A sequence that is uniformly bounded on a compact interval is equicontinuous

Question

Problem Statement: If $\left\{f_{n}\right\}_{n\in\mathbb{N}}$ is a sequence of polynomials of degree $\leq m$ this is uniformly bounded on a compact interval $I$, then $\left\{f_{n}\right\}$ is equicontinuous on $I$.

I am trying to work through this problem, but I am having trouble determining my choices of $\delta$ and $\epsilon$ to show equicontinuity.

So we know that $\exists M>0$ such that $\lVert f_{n}(x)\rVert \leq M$, $\forall x\in X, n\in \mathbb{N}$.

Then to prove equicontinuity, I must show that $\forall \epsilon >0$, $\exists\delta=\delta(\epsilon)>0$ so that $$\lVert x-y\rVert <\delta \Rightarrow \lVert f_{n}(x)-f_{n}(y)\rVert <\epsilon,\ \forall x,y\in I, n\in\mathbb{N}.$$

One approach: Since polynomials are continuous, taking $\epsilon >0$, we can find a $\delta=\delta(\epsilon)$ so that for any $y\in B_{\delta}(x)$, we have $$\lVert x-y\rVert <\delta \Rightarrow \lVert f_{n}(x)-f_{n}(y)\rVert <\epsilon,\ \forall x,y\in I, n\in\mathbb{N}.$$ But we must show that this is true for any $x,y\in I$. Then since $I$ is compact, $\exists$ a Lebesgue number $r>0$ so that $\lVert x-y\rVert <r$ implies that $x,y\in B_{r}(z)$ for some $z\in I$. Then $\lVert f_{n}(x)-f_{n}(z)\rVert<\epsilon$ and $\lVert f_{n}(y)-f_{n}(z)\rVert<\epsilon$, so $$\lVert f_{n}(x)-f_{n}(y)\rVert\leq \lVert f_{n}(x)-f_{n}(z)\rVert+\lVert f_{n}(y)-f_{n}(z)\rVert<2\epsilon.$$

I know I must be missing an important piece of the proof, because i did not use the fact that $f_{n}$ has degree less than $m$, nor the boundedness of the sequence.

Alternatively: I was trying to define $$f_{n}(x)=c_{m}x^{m}+\cdots c_{1}x+c_{0}$$ so then $$\lVert f_{n}(x)-f_{n}(y)\rVert=\lVert c_{m}(x^{m}-y^{m})+\cdots +c_{1}(x-y)\rVert=\lVert c_{m}\frac{x^{m}-y^{m}}{x-y}+\cdots+ c_{2}(x+y)+c_{1}\rVert \lVert x-y\rVert$$

From there I was trying to use boundedness of the polynomials to show $\lVert f_{n}(x)-f_{n}(y)\rVert\rightarrow 0$, but in this case, the normed difference would depend on delta rather than epsilon.

Should i define $\delta$ in terms of epsilon to get the result I want? I was having trouble doing this, unless i just let $\lVert x-y\rVert<\epsilon$, but I think it should not be that easy? Also, in this approach, I have not used compactness of $I$!

Any tips on how to solve this problem are appreciated!

Answer 1

A hint: from this sequence one may obtain the sequence of derivatives $(f_n')_{n=1}^\infty$. You may show the $f_n'$ are uniformly bounded. From there, the mean value theorem should do the trick.

Answer 2

Sketch of continuation:

The key observation is that, if $S$ is your uniformly bounded family of polynomials, then the family of vectors $\{(c_m^f,\cdots,c_0^f)\,:\, f\in S\}$ is bounded in $\Bbb R^{m+1}$. You might need two cases to do it: one where $0\in I$ (and here you need a trick) and one where you can safely use the approximation $\lvert x\rvert^k\ge \min\{1,(\inf\{\lvert x\rvert\,:\,x\in I\})^m\}=\beta>0$.

Once you have a bound for the coefficients of the polynomial, you can bound the expression in your proof.

I propose now a more abstract point of view.

Consider $$\Bbb R^I_{ m}[x]:=\{f\in C^0(I)\,:\, \exists g\text{ polynomial of degree }\le m\ \text{ s.t. }\left.g\right\rvert_I=f\}$$

This subspace of $C^1(I)\subseteq C^0(I)$ can be canonically identified with the vector space of polynomials of degree $\le m$ via the restriction map. Hence, $\dim\Bbb R^I_m[x]=\dim\Bbb R_m[x]=m+1$.

Now, consider the norm $\lVert f\rVert_{C^0}:=\max\{\lvert f(x)\rvert\,:\,x\in I\}$ on $C^0(I)$ and the norm $\lVert f\rVert_{C^1}:=\max\{\lVert f\rVert_{C^0},\lVert f'\rVert_{C^0}\}$ on $C^1(I)$.

Since $\Bbb R^I_{m}[x]$ is a vector subspace of both of them, the restrictions of these norms to it are norms on $\Bbb R^I_m[x]$.

But, since $\dim\Bbb R^I_m[x]=m+1$, the aforementioned restrictions must be bequivalent norms. In particular, a family $S\subseteq \Bbb R^I_m[x]$ is $\lVert \bullet\rVert_{C^0}$-bounded if and only if it is $\lVert \bullet\rVert_{C^1}$-bounded. But, by Lagrange's MVT, a $\lVert \bullet\rVert_{C^1}$-bounded family of functions is uniformly Lipschitz (hence equicontinuous).