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edit approved Jan 29 at 2:50
The Amplitwist
edit approved Jan 29 at 2:50
The Amplitwist
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If a real analytic function $f:U\subset\mathbb R^n\to\mathbb R^m$ is zero on a set $Z$ of positive measure (and $U$ is connected), then $f\equiv 0$.

Indeed, almost every point of $Z$ is a density pointdensity point. It is easy to see that the derivative at a density point is zero. Therefore $df=0$ a.e. on $Z$. Applying the same argument to $df$, conclude that the second derivative vanishes a.e. on $Z$ too. And so on. Thus $f$ has zero Taylor expansion at some point, hence $f\equiv 0$.

If a real analytic function $f:U\subset\mathbb R^n\to\mathbb R^m$ is zero on a set $Z$ of positive measure (and $U$ is connected), then $f\equiv 0$.

Indeed, almost every point of $Z$ is a density point. It is easy to see that the derivative at a density point is zero. Therefore $df=0$ a.e. on $Z$. Applying the same argument to $df$, conclude that the second derivative vanishes a.e. on $Z$ too. And so on. Thus $f$ has zero Taylor expansion at some point, hence $f\equiv 0$.

If a real analytic function $f:U\subset\mathbb R^n\to\mathbb R^m$ is zero on a set $Z$ of positive measure (and $U$ is connected), then $f\equiv 0$.

Indeed, almost every point of $Z$ is a density point. It is easy to see that the derivative at a density point is zero. Therefore $df=0$ a.e. on $Z$. Applying the same argument to $df$, conclude that the second derivative vanishes a.e. on $Z$ too. And so on. Thus $f$ has zero Taylor expansion at some point, hence $f\equiv 0$.

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Sergei Ivanov
Sergei Ivanov
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If a real analytic function $f:U\subset\mathbb R^n\to\mathbb R^m$ is zero on a set $Z$ of positive measure (and $U$ is connected), then $f\equiv 0$.

Indeed, almost every point of $Z$ is a density point. It is easy to see that the derivative at a density point is zero. Therefore $df=0$ a.e. on $Z$. Applying the same argument to $df$, conclude that the second derivative vanishes a.e. on $Z$ too. And so on. SinceThus $f$ is analytic and thehas zero Taylor expansion at some point vanishes, we havehence $f\equiv 0$.

If a real analytic function $f:U\subset\mathbb R^n\to\mathbb R^m$ is zero on a set $Z$ of positive measure (and $U$ is connected), then $f\equiv 0$.

Indeed, almost every point of $Z$ is a density point. It is easy to see that the derivative at a density point is zero. Therefore $df=0$ on $Z$. Applying the same argument to $df$, conclude that the second derivative vanishes on $Z$ too. And so on. Since $f$ is analytic and the Taylor expansion at some point vanishes, we have $f\equiv 0$.

If a real analytic function $f:U\subset\mathbb R^n\to\mathbb R^m$ is zero on a set $Z$ of positive measure (and $U$ is connected), then $f\equiv 0$.

Indeed, almost every point of $Z$ is a density point. It is easy to see that the derivative at a density point is zero. Therefore $df=0$ a.e. on $Z$. Applying the same argument to $df$, conclude that the second derivative vanishes a.e. on $Z$ too. And so on. Thus $f$ has zero Taylor expansion at some point, hence $f\equiv 0$.

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Sergei Ivanov
Sergei Ivanov
  • 32.4k
  • 2
  • 99
  • 154

If a real analytic function $f:U\subset\mathbb R^n\to\mathbb R^m$ is zero on a set $Z$ of positive measure (and $U$ is connected), then $f\equiv 0$.

Indeed, almost every point of $Z$ is a density point. It is easy to see that the derivative at a density point is zero. Therefore $df=0$ on $Z$. Applying the same argument to $df$, conclude that the second derivative vanishes on $Z$ too. And so on. Since $f$ is analytic and the Taylor expansion at some point vanishes, we have $f\equiv 0$.