Equicontinuity and Uniform Boundedness If we have a sequence of smooth functions $\{f_{n}\}_{n}$ where $f_{n}: U \rightarrow \mathbb{R}$, where
$U \subset \mathbb{R}^{n}$.
We are given the following two results:
For $x \in U$ we have $|f_{n}(x)| < \infty$ for all $n=1,2,...$
and also similarly $|Df_{n}(x)| < \infty$ for all $n=1,2,...$ then how does it
follow that $\{f_{n}\}_{n}$ is uniformly bounded and equicontinuous. Note that $Df_{n}$ is the gradient vector.
The uniform boundedness seems to follow directly from $|f_{n}(x)| < \infty$
for all $n=1,2,...$ and all $x$, but I can't see how equicontinuity follows, maybe
I'm missing some result that is used?
Thanks for any assistance, let me know if something is unclear.
 A: So this is my first answer at all on this website, I hope it's not 'over-extensive', but I wanted to provide all the details.
The following claim is false:
False claim 1:
Let $d \in \mathbb N$, let $U \subseteq \mathbb R^d$ be open and bounded and let $(f_l)_{l \in \mathbb N}$ be a sequence of totally differentiable functions such that


*

*$\forall l \in \mathbb N : \exists b_l > 0 : \forall x \in \mathbb R^d : \|\nabla f(x)\|_2 < b_l$

*$\forall l \in \mathbb N : \exists c_l > 0 : \forall x \in \mathbb R^d : |f(x)| < c_l$


, where the $b_l$ and $c_l$, as indicated by the subscript $l$, depend on $l$. Then the set of functions $\{f_l | l \in \mathbb N\}$ is always uniformly bounded or equicontinuous.
Proof of falsity (as suggested by Tom in the comments of the question):
Let $U = B_1(0)$. We define
$$
\eta(x) := \begin{cases}
\frac{e^{-\frac{1}{1 - \|x\|_2^2}}}{\int_{B_1(0)}e^{-\frac{1}{1 - \|y\|_2^2}} dy} & \|x\|_2 < 1 \\
0 & \text{else}
\end{cases}
$$
and $f_l(x) := l \eta(lx), l \in \mathbb N$. The integral in the denominator in the def. of $\eta$ exists because the function which is integrated is dominated by a constant function which is constantly 1. Since the functions
$$
\mathbb R \ni \lambda \mapsto \begin{cases}
e^{-\frac{1}{\lambda}} & \lambda > 1 \\
0 & \lambda \le 1
\end{cases}
$$
and
$$
\mathbb R^d \ni x \mapsto 1 - \|x\|_2^2
$$
are infinitely often continuously differentiable, so are (by the chain rule and the fact that compositions of continuous functions are continuous again) all the $f_l$. Further, since thus both functions $x \mapsto f_l(x)$ and $x \mapsto \|\nabla f_l(x)\|_2$ are continuous as well as supported inside $\overline{B_{1/l}(0)}$, with the extreme value theorem follows that both have maximum and minimum. Thus all the conditions from the false claim are met, but from
$$
f_l(0) = \frac{l e^{-1}}{\int_{B_1(0)}e^{-\frac{1}{1 - \|y\|_2^2}} dy}
$$
we see that $\{f_l | l \in \mathbb N\}$ is not uniformly bounded, and from
$$
f_l(\frac{1}{l}e_1) = 0
$$
, where $e_1$ is the first unit vector, we see that $\{f_l | l \in \mathbb N\}$ is not equicontinuous. $\Box$
Instead, theorems 3 & 4 of the following give sufficient conditions for sets of functions to be uniformly bounded and equicontinuous:
Lemma 2:
Let $F$ be a set of totally differentiable functions defined on an open set $U \subseteq \mathbb R^d$ such that
$$
\exists b > 0 : \forall l \in \mathbb N : \forall x \in U : \|\nabla f_l (x)\|_2 \le b
$$
Then $F$ is equicontinuous.
Proof:
Let $x \in U$. Since $U$ is open, there exists a $\delta_1 > 0$ such that $B_{\delta_1}(x) \subseteq U$. We show that if we choose $\delta_2 := \frac{\epsilon}{b}$ and $\delta := \min\{\delta_1, \delta_2\}$, then for $y \in B_\delta(x)$ we have 
$$
\forall f \in F : |f(x) - f(y)| < \epsilon
$$
Suppose otherwise, i. e. $\exists f \in F : \exists y \in B_\delta(x) : |f(x) - f(y)| \ge \epsilon$. We define an auxiliary function:
$$
\mu(\xi) := f(x - \xi(y - x))
$$
Then by definition of $\mu$, $|\mu(0) - \mu(1)| \ge \epsilon$. Therefore, by the mean value theorem, there exists an $\upsilon \in (0, 1)$ such that $|\mu'(\upsilon)| \ge \epsilon$. But on the other hand, by the chain rule
$$
\mu'(\upsilon) = \nabla f(x + \upsilon(y - x)) \cdot (y - x)
$$
and further, by the Cauchy-Schwarz inequality, $|\nabla f(x + \upsilon(y - x)) \cdot (y - x)| \le \|\nabla f(x + \upsilon(y - x))\|_2 \cdot \|y - x\|_2 < b \frac{\epsilon}{b} = \epsilon$. Therefore
$$
|\mu'(\upsilon)| \ge \epsilon \wedge |\mu'(\upsilon)| < \epsilon
$$
, which is a contradiction.
Theorem 3:
Let $d \in \mathbb N$, let $U \subset \mathbb R^d$ be open and let $(f_l)_{l \in \mathbb N}$ be a sequence of totally differentiable functions defined on $U$ such that


*

*(1) $\exists b > 0 : \forall l \in \mathbb N : \forall x \in U : \|\nabla f_l (x)\|_2 \le b$

*(2) $\exists c > 0 : \forall l \in \mathbb N : \forall x \in U : |f_l(x)| \le c$


Then $\{f_l | l \in \mathbb N\}$ is uniformly bounded and equicontinuous.
Proof:
Uniform boundedness follows directly from (2), since (2) just states that all the $f_l$ are bounded by $c$. Equicontinuity follows from (1) and lemma (2). $\Box$
Theorem 4:
Let $d \in \mathbb N$, let $B \subset \mathbb R^d$ be a set such that


*

*$\exists R > 0 : \forall x \in B : \exists y \notin B : \|x - y\|_2 \le R$


and let $F$ be a set of totally differentiable functions such that


*

*$\exists b > 0 : \forall f \in F : \forall x \in B : \|\nabla f (x)\|_2 \le b$

*$\forall f \in F : \text{supp } f \subset B$


Then the set of functions $F$ is uniformly bounded and equicontinuous.
Proof:
1.
We show that $\forall f \in F : \|f\|_\infty \le R b$. Assume the contrary, i. e. $\exists f \in F : \exists x \in B : |f(x)| > Rb$. By (3), there exists $y \notin B : \|x - y\| < R$. We choose an auxiliary function:
$$
\mu(\xi) := f(x + \xi(y - x))
$$
Since $\mu(1) = f(y)$, $\text{supp } f \subset B$ and $y \notin B$, $\mu(1) = 0$. Further, $\mu(0) = f(x)$ and thus $|\mu(0)| > Rb$. By the mean value theorem, there exists a $\iota \in [0, 1]$ such that $|\mu'(\iota)| > Rb$. But by the chain rule,
$$
\mu'(\iota) = \nabla f(x + \iota(y - x)) \cdot (y - x)
$$
and further, by the Cauchy-Schwarz inequality, $|\nabla f(x + \iota(y - x)) \cdot (y - x)| \le \|\nabla f(x + \iota(y - x))\|_2 \|y - x\|_2 \le \|\nabla f(x + \iota(y - x))\|_2 R \le Rb$ and thus
$$
|\mu'(\iota)| > Rb \wedge |\mu'(\iota)| \le Rb
$$
, a contradiction.
2.
Equicontinuity follows from lemma 2. $\Box$
