Rudin's PMA: Trigonometric function

Question

Cited from Rudin:

Let us define $(46)$ $$C(x) = 2 [E(ix) + E( - ix)]$$

$$S(x) = \frac{1}{2i} [E(ix) - E(-ix)]$$ We shall show that $C(x)$ and $S(x)$ coincide with the functions $cos(x)$ and $sin(x)$, whose definition is usually based on geometric considerations. By $(25)$ $$E(z)=\sum_{n=0}^{\infty}\frac{z^n}{n!}$$ $\overline{E(z)} = > E(\overline{z})$. Hence $(46)$ shows that $C(x)$ and $S(x)$ are real for real $x$. Also,$(47)$ $$E(ix) = C(x) + iS(x)$$ Thus $C(x)$ and $S(x)$ are the real and imaginary parts, respectively, of $E(ix)$, if $x$ is real. By $(27)$ $$E(z)E( - z) = E(z - z) = E(0) = 1$$, $(48)$ $$|E(ix)|= E(ix)\overline{E(ix)} = E(ix)E( -ix) = 1$$ so that $$|E(ix)| = 1$$ ($x$ real).From $(46)$ we can read off that $C(0) = > 1$, $S(0) = 0$, and $(28)$ shows that$(49)$ $$C'(x) = -S(x)$$ $$ S'(x) > = C(x)$$ We assert that there exist positive numbers $x$ such that $C(x) = 0$. For suppose this is not so. Since $C(0) = 1$, it then follows that $C(x) > 0$ for all $x > 0$, hence $S'(x) > 0$

After $(49)$, rudin said "Since $C(0)=1$, it then follows that $C(x)>0$ for all $x>0$", I don't understand how rudin derived it.

Thanks in advance!

Answer 1

Thanks for the help, now I post the answer.

Because $E$ is a continuous function, $C$ is a continuous function. We also know that $C(0)=1>0$. If there exists $C(x) < 0$ for some $x > 0$, by the intermediate value theorem, there exists a $x>0$, $C(x)=0$ contradicts to the assumption.