Recursive definition of hyperbolic functions I have a question about the recursive definition of functions, given as follows.

Two sequences of functions, $\{C_k\}, \{S_k\}$ satisfy the following.
$$S_k(x)=\int_0^x C_k(t)\,\mathrm dt$$
$$C_{k+1}(x)=1+\int_0^x S_k(t)\,\mathrm dt$$
with $C_1(x)=1$. Prove that there exist two functions $C$ and $S$ on $\mathbb{R}$, such that $\{C_k\}$ converges to $C$, and $\{S_k\}$ converges to $S$.

As $C_1(x)$ is a polynomial, $C_k, S_k$ are polynomials for all $k$. I wrote down some terms, and I got the following.




$k$
$C_k(x)$
$S_k(x)$




$1$
$1$
$x$


$2$
$1+\frac{1}{2}x^2$
$x+\frac{1}{6}x^3$


$3$
$1+\frac{1}{2}x^2+\frac{1}{24}x^4$
$x+\frac{1}{6}x^3+\frac{1}{120}x^5$




I noticed that there are only even degrees in $C_k(x)$, and odd degrees in $S_k(x)$, so they are each even functions and odd functions, respectively. Given the pattern, I suppose $C_k$ and $S_k$ are given as
$$C_k(x)=\sum_{n=0}^{k-1}\frac{x^{2n}}{(2n)!}$$
$$S_k(x)=\sum_{n=0}^{k-1}\frac{x^{2n+1}}{(2n+1)!}$$
(I can prove it by induction, but I haven't tried yet. I'm pretty sure that this is the right expression.) This is exactly the Taylor polynomial for $\cosh x$ and $\sinh x$. So I suspect that $C=\cosh x$, and $S=\sinh x$.
At this point, I have two questions.

*

*If I could show that $C_k$ is the $(k-1)$th Taylor polynomial of $\cosh x$ for all $k$, would that mean that $C=\cosh x$? (The same question goes to $S_k$, too.)


*Whether this is a valid way or not, I think this way can give great insight, but not an elegant proof. I would like to prove the existence of $C$ and $S$ by only using the recursion expression, and that we can make $\|C_k-C\|_\infty <\epsilon$ for all $k>k_0$ (i.e the definition of uniform convergence). How could I do that?
Any other proofs are welcome. Thanks in advance.
 A: 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.
