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Let $\{x_n\}$ be a sequence in $\mathbb R^d$ such that $x_n \to A$ where $A$ is convex compact. If $$\frac{1}{n}\sum_{k=1}^{n}x_k \to x$$

then is it true that $x \in A$.

I am not able to prove it nor getting counter example.

Anonymous
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  • What's a realtion? – Rudy the Reindeer Jan 16 '15 at 06:30
  • It is not always clear to an unsuspecting person whether $x_n$ refers to an individual element of a sequence or the entire sequence. Context helps. Sometimes one can suspect an unfamiliar context and walk away, but this is not always clear. Notation is often overloaded in math, so if you see that a fellow user misunderstood, surely there's no harm explaining what the notation means. – Jyrki Lahtonen Jan 16 '15 at 06:34

1 Answers1

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Let $f(y)=\operatorname{dist}(y,A)$. Observe that $f$ is a nonnegative continuous convex function and its zero set is precisely $A$.

By assumption, $f(x_n)\to 0$. It is a standard fact that the means of the sequence $\{f(x_n)\}$ converge to $0$. Since $$ f\left(\frac{1}{n}\sum_{k=1}^{n}x_k \right)\le \frac{1}{n}\sum_{k=1}^{n}f(x_k) \to 0 $$ it follows that $f(x)=0$, as desired.

Note that compactness is not needed: being closed and convex is enough.

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