# Equicontinuity if the sequence of derivatives is uniformly bounded.

I would really appreciate if someone could look over this proof for me.

Let $\left\{ g_m \right\}$ be a sequence of functions defined on an interval $[a,b] \subset \mathbb{R}^n$. Let $\left\{ g’_m \right\}$ be uniformly bounded on $[a,b]$. Show that $\left\{ g_m \right\}$ is equicontinuous on $[a,b]$.

My proof goes like this:

If $g’_m$ is uniformly bounded, then that means that $\lvert g’_m \rvert \leq M$ for some non-negative $M$. That means that for all $m$,

$$\left\lvert \frac{g_m(x)-g_m(y)}{x-y} \right\rvert\leq M$$ as $x\rightarrow y$.

This means that the sequence of functions $\left\{ g_m \right\}$ all have the same Lipshitz constant, and that means $\left\{ g_m \right\}$ is equicontinuous.

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This is correct.

Uniform equicontinuity means that, for every $\varepsilon>0$, there exists a $\delta>0$, which depends ONLY on $\varepsilon$ and not of $g_n$ or on $x$.

$$\delta=\frac{\varepsilon}{M}.$$
$$\lvert x-y\rvert<\delta=\frac{\varepsilon}{M},$$
$$\lvert g_n(x)-g_n(y)\rvert \le M\lvert x-y\rvert<M\delta={\varepsilon}.$$