One of the beneficial properties of persistent homology is its stability results (so called robustness to noise).

Usually the referenced paper is this paper: http://pub.ist.ac.at/~edels/Papers/2010-J-01-LpStablePersistence.pdf titled "Lipschitz Functions Have $L_p$-stable Persistence".

According to the authors, the Lipschitz condition is crucial (otherwise the results don't hold).

My understanding is that the Lipschitz function $f$ is used to construct the filtration $X_a = f^{−1}(−\infty, a] $ in the context of persistent homology.

My question is how "universal" is this assumption of Lipschitz-ness?

That is, for a "typical" filtration $K_1\subseteq K_2 \subseteq \dots K_m$, can it be expressed in the form of $K_i=f^{−1}(−\infty, a] $ for some Lipschitz function $f$?

If no, doesn't the oft-cited statement that persistent homology is "robust to noise" fail to hold?

Thanks a lot. That is my main question after reading the referenced paper since it is written in a very general viewpoint (more general than the context of persistent homology).

One of the beneficial properties of persistent homology is its stability results (so called robustness to noise).

Usually the referenced paper is this paper titled "Lipschitz functions have $L_p$-stable persistence".

According to the authors, the Lipschitz condition is crucial (otherwise the results don't hold).

My understanding is that the Lipschitz function $f$ is used to construct the filtration $X_a = f^{−1}(−\infty, a] $ in the context of persistent homology.

My question is how "universal" is this assumption of Lipschitz-ness?

That is, for a "typical" filtration $K_1\subseteq K_2 \subseteq \dots K_m$, can it be expressed in the form of $K_i=f^{−1}(−\infty, a] $ for some Lipschitz function $f$?

If no, doesn't the oft-cited statement that persistent homology is "robust to noise" fail to hold?

Thanks a lot. That is my main question after reading the referenced paper since it is written in a very general viewpoint (more general than the context of persistent homology).

One of the beneficial properties of persistent homology is its stability results (so called robustness to noise).

Usually the referenced paper is this paper: http://pub.ist.ac.at/~edels/Papers/2010-J-01-LpStablePersistence.pdf titled "Lipschitz Functions Have $L_p$-stable Persistence".

According to the authors, the Lipschitz condition is crucial (otherwise the results don't hold).

My understanding is that the Lipschitz function $f$ is used to construct the filtration $X_a = f^{−1}(−\infty, a] $ in the context of persistent homology.

My question is how "universal" is this assumption of Lipschitz-ness?

That is, for a "typical" filtration $K_1\subseteq K_2 \subseteq \dots K_m$, can it be expressed in the form of $K_i=f^{−1}(−\infty, a] $ for some Lipschitz function $f$?

Thanks. That is my main question after reading the referenced paper since it is written in a very general viewpoint (more general than the context of persistent homology).

One of the beneficial properties of persistent homology is its stability results (so called robustness to noise).

Usually the referenced paper is this paper: http://pub.ist.ac.at/~edels/Papers/2010-J-01-LpStablePersistence.pdf titled "Lipschitz Functions Have $L_p$-stable Persistence".

According to the authors, the Lipschitz condition is crucial (otherwise the results don't hold).

My understanding is that the Lipschitz function $f$ is used to construct the filtration $X_a = f^{−1}(−\infty, a] $ in the context of persistent homology.

My question is how "universal" is this assumption of Lipschitz-ness?

That is, for a "typical" filtration $K_1\subseteq K_2 \subseteq \dots K_m$, can it be expressed in the form of $K_i=f^{−1}(−\infty, a] $ for some Lipschitz function $f$?

If no, doesn't the oft-cited statement that persistent homology is "robust to noise" fail to hold?

Thanks a lot. That is my main question after reading the referenced paper since it is written in a very general viewpoint (more general than the context of persistent homology).