that may be of more use, as I cannot figure out how to get started for this problem. Part (b) I think I have some idea. A set $A$ is dense in $B$ if $B \subset [A]$. The polynomials with rational coefficients are given to be dense in $C_{[a,b]}$, and since the rationals are dense in the reals, it must follow that polynomials with real coefficients are dense in $C_{(a,b)}$. I think the result follows from the fact that if $M$ were closed, we wouldn't be able to find a dense proper subset of $M$ that doesn't satisfy the Lipschitz condition. (I hope I am not misstating this.)

For part (c), it seems possible to proceed by contradiction, supposing that the space is smaller than $C_{[a,b]}$. It would need to, in cardinality, lie between $M$ and $C_{[a,b]}$, and by denseness such a proper subset of $C_{[a,b]}$ doesn't exist.

Any help would be greatly appreciated.

Answer 1

Here is (a) to get you started:

Suppose $f_n \to f$ in $C[a,b]$ and $|f_n(t_1)-f_n(t_2)| \le M |t_1-t_2|$ for all $n$, then taking limits we have $|f(t_1)-f(t_2)| \le M |t_1-t_2|$.

For (b) take a function like $x \mapsto \sqrt{|x|}$ which is not Lipschitz and find a sequence o Lipschitz approximations.

For (c) note that the polynomials are dense in $C[a,b]$.

Addendum:

I'm not sure if this was Kolmogorov adding extra information of if you wanted an answer.

Using the mean value theorem it is straightforward to show that if $f_n \to f$ and $f_n$ is differentiable with $\|f_n'\|_\infty \le M$, then $f$ is Lipschitz with rank $M$.

For the other direction, suppose $f$ is Lipschitz with rank $M$. For convenience extend $f$ to the 'right' with the value $f(b)$. Define $f_n(x) = f(a)+\int_a^x { f(t+{1 \over n})-f(t) \over {1 \over n}} dt $. It is easy to see that $f'_n(x) = { f(x+{1 \over n})-f(x) \over {1 \over n}}$ which is bounded by $M$ hence $f_n$ is differentiable and Lipschitz with rank $M$.

It is straightforward to see that $f_n(x) = f(a) -n \int_a^{a+{1 \over n}} f(s) ds +n \int_x^{x+{1 \over n}} f(s) ds $ and so \begin{eqnarray} |f_n(x)-f(x)| &\le& |f(a)-n \int_a^{a+{1 \over n}}f(s)ds|+n \int_x^{x+{1 \over n}}|f(s)-f(x)|ds \\ &\le& n\int_a^{a+{1 \over n}} |f(a)-f(s)|ds|+n \int_x^{x+{1 \over n}}|f(s)-f(x)|ds \\ &\le& {1 \over n} M + {1 \over n} M \end{eqnarray} and so $f_n \to f$ in $C[a,b]$.

Answer 2

Hint

(a) If $f_n \to f$ and $|f_n(a)-f_n(b)| \leq K|a-b|$ you need to show that $f$ also satisfies this inequality... Take limits.

(b) You need to find a sequence $f_n$ and som $K_n$ such that $|f_n(a)-f_n(b)| \leq K_n|a-b|$, but the limit doesn't satisfy this inequality for any $K_n$.

If $K_n$ are bounded, then you can show that $K:= \sup \{ K_n \}$ would work for the limit, so try to find a sequence such that $K_n \to \infty$.

(c): Show that any polynomial belongs to some $M_K$.

For the second part of (a), try approximating $f$ by trig polynomials $P_n$. Now, the issue is that $K_n:= \sup \{ P_n'(t) \}$ may be larger than $K$.

Show that $K_n \to K$ and $\frac{K}{K_n}P_n$ is the approximation you want.

Note here that the case $K=0$ is trivial, and if $K \neq 0$, then $K_n \neq 0$ for $n$ large enough, so the above approximation works.