The answer by @Jair Taylor addresses the case of bounded operators of norm strictly less than $1$ (hence the restriction to the interval $[0,1-\epsilon]$), and justifies rigorously why the series works. I’m not sure if it’s possible to generalize that argument directly to work for larger domains. Hence, what follows is really more of an excuse to illustrate/apply some of the ideas in functional analysis, particularly around Banach’s fixed point theorem.
Your problem really is a fixed point problem, because $f-\int f=1$ rearranged gives $f=1+\int f$, i.e the function $f$ equals some other operator applied to $f$, namely $\Phi(f)=1+\int f$, i.e we have to solve the equation $f=\Phi(f)$. This is actually a very common theme when solving (nonlinear) integral equations, or even differential equations, or simply to prove many types of existence results: reformulate your original problem as a fixed-point problem (or whatever the case may be), and try to solve that new problem using the ‘fancy’ tools of functional analysis.
Some generalities.
First, a simple lemma.
Lemma 1.
Let $X$ be any set, suppose $\Phi:X\to X$ is a map such that for some $n\in\Bbb{N}$, the $n$-fold composition $\Phi^n$ has a unique fixed point $x$. Then, $x$ is also the unique fixed point of $\Phi$.
To prove this, note first that $\Phi^n(\Phi(x))=\Phi(\Phi^n(x))=\Phi(x)$, meaning that $\Phi(x)$ is a fixed point of $\Phi^n$, so by uniqueness of fixed points of $\Phi^n$, we must have $\Phi(x)=x$, i.e $x$ is a fixed point of $\Phi$. To show uniqueness, suppose $y$ is a fixed-point of $\Phi$. Then, $\Phi^n(y)=\Phi^{n-1}(y)=\dots=\Phi(y)=y$, so $y$ is a fixed point of $\Phi^n$, hence by uniqueness, $y=x$.
Next, we have a simple result about calculating fixed points in complete metric spaces; it’s just a mild (but useful) variant of Banach’s fixed point theorem.
Theorem.
Let $X$ be a non-empty complete metric space, and $\Phi:X\to X$ a map with the following property: for each $n\in\Bbb{N}$, $\Phi^n$ is a Lipschitz map, with constant $C_n$ such that $\sum_{n=0}^{\infty}C_n<\infty$. Then, $\Phi$ has a unique fixed point $\xi$; furthermore, for every $x\in X$, the sequence $\{\Phi^n(x)\}_{n=0}^{\infty}$ converges in $X$ to $\xi$.
The utility of this theorem is that it first tells us that a fixed point exists and is unique, and secondly it tells us how to calculate it: start with any arbitrary guess $x\in X$, and then successively refine your guess by calculating $\Phi(x),\Phi^2(x),\Phi^3(x),\dots$, and this sequence has a limit, and this is limit equals the sought after $\xi$.
To prove the theorem, note that since $\sum_{n=0}^{\infty}C_n<\infty$, we must have $C_n\to 0$ as $n\to\infty$, so there exists an $n_0$ such that $C_{n_0}<1$, and so $\Psi:=\Phi^{n_0}$ is a contraction in $X$, hence by Banach’s fixed point theorem, $\Psi$ has a unique fixed point, call it $\xi$; and furthermore, Banach’s theorem tells us that to calculate this fixed point, we can start with any $x\in X$, and then $\{\Psi^n(x)\}_{n=0}^{\infty}$ will have a limit and the limit will equal $\xi$.
We’re almost done, but not quite, because the theorem talks about $\Phi$, not $\Psi$. But now first applying lemma 1, we see that $\xi$ is also the unique fixed point of $\Phi$. Next, note that $\{\Phi^n(x)\}_{n=0}^{\infty}$ is a Cauchy sequence, because for any $n>m$, we have
\begin{align}
d(\Phi^n(x),\Phi^m(x))\leq\sum_{j=m}^{n-1}d(\Phi^{j+1}(x),\Phi^j(x))\leq\sum_{j=m}^{n-1}C_jd(\Phi(x),x).
\end{align}
Since $\sum_{k=0}^{\infty}C_k<\infty$, by making $n,m$ large enough, the RHS can be made as small as desired. Lastly, keep in mind that $\{\Psi^n(x)\}_{n=0}^{\infty}$ is actually a subsequence of $\{\Phi^n(x)\}_{n=0}^{\infty}$, which in fact converges to $\xi$. Recall that if we have a Cauchy sequence, and some subsequence converges, then the whole sequence converges to the same limit. This proves that $\Phi^n(x)\to \xi$ as $n\to\infty$, and hence completes the proof of the theorem.
Applying to the question at hand.
We fix an $R>0$, and consider the (complex) Banach space $X=C([-R,R];\Bbb{C})$, and fix any function $h\in X$, and consider the map $\Phi_h:X\to X$ defined by $(\Phi_h f)(x):= h(x)+\int_0^xf(t)\,dt$. Then, a simple exercise in induction shows that for all $f,g\in X$, all $n\in\Bbb{N}$ and all $x\in [-R,R]$, we have
\begin{align}
|(\Phi_h^nf)(x)-(\Phi_h^ng)(x)|&\leq \|f-g\|\cdot\frac{|x|^n}{n!},
\end{align}
and so $\|\Phi_h^nf-\Phi_h^ng\|\leq \|f-g\|\cdot\frac{R^n}{n!}$, meaning that $\Phi_h^n$ is Lipschitz continuous with Lipschitz constant at most $\frac{R^n}{n!}$; the ratio test now implies that $\sum_{n=0}^{\infty}\frac{R^n}{n!}<\infty$, and so by the theorem, $\Phi_h$ has a unique fixed point. Having a unique fixed point means that for every function $h\in X$, there exists a unique function $f\in X$ such that for all $x\in [-R,R]$,
\begin{align}
f(x)=h(x)+\int_0^xf(t)\,dt,
\end{align}
or in other words, the operator $I-\int_0^x:X\to X$ is bijective.
In your particular case, fix $h(x)\equiv \lambda$, and let us start off with the function $f_0\equiv 1$. Then, (calling it $\Phi$ instead of $\Phi_{\lambda}$ for simplicity)
- $(\Phi f_0)(x):=\lambda+\int_0^x1\,dt=\lambda+x$.
- $(\Phi^2f_0)(x):=\lambda +\int_0^x(\Phi f)(t)\,dt=\lambda+\int_0^x(\lambda +t)\,dt=\lambda+\lambda x+\frac{x^2}{2}$.
- We can keep going by induction to prove that for all $n\in\Bbb{N}$, $(\Phi^nf_0)(x)=\lambda\sum_{k=0}^{n-1}\frac{x^k}{k!}+\frac{x^n}{n!}$.
Note that we already know $\{\Phi^n(f_0)\}_{n=0}^{\infty}$ converges in $X$ (i.e converges uniformly) to some unique function $f$, which is a fixed point of $\Phi$, i.e satisfies the following equation for all $x\in [-R,R]$:
\begin{align}
f(x)&=\lambda+\int_0^xf(t)\,dt.
\end{align}
On the other hand, from the explicit formula for $\Phi^n(f_0)$, we can immediately see that it converges pointwise to $\lambda e^x$. Therefore, we see that $f(x)=\lambda e^x$ is the unique continuous function on $[-R,R]$ which satisfies the above integral equation. And finally, since $R>0$ was chosen arbitrarily, the above equality holds for all $x\in\Bbb{R}$.
In fact, if you think about it, solving this integral equation (via Banach’s fixed point method) is precisely one of the ways of defining the exponential function in the first place: one of the definitions is that $\exp$ is the unique function satisfying $f’=f$ with $f(0)=1$. If you integrate both sides of the ODE, you get the above integral equation with $\lambda=1$ (and indeed, the existence and uniqueness theorems for ODEs go through this route of recasting the ODE into an integral equation which is solvable by such approximating arguments… the underlying principle here is that, as in much of analysis, integrals are better behaved than derivatives). Thus, even if you didn’t know that the pointwise limit of $\Phi^n(f_0)(x)$ was $\lambda e^x$, you could have simply defined $e^x$ to be the limit (which we already proved exists).
Edit: More in line with the question.
The comment by @jd27 made me realize that we can definitely apply the geometric series in general. Again, fix the Banach space $X=C[-R,R]$ of continuous function $[-R,R]\to\Bbb{C}$, and consider the linear operator $T:X\to X$ given by $(Tf)(x):=\int_0^xf(t)\,dt$. Then, by induction, one can show that for all $n\in\Bbb{N}_0$, and all $x\in [-R,R]$,
\begin{align}
|(T^nf)(x)|\leq\|f\|\cdot\frac{|x|^n}{n!},
\end{align}
and hence $\|T^n\|\leq \frac{R^n}{n!}$. So, the series $\sum_{n=0}^{\infty}\|T^n\|$ is finite. Since in any Banach space (completeness is key), absolute convergence implies convergence in norm, it follows that there is a well-defined operator $S=\sum_{n=0}^{\infty}T^n$. Finally, simple manipulations with convergent power series shows that $S$ is the inverse operator of $I-T$. Therefore, $(I-T)^{-1}:X\to X$ exists and can be calculated as $\sum_{n=0}^{\infty}T^n$. Evaluating this series on the constant function with value $\lambda$ shows that $(I-T)^{-1}(\lambda)=\lambda\cdot \exp|_{[-R,R]}$.
Finally, since $R>0$ was arbitrary, we have shown that $f(x)=\lambda e^x$ is the unique solution of the integral equation $f(x)-\int_0^xf(t)\,dt=\lambda$ on all of $\Bbb{R}$.
So, tldr is there’s lots of cool math behind the scenes.
Ais either an empty string""(call itϵ), or itself followed by "a", then we have:A = Aa + ϵ, orA - Aa = ϵ, orA(ϵ - a) = ϵ, orA = ϵ / (ϵ - a), or (by Taylor expansion)A = ϵ + a + aa + aaa + aaaa + ...– user541686 Mar 01 '24 at 05:29