Prove that for each $\lambda \in [0,1]$ there exists a unique continuous $f: [0,1] \to [0, \infty)$ that solves the integral equation $f(x) = \lambda \int_0^1e^{-xf(t)}dt$ for all $x \in [0,1]$. Hint: for all $a, b \geq 0, |e^{-a} - e^{-b}| \leq e^{-\min\{a,b\}}|a-b|$.
Here is my work on uniqueness: Suppose $f, g$ are two solutions with $f(x) \neq g(x)$ for some $x \in [0,1]$. Since $[0,1]$ is compact and $|f(x) - g(x)|$ is continuous, we may let $x_0 := \max\{x: |f(x) - g(x)| > 0\}$. Then $|f(x_0) - g(x_0)| \leq \lambda \int_0^1e^{-x_0\min\{f(t),g(t)\}}|f(t) - g(t)|dt \leq \lambda |f(x_0) - g(x_0)| \int_0^1e^{-x*0}dx = \lambda |f(x_0) - g(x_0)|$ (here, the last inequality is strict since $\min\{f(t),g(t)\} = \frac{f+g-|f-g|}{2}$ is a continuous function, since the constant function $0$ is not the solution because $f(0) = \lambda$, and since the integrand thus is strictly positive. thus $1 < \lambda$, a contradiction.
Hence, it now remains to show that there exists a continuous solution. I tried guessing various solutions, but all functions of the form $ax + \lambda$ and $\lambda e^{ax}$ seemed to fail. How can I show that there exists a continuous solution?
Consider the metric space $C^+[0,1]$ of nonnegative and continuous functions on $[0,1]$ and the operator $$(Af)(x)=\lambda \int\limits_0^1e^{-xf(t)}\,dt$$ Observe that $(Af)(x)\le \lambda. $ Hence $$(A^2f)(x)=\lambda \int\limits_0^1e^{-t(Af)(x)}\,dt\ge \lambda\int\limits_0^1e^{-t\lambda}\,dt=1-e^{-\lambda}$$ Let $$X=\{f\in C[0,1]\,:\,1- e^{-\lambda}\le f\le 1\}$$ If the operator $A$ admits a fixed point, i.e. there exists $f$ such that $Af=f,$ then $A^2f=Af=f,$ hence $f\in X.$ Thus we may consider $A:X\to X.$ We are going to apply the Banach fixed point theorem, to show that $A$ admits a unique fixed point in $X$ (therefore in $C[0,1]$). i.e. a function $f$ satisfying $Af=f.$ To this end we will show that $A$ is a contraction.
We have $$|e^{-xf(t)}-e^{-xg(t)}|\le x e^{-x\min\{f(t),g(t)\}}|f(t)-g(t)|\le xe^{-x[1-e^{-\lambda}]}|f(t)-g(t)|$$ Thus $$|(Af)(x)-(Ag)(x)|\le \lambda xe^{-x[1-e^{-\lambda}]}\|f-g\|_\infty $$ The function $[0,1]\ni u\mapsto \lambda ue^{-u[1-e^{-\lambda}]}$ is increasing. Hence $$|(Af)(x)-(Ag)(x)|\le \lambda e^{-[1-e^{-\lambda}]}\|f-g\|_\infty $$ The function $v(\lambda)=\lambda e^{-[1-e^{-\lambda}]}$ is increasing and $$v(1)=e^{-[1-e^{-1}]}<1$$ Therefore $$|(Af)(x)-(Ag)(x)|\le e^{1-e^{-1}}\|f-g\|_\infty $$ and $$\|Af-Ag\|_\infty \le e^{1-e^{-1}}\|f-g\|_\infty$$ Thus $A$ is a contraction, which completes the proof.