Why are these two definitions of distribution equivalent?

Question

On Wikipedia, there are two criterions for determining whether a linear functional is a distribution, and it is stated that they are in fact equivalent:

If $T$ is a linear functional on $C_{c}^{\infty}(U)$ then $T$ is a distribution if and only if the following equivalent conditions are satisfied:

1. For every compact subset $K\subseteq U$ there exist constants $C>0$ and $N\in \mathbb {N}\cup\{0\}$ such that for all $f\in C^{\infty}(K;U)$ $$|T(f)|\leq C\sup\{|\partial ^{\alpha }f(x)|:x\in U,|\alpha |\leq N\}$$

2. For any compact subset $K\subseteq U$ and any sequence $\{f_{i}\}_{i=1}^{\infty }$ in $C^{\infty }(K;U)$, if $\{\partial ^{\alpha }f_{i}\}_{i=1}^{\infty }$ converges uniformly to zero on $K$ for all multi-indices $\alpha$, then $T(f_{i})\to 0$.

Notations:

1. $U$ is a non-empty open subset on $\mathbb R^n$.
2. $C^{\infty}_c(U)$ denotes the vector space of all smooth real-valued functions on $U$ with compact support in $U$.
3. $C^{\infty}(K;U)$ denotes the vector space of all smooth real-valued functions on $U$ with compact support in $K$.
4. $\alpha$ denotes an $n$-tuple of non-negative integers $(\alpha_1,\alpha_2,\cdots,\alpha_n)$.
5. $|\alpha|:=\alpha_1+\cdots+\alpha_n$.
6. $\partial^\alpha f:=\displaystyle\frac{\partial^{|\alpha|}f}{\partial x_1^{\alpha_1}\cdots\partial x_n^{\alpha_n}}$.

It is easy to show $1\implies 2$, but how to prove $2\implies 1$?

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$1\implies 2$:

As the supremum is the least upper bound, there exists a sequence $\{|\partial^{\alpha^{(i)}_k}f_i(x^{(i)}_k)|\}_{k=1}^\infty$ which tends to $\sup\{|\partial ^{\alpha }f_i(x)|:x\in U,|\alpha |\leq N\}$, where $|\alpha^{(i)}_k|\le N$ and $x^{(i)}_k\in U$.

Also, we have $\{\partial ^{\alpha }f_{i}\}_{i=1}^{\infty }$ converges uniformly to zero on $K$ for all multi-indices $\alpha$. Thus for every $\epsilon>0$ and for each $\alpha$, there exists $I_\alpha>0$ such that $$i>I_\alpha\implies|\partial ^{\alpha }f_{i}(x)|<\epsilon\qquad\forall x\in U$$

Let $I:=\max\{I_\alpha:|\alpha|\le N\}$. Then, $$i>I\implies|\partial ^{\alpha }f_{i}(x)|<\epsilon\qquad\forall x\in U,|\alpha|\le N$$

Hence, when $i>I$, $$\begin{align} |T(f_i)| &\leq C\sup\{|\partial ^{\alpha }f_i(x)|:x\in U,|\alpha |\leq N\} \\ &= C\lim_{k\to\infty}|\partial^{\alpha^{(i)}_k}f_i(x^{(i)}_k)| \\ &\leq C\lim_{k\to\infty}\epsilon \\ &=C\epsilon \end{align} $$ implying that $T(f_i)\to 0$. $\blacksquare$

Answer 1

(2) implies (1) is an easy proof by contradiction. If (1) is fasle then there exist a compact set $K$ and a sequence $(f_n) \in C^{\infty}(K,U)$ such that $|Tf_n| >n \sup \{|\partial^{\alpha} f(x)| : x \in U, |\alpha| \leq n\}$. Define $g_n=\frac {f_n} {\sqrt n C_n}$ where $C_n =\sup \{|\partial^{\alpha} f(x)| : x \in U, |\alpha| \leq n\}$. Now apply (2) to this sequnce to get a contradiction.

Answer 2

If $K$ is a compact subset of $U$, and $f_i \in C^{\infty}(K;U)$ is a sequence of functions, and $g \in C^{\infty}(K;U)$, $\partial^{\alpha} f_i$ converges to $g$ for every multi-index $\alpha$ iff $\delta(f,g):=\sum_{\alpha}{2^{-|\alpha|}\min(\|\partial^{\alpha}(f-g\|_{\infty,K},1)}$ converges to zero.

So if $T$ satisfies the definition of 2., it means that $T: (C^{\infty}(K;U),\delta) \rightarrow \mathbb{C}$ is a continuous map, and thus the image of some $B_{\delta}(0,\epsilon)$ is contained in $B(0,1)$ (where $\epsilon <1$).

Let $C=\sum_{\alpha}{2^{-|\alpha|}} >1$. Let $N >0$ be such that $\sum_{|\alpha|\geq N}{2^{-|\alpha|}} \leq \epsilon/2$.

Now, let $f \in C^{\infty}(K;U)$ be such that for every $\alpha$ such that $|\alpha| < N$, $\|\partial^{\alpha}f\|_{\infty,K} \leq \frac{\epsilon}{2C}$. Then it is easy to see that $\delta(f,0) \leq \epsilon$ thus $|T(f)| \leq 1$.

Thus if $f \in C^{\infty}(K;U)$, $|T(f)| \leq \frac{2C}{\epsilon} \sup_{|\alpha| < N} \,\|\partial^{\alpha}f\|_{\infty,K}$.