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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.

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    $\begingroup$ In the one-variable case, you can use mean value theorem to proceed. Can you see how that generalizes? $\endgroup$ – user27126 Nov 16 '13 at 13:18
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    $\begingroup$ @John Would not need the compactness of the $U$? $\endgroup$ – Elias Costa Nov 16 '13 at 13:23
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    $\begingroup$ Why isn't $U = [0,1]$ and $f_n(x) = x^n$ a counterexample? You have $|f_n(x)|\leq 1$ for every $n$ and $|Df_n(x)|\leq n < \infty$ for all $n = 1, 2, ...$ and all $x$. However, $\{f_n\}_n$ is not equicontinuous at $1$. $\endgroup$ – Tom Nov 16 '13 at 13:34
  • $\begingroup$ @Elias yes you are right, I forgot to state that each $f_{n}$ has compact support in $U$.@Sanchez Yes I think I see that, is the following fine to show equicontinuity at $x_{0}$ using general Mean Value Theorem: $(f_{n}(x)-f_{n}(x_{0})) = |Df(c)|(x-x_{0}) \leq M(x-x_{0})$ for some $c \in \text{supp}f_{n}$ and constant $M > 0$. Then equicontinuity follows simply from definition. $\endgroup$ – user100431 Nov 16 '13 at 13:34
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    $\begingroup$ @Elias I think that there may need to be more of a uniform boundedness assumption on the $Df_n$. take another example of $U=(-2,2)$ and $f_n(x)$ to be smooth approximations to the delta function such that $\operatorname{supp}(f_n) = [-1,1]$ and $f_n \to \delta_0$ (in the distributional sense). Then $\{f_n\}_n$ are not uniformly bounded, nor equicontinuous (at the point $0$). However, if the assumption was that: There exists an $M > 0$ such that for all $x$ and $n$, $|Df_n(x)| \leq M$ then the MVT method discussed above would likely pan out. $\endgroup$ – Tom Nov 16 '13 at 13:57
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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. $\Box$

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