Series of series of integrals                           $f_1(x)=\frac{M}{C}$, where M and C are constants
$$h_1(x)=\frac{\int_0^xf_1(y)dy}C + \frac{\int_0^x\int_0^yf_1(z)dzdy}{C^2} + \frac{\int_0^x\int_0^y\int_0^zf_1(t)dtdzdy}{C^3} + \cdots$$
$$f_2(x)=h_1(C)-h_1(x)$$
$$h_2(x)=\frac{\int_0^xf_2(y)dy}C + \frac{\int_0^x\int_0^yf_2(z)dzdy}{C^2} + \frac{int_0^x\int_0^y\int_0^zf_2(t)dtdzdy}{C^3} + \cdots$$
$$f_3(x)=h_2(C)-h_2(x)$$
$$\cdots$$

$$h_{i}(x)=\frac{\int_0^xf_i(y)dy}C + \frac{\int_0^x\int_0^yf_i(z)dzdy}{C^2} + \frac{\int_0^x\int_0^y\int_0^zf_i(t)dtdzdy}{C^3} + \cdots$$
$$f_{i+1}(x)=h_i(C)-h_i(x)$$

$$h_{\infty}(x) = ? $$
I want to examine the convergence of the function of $h_{\infty}(x)$. 
Each function $h_i(x)$ can be represented with $e^{\frac{x}C}$ function as a shorter version by using the maclaurin Series. 
$h_1(x)$ becomes $ \frac{M}{C} \left( e^{\frac{x}C}-1 \right) $ when the infinite series is arranged by using the Maclaurin Series.  
$h_2(x)$ becomes $\frac{M}{C} \left( e^{\frac{x}C} \left(-\frac{x}C+e \right) - e \right)$. 
$h_3(x)$ becomes $\frac{M}{C} \left( e^x\left(\frac{1}{2}{\frac{x}{C}}^{2}-e{\frac{x}{C}}+e^2-e\right)-\left(e^2-e\right) \right)$

By writing program codes, I calculated and found that $h_i(x)$ function is getting closer to the function $2\frac{M}{C}\left(\frac{x}{C}\right)$ with increasing i,
I want to mathematically prove this convergence. 
$$h_{\infty}(x)=2\frac{M}{C}\left(\frac{x}{C}\right)$$
Any tip will be appreciated. Thank you for reading this.

For the recently posted question, the $f(i)$ mentioned in the question is the same as the following variant of $h_i(x)$ when x = 1. $$ f(i) = h_i(1) $$
The variant of $h_i(x)$ is defined as follows. 
$$g_1(x)=1$$ 
$$h_1(x)=\int_0^xg_1(y)dy + \int_0^x\int_0^yg_1(z)dzdy + \int_0^x\int_0^y\int_0^zg_1(t)dtdzdy + \cdots$$
$$g_2(x)=h_1(1)-h_1(x)$$
$$h_2(x)=\int_0^xg_2(y)dy + \int_0^x\int_0^yg_2(z)dzdy + \int_0^x\int_0^y\int_0^zg_2(t)dtdzdy + \cdots$$
$$g_3(x)=h_2(1)-h_2(x)$$
$$\cdots$$
$$h_{i}(x)=\int_0^xg_i(y)dy + \int_0^x\int_0^yg_i(z)dzdy + \int_0^x\int_0^y\int_0^zg_i(t)dtdzdy + \cdots$$
$$g_{i+1}(x)=h_i(1)-h_i(x)$$
In the variant of $h_i(x)$, the constant M is removed and C is substituted with 1 from the original $h_i(x)$.
 A: The constants $M$ and $C$ can be absorbed by scaling the functions by $M$ and the independent variables by $C$, so I'll use $M=C=1$. 
Then for any $a\in\mathbb R$, the functions $h(x)=ax$ and $f(x)=a(1-x)$ form a fixed point of the iteration. To show that these are the only fixed points, we merely need to correct a small mistake in Didier's answer. The correct equation is
$$h_i(x)=\int_0^x(f_i(y)+h_i(y))\,\mathrm dy\;,$$
and with $f_{i+1}(x)=h_i(1)-h_i(x)$ this yields, 
$$h_{i+1}(x)=\int_0^x(h_i(1)-h_i(y)+h_{i+1}(y))\,\mathrm dy\;,$$
from which
$$h(x)=\int_0^xh(1)\,\mathrm dy=h(1)x=:ax$$
and
$$f(x)=a-h(x)=a(1-x)\;.$$
Thus, if the sequence converges, it converges to one of these fixed points, and it only remains to determine $a$. To do so, note that the iteration preserves the normalization of $f$:
$$
\begin{align}
\int_0^1f_{i+1}(x)\,\mathrm dx
&=
\int_0^1(h_i(1)-h_i(x))\,\mathrm dx
\\
&=
h_i(1)-\int_0^1h_i(x)\,\mathrm dx
\\
&=
\int_0^1(f_i(x)+h_i(x))\,\mathrm dx-\int_0^1h_i(x)\,\mathrm dx
\\
&=
\int_0^1f_i(x)\,\mathrm dx\;.
\end{align}
$$
Thus, since $\int_0^1f_1(x)\,\mathrm dx=1$ and $\int_0^1f(x)\,\mathrm dx=a/2$, we must have $a=2$, and thus, if the sequence converges, it converges to $h(x)=2x$ and $f(x)=2(1-x)$.
Here's a plot of the first three iterates and the fixed point, showing quite rapid convergence.
