Recursive definition of hyperbolic functions

Preliminary remarks:

• Proving the convergence of a sequence without a-priory knowledge of the limit usually somehow uses the completeness of the underlying space: For sequences of real numbers we have the Cauchy criterion, for sequences of functions we have the Banach fixed-point theorem.
• Due to the even/odd symmetry of the functions $C_k$ and $S_k$ it suffices to investigate the convergence on $[0, \infty)$.
• The following argument mimics the Picard-Lindelöf iteration for the solution of differential equations.

We have, using integration by parts, $$ C_{k+1}(x)=1+\int_0^x \int_0^t C_k(u) \, \mathrm du \mathrm dt = 1 + \int_0^x (x-t) C_k(t) \, \mathrm dt. $$ That suggests to define an operator $T$ on the space $\mathcal C \bigl( [0, \infty), \mathbb R \bigr)$ of continuous real-valued functions on $[0, \infty)$ by $$ T(f)(x) = 1 + \int_0^x (x-t) f(t) \, \mathrm dt \, . $$ If we can find a norm on $\mathcal C \bigl( [0, \infty), \mathbb R \bigr)$ such that $T$ is a contraction, then the Banach fixed-point theorem states that $T$ has a (unique) fixed-point , and the sequence defined by $C_{k+1} = T(C_k)$ converges to that fixed-point.

A suitable norm is $$ \| f \| = \sup \left\{ e^{-2x} |f(x)| : x \ge 0 \right\} \, . $$ Then for $f, g \in \mathcal C \bigl( [0, \infty), \mathbb R \bigr)$ and all $x \ge 0$, $$ \left|e^{-2x}T(f)(x) - e^{-2x}T(g)(x) \right| \le \int_0^x (x-t)e^{2(t-x)}e^{-2t}|f(t)-g(t)| \, \mathrm dt \\ \le \| f-g \| \int_0^x (x-t)e^{2(t-x)} \, \mathrm dt \le \frac 14 \| f-g \| $$ since $$ \int_0^x (x-t)e^{2(t-x)} \, \mathrm dt = \int_0^x te^{-2t} \, \mathrm dt \le \int_0^\infty te^{-2t} \, \mathrm dt = \frac 14 \, , $$ so that $$ \| T(f) - T(g) \| \le \frac 14 \| f-g \| \, . $$

It follows that there is a continuous function $C$ such that $\| C_k - C\| \to 0$ for $k \to \infty$, and that implies the uniform convergence of $C_k$ to $C$ on every compact interval $[0, a]$.

Now one can conclude that the $S_k$ also converge locally uniformly to some continuous function $S$, and that the limit functions satisfy $$ S(x) = \int_0^x C(t) \, \mathrm dt \\ C(x) = 1 + \int_0^x S(t) \, \mathrm dt \, . $$

Finally, $C = \cosh$ (and consequently, $S=\sinh$) follows from the fact that $T(\cosh) = \cosh$ and that the fixed-point of a contraction is unique.