Let $X\subseteq \mathbb{R}^n$, $f: X \rightarrow \mathbb{R}$: what further conditions on $X$ and $f$ do we need to ensure that $\text{Int}\{x|f(x)\leq 0\}=\{x|f(x)<0\}$?

Context: I've encountered this kind of question when dealing with open bounded $X$ with $C^1$ boundary, in partial differential equations, but I'm sure that one could benefit from such a fact in optimization, for instance.

Clearly, continuity is not enough, since $f:[0,1]\rightarrow \mathbb{R} , f(x)=0$ is such that $[0,1]=\{f\leq 0\}$, but the interior of this set is not the empty set.


1 Answer 1


If $f$ is continuous, it is easy to see that $$\tag {1}\{x:f(x)<0\}\subseteq \text{Int}\{x:f(x)\le 0\}$$

This holds because the LHT is open and the RHT is, by definition, the biggest open subset of $\{x:f(x)\le 0\}$.

Let us first suppose $X=\mathbb{R}^n$:

One necessary and sufficient condition for equality, provided that $f$ is differentiable, is that $\nabla f\neq 0$ on $f^{-1}(0)$.

To prove it, let $x$ be a point inside $\text{Int}\{x:f(x)\le 0\}\backslash \{x:f(x)<0\}$. This means that $x$ is locally a maximum, and thus $\nabla f=0$

This clearly generalizes to open subsets of $\mathbb{R}^n$. It is now easy to see that, for general $X$ and continuous $f$, a necessary and sufficient condition is that $f^{-1}(0)$ does not contain local maxima (although this is not a very useful condition)

Note that some kind of regularity hypotesis is necessary: as Whitney proved, given a closed subset $K$ of $\mathbb{R}^n$, there's always a $\mathcal{C}^\infty(\mathbb{R}^n)$ function with $f^{-1}(0)=K$. Given such an $f$, considering $f^2$ gives you a class of smooth examples with strict (or trivial) inclusion.

  • $\begingroup$ Thank you. Why is there such a neighborhood? $\endgroup$
    – Leonardo
    Nov 17, 2020 at 13:14
  • $\begingroup$ @warm_fish My argument was faulty, I modified it. Indeed, there does not need to be such a neighbourhood as the one I was describing, as $f(x,y)=-y^2$ shows $\endgroup$
    – user515010
    Nov 17, 2020 at 13:24
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    $\begingroup$ You only need $\nabla f$ to be nonzero at the points where $f(x) = 0$. $\endgroup$ Nov 17, 2020 at 13:28
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    $\begingroup$ From a slightly different perspective, this is also essentially the condition that $f(x)=0$ defines a submanifold. $\endgroup$
    – Toffomat
    Nov 18, 2020 at 13:01

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