Equicontinuity of family of polynomial approximations

Question

Suppose $f:[0,1]\to \mathbb{R}$ is uniformly continuous, and $(p_n)_{n\in\mathbb{N}}$ is a sequence of polynomial functions converging uniformly to $f$.

Does it follow that $\mathcal{F}=\{p_n\mid n\in\mathbb{N}\}\cup \{f\}$ is equicontinuous?

Also, if $C_n$ are the Lipschitz constants of the polynomials $p_n$, does it follow that $C_n<\infty$ for all $n$, and $\lim_{n\to\infty} C_n=\infty?$

I'm preparing for a test, but I'm not sure how to go about answering these two question. Any hints or tips as to what to look for would be appreciated.

Answer 1

Hint: $p_n(y) - p_n(x) = p_n(y)-f(y) + f(y)-f(x) + f(x) - p_n(x).$

Answer 2

Yes, $\mathcal F$ is equicontinuous (and the fact that the $p_n$'s are polynomials doesn't matter). Let $\varepsilon>0$. Since $f$ is uniformly continuous, there is a $\delta>0$ such that$$(\forall x,y\in[0,1]):|x-y|<\delta\implies\bigl|f(x)-f(y)\bigr|<\frac\varepsilon2.$$Take $p\in\mathbb N$ such that $n\geqslant p\implies\sup|f-p_n|<\frac\varepsilon4$. Then, if $n\geqslant p$ and $x,y\in[0,1]$,$$|x-y|<\delta\implies|p_n(x)-p_n(y)|\leqslant|p_n(x)-f(x)|+|f(x)-f(y)|+|f(y)-p_n(y)|<\varepsilon.$$This $\delta$ only works for the set $\{f\}\cup\{p_n\,|\,n\geqslant p\}$ but since thare are only finitely many elements of $\mathcal F$ outside this set, it is easy to see that there is a $\delta$ that works for the whole family $\mathcal F$.

Answer 3

I am also preparing for a test. This is my attempt for the second part of the question.

For every $n$, we have that $p_n$ is Lipschitz (because polynomials are Lipschitz), i.e, $$|p_n(x)-p_n(y)|<C_n|x-y|$$ so, $$|p_n(x)-p_n(y)|<\epsilon$$ for $|x-y|<\delta$ where $\delta = \frac{\epsilon}{C_n}$, and $\delta > 0$.

This implies that $C_n<\infty$, or $\exists M>0$ such that $|Cn|<M$, for otherwise $\delta=0$. This is $\forall n$.

Now, put $$\lim_{n\to \infty} C_n = C.$$

If $f$ is Lipschitz, $|f(x)-f(y)|<\epsilon$ whenever $|x-y|<\delta$ for $\delta = \frac{\epsilon}{C}$ and $\delta>0$. $(*)$

In this case also, $C_n<\infty$, or $\exists M>0$ such that $|Cn|<M$, because otherwise $\delta=0$.

So if $f$ is $Lipschitz$, $$\lim_{n \to \infty}C_n \neq \infty.$$

If $f$ is not $Lipschitz$, then $(*)$ doesn't hold for the choice of $\delta$ (because its not $Lipschitz$). But since we have $\forall n,$ $$|p_n(x) - p_n(y) |<C_n|x- y|$$ then $$\lim_{n\to \infty}|p_n(x) - p_n(y) |<\lim_{n\to \infty}C_n|x- y|$$ and we get $$|f(x) - f(y)|<C|x-y|$$ which must hold.

But since $f$ is not $Lipschitz$, it must be that $\forall M>0$, $C>M$.

Which means $\lim_{n \to \infty}C_n = \infty.$