Suppose $f:[0,1]\to\mathbb R$ belongs to the Holder space of order $k\in\mathbb N$, meaning that $f$ is $k$-times differentiable and that $\sum_\limits{\ell=0}^k{\|f^{(\ell)}\|_{\infty}} \leq 1$. Consider the level set function $S_f:\mathbb R^*\to\mathbb R^*$ defined as $S_f(\epsilon) := \mu(\{x\in[0,1]: f(x)\leq f^*+\epsilon\})$, where $f^*=\inf_{0\leq x\leq 1}f(x)$ and $\mu(\cdot)$ is the Lebesgue measure on $[0,1]$. Clearly $f$ is measurable by our assumptions and hence $S_f$ is well-defined. It is also obvious that $S_f$ is a monotonically non-decreasing function in $\epsilon$.

My question is the following:

Under what conditions on $f$ is $S_f(\epsilon)$ a continuous function in $\epsilon$?

Some simple remarks:

  1. It is clear that without additional assumptions $S_f$ is discontinuous for any possible values of $k$. For example, take $f$ to be a constant function with a small "dip" somewhere in the middle, and suppose the regions that connecting the "dip" and the constant line is sufficiently smooth. Then $S_f$ would have a jump somewhere.
  2. For the special case of $k=2$, it seems that assuming "strong convexity" $\inf_{0\leq x\leq 1}f''(x)>0$ is sufficient to guarantee continuity of $S_f$. I'm wondering whether there's a weaker condition? For example, without assuming convexity $f''\geq 0$.
  3. And most interestingly, with high-order Holder assumptions $k>2$.

$S_f$ is usually called the distribution function of $f$. It is continuous if and only if every level set $f^{-1}(t) = \{x\in[0,1]: f(x) = t\}$ has zero measure. So you have to find a contradiction between $f$ being constant on some set $A$ of positive measure, and the differentiability properties you impose on $f$.

As you observed, smoothness by itself does not help. Indeed, for any closed set $A$ one can construct a $C^\infty$ smooth function such that $f^{-1}(0)=A$.

By Lebesgue density theorem, almost every point of $A$ is a point of density. Let $B$ be the set of such points. If $f$ is differentiable everywhere, then $f'\equiv 0$ on $B$. Also, if $f$ is differentiable $k$ times, then by the same logic, $f^{(k)}\equiv 0$ on $B$.

So, it suffices to assume that some derivative is nonzero almost everywhere. You don't need strong convexity; just having $f''\ne 0$ almost everywhere is enough. Same for any other order of derivative.


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