Series of series of integrals

Question

                          $f_1(x)=\frac{M}{C}$, 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.

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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 $h_i(x)$.

Answer 1

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.