Let $\{f_{n}\}_{n\geq 1}$ be a sequence of Lebesgue-integrable functions defined on the interval $K:=[-1,1]$ such that, for every function $\phi\in C^{\infty}$ supported in $K$, the limit \begin{align*} \lim_{n\to\infty}\,\int_{-1}^{1}\,f_{n}\phi &\,=:\,\Lambda(\phi) \end{align*} exists. Exercise 3 of Chapter 2 of Rudin’s Functional Analysis asks the reader to prove that the limit functional $\Lambda$ is continuous and that there exist a positive constant $M>0$ and a nonnegative integer $a$ such that \begin{align*} \left|\int_{-1}^{1}\,f_{n}\phi\,\right| & \,\leq\, M\| D^{a}\phi\|_{\infty} \end{align*} for all $n$ and all $\phi$.
How is it possible that the uniform bound on the sequence of functionals $f_{n}$ depend on only one derivative of the argument $\phi$, when the typical basic neighborhood involves an arbitrary number of them?
Suppose we have a function $f$ on a compact interval $[a,b]$ in $\mathbb R$ and $f$ vanishes at the end points. Then $|f(x)|=|\int_a ^{x} f'(x) \, dx| \leq (b-a)||f'||_{\infty}$ for all $x$ so $||f||_{\infty} \leq (b-a)||f'||_{\infty}$. Repeated use of this argument tells you why it is enough to have one derivative in the inequality instead of a finite number of them.