Showing a set is equicontinuous

Question

I have the subset $S=\{ f:[0,1]\rightarrow R:f(x) = \alpha x^2 + \beta x +\gamma , |\alpha|\leq a, |\beta| \leq b, |\gamma| \leq c \}$ of $C[0,1] $ where $a,b,c,>0 $ are fixed constants. I want to show S is equicontinuous.

Here's where I've got: $\forall \epsilon>0 , \exists\delta=?$(will choose at the end) such that if $|x-y|<\delta $ then for $ x,y\in C[0,1]$, $ |f(x)-f(y)| = | \alpha x^2 + \beta x +\gamma - \alpha y^2 - \beta y -\gamma| = |\alpha(x^2-y^2) + \beta(x-y)| = |(x-y)(\alpha(x+y)+\beta|=|x-y|.|\alpha x + \alpha y + \beta| < \delta . |\alpha x + \alpha y + \beta| $

I don't know where to go from here.

Answer 1

Note that for any $f \in S$, $$\lvert f'(x)\rvert = \lvert 2\alpha x + \beta \rvert \le 2a + b, \,\,\, \forall x \in [0,1]. $$ Since the derivatives of all functions in $S$ are uniformly bounded, the set is equicontinuous by the mean value theorem. Indeed, for any $f\in S$ and any $x,y \in [0,1]$, there is $c$ between $x$ and $y$ such that $$\lvert f(x) - f(y) \rvert = \lvert f'(c) \rvert \lvert x - y \rvert \le (2a+b) \lvert x - y \rvert.$$ Thus for any $\epsilon > 0$, you can take $\delta = \epsilon / (2a+b)$ and when $\lvert x - y \rvert < \delta$, we will have $\lvert f(x) - f(y) \rvert < \epsilon$ for all $f \in S$.

Answer 2

Your approach would have worked just fine: The first inequality below is yours, and I'll just follow through.

$$|f(x)-f(y)| \le|(x-y)(\alpha(x+y)+\beta)| \le |x-y|(a(|x|+|y|) + b)$$ $$ \le |x-y|(a(1+1) + b) = (2a+b)|x-y|.$$

This holds for all $f\in S,$ and you're done.

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Added later You might like to try your hand at a more general result that may be no harder to prove: Suppose $f,g,h\in C[0,1]$ and $a,b,c>0.$ Then

$$S=\{\alpha f+ \beta g + \gamma h: |\alpha|\le a, |\beta|\le b, |\gamma|\le c\}$$

is equicontinuous.