Prove uniform equicontinuity and boundness

Question

Suppose $K = \{x(t) \in C[a,b], \forall x \ x(t) = 0 \ has \ a \ solution \}$.

Prove that $K$ is uniformly bounded and uniformly equicontinuous.

Could anyone spare a hint on how to use the fact every function in the set has at least one root?

Answer 1

In fact, the set $K$ is not uniformly bounded or equicontinuous.

Consider the family $x_n \in C[a,b]$ defined by $$x_n(t) = n\left(t - \tfrac{a+b}2 \right), \,\,\,\,\,\,\,\,\, t \in [a,b].$$ These are just increasingly steep linear functions which are zero at the midpoint $t = \frac{a+b}2$. The maximum then occurs at $t =a$ or $t = b$ , and has a value $$\lvert x_n(a)\rvert = \lvert x_n(b) \rvert = \frac{n(b-a)}{2}$$ which is unbounded as $n$ gets large. Therefore $K$ is not uniformly bounded. Likewise, since each function $x_n$ has increasingly large slope, the family is not equicontinuous.