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.

Solutions Collecting From Web of "Equicontinuity if the sequence of derivatives is uniformly bounded."

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 $x$.

Indeed, in your case
$$
\delta=\frac{\varepsilon}{M}.
$$
For
$$
\lvert x-y\rvert<\delta=\frac{\varepsilon}{M},
$$
then
$$
\lvert g_n(x)-g_n(y)\rvert \le M\lvert x-y\rvert<M\delta={\varepsilon}.
$$