## 3 Reconstruction Algorithms
### Algorithm
From Proposition 2:

\[\text{A}^{(n+1)} \leq\text{A}^{(n)}-\frac{\text{A}^{(n)}}{2(I^{(n)}-1)}\]5

thus \(\mathrm{A}^{(n)}\to 0\) and \(\mathrm{WA}^{(n)}\to 0\) as \(n\to\infty\) since \(q_{-}^{(n)}\) is kept unchanged, which is a contradiction. We now provide three results on the convergence of Algorithm 1. As is to be expected, the algorithm can only be shown to converge uniformly when the target response function \(F^{\ddagger}\) is sufficiently smooth; otherwise, the convergence is at best pointwise or in mean. Theorem 3.1 (Algorithm convergence): _Assume that \(F^{\ddagger}\) is strictly increasing. Then, for any choice of \(\mathcal{E}>0\), Algorithm 1 is convergent in the following senses:_

* _If_ \(F^{\ddagger}\) _is piecewise continuous on_ \([a,b]\)_, then_ \(\lim_{n\to\infty}F^{(n)}(x)=F^{\ddagger}(x)\) _at all points_ \(x\in[a,b]\) _where_ \(F^{\ddagger}\) _is continuous;_
* _If_ \(F^{\ddagger}\) _is continuous on_ \([a,b]\)_, then convergence holds uniformly:_ \(\|F^{(n)}-F^{\ddagger}\|_{\infty}\xrightarrow[n\to\infty]{}0\)_._

Proof: Let \(\mathcal{E}>0\). We know from Propositions 1 and 3 that \(\mathrm{WA}^{(n)}\) will oscillate about \(\mathcal{E}\) in the iterating process as \(n\to\infty\), while \(\lim_{n\to\infty}q_{-}^{(n)}=+\infty\) from Assumption 3. Furthermore, let

\[\text{A}^{(n+1)} \leq\text{A}^{(n)}-\frac{\text{A}^{(n)}}{2(I^{(n)}-1)}\]6

Assuming for example that for some \(j\), \(s_{j}^{(n)}=[x_{j}^{(n)},x_{j+1}^{(n)})\) is never divided in two in the iteration process and is thus independent of \(n\), it turns out that \(a_{j}^{(n)}\to(x_{j+1}-x_{j})(F^{\ddagger}(x_{j+1})-F^{\ddagger}(x_{j}))>0\) as \(n\to\infty\), which is impossible because \(\mathrm{A}^{(n)}\) goes to 0 as \(n\to\infty\) from Proposition 2. Therefore there exists some \(m\in\mathbb{N}^{*}\) (depending on \(n\)) such that \(\Delta^{(n+m)}\leq\frac{1}{2}\Delta^{(n)}\); also the sequence \(\Delta^{(n)}\) is decreasing, hence \(\Delta^{(n)}\to 0\) as \(n\to\infty\). Now let \(x\in[x_{i}^{(n)},x_{i+1}^{(n)})\). Then:

\[\text{A}^{(n+1)} \leq\text{A}^{(n)}-\frac{\text{A}^{(n)}}{2(I^{(n)}-1)}\]7 \[\text{A}^{(n+1)} \leq\text{A}^{(n)}-\frac{\text{A}^{(n)}}{2(I^{(n)}-1)}\]8

However, \(x_{i}^{(n)}\to x\) as \(n\to\infty\) because \(\Delta^{(n)}\to 0\); thus if \(F^{\ddagger}\) is continuous at \(x\), the second term on the right hand side above goes to 0 as \(n\to\infty\). However, if \(F^{\ddagger}\) is continuous everywhere on \([a,b]\), it is in addition uniformly continuous on \([a,b]\) by [PERSON]'s theorem, and the second term goes to 0 as \(n\to\infty\) uniformly on \([a,b]\). Finally, invoking Assumption 1, the first term on the right hand side above also tends to 0 as \(n\to\infty\). This completes the proof. Proposition 4 (Convergence in mean): _Let \(F^{\ddagger}\colon[a,b]\to\mathbb{R}\) be piecewise continuous. Then Algorithm 1 is convergent in mean in the sense that_

\[\text{A}^{(n+1)} \leq\text{A}^{(n)}-\frac{\text{A}^{(n)}}{2(I^{(n)}-1)}\]9

Proof: We can check that the sequence \(F^{(n)}\) is monotone. Indeed, if \(\mathrm{WA}^{(n)}<\mathcal{E}\), then by construction we have \[\leq\text{A}^{(n)}\!\left(\frac{2(I^{(n)}-1)-1}{2(I^{(n)}-1)} \right),\]0

where \(s^{(n)}_{-}=\left[x^{(n)}_{i^{(n)}_{-}}x^{(n)}_{i^{(n)}_{-}+1}\right)\). However, if WA\({}^{(n)}>\mathcal{E}\), then consistency implies that

\[\leq\text{A}^{(n)}\!\left(\frac{2(I^{(n)}-1)-1}{2(I^{(n)}-1)} \right),\]1

where \(s^{(n+1)}_{+}=\left[x^{(n+1)}_{i^{(n)}_{+}+1},x^{(n+1)}_{i^{(n)}_{+}+2}\right)\). The claim now follows from the monotone convergence theorem and the fact that \(F^{(0)}\) is integrable. ## 4 Test Cases

To show the effectiveness of Algorithm 1, we try it on two cases, in which \(F^{\dagger}\) is a continuous function and a discontinuous function respectively. For both cases, the error between the numerical estimate and the ground truth function is modelled as a random variable following a Log-normal distribution.