# Why are these two definitions of distribution equivalent?

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 $$f\in C^{\infty}(K;U)$$ $$|T(f)|\leq C\sup\{|\partial ^{\alpha }f(x)|:x\in U,|\alpha |\leq N\}$$
1. 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 $$2\implies 1$$?

$$1\implies 2$$:

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 $$T(f_i)\to 0$$. $$\blacksquare$$

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