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Suppose $C$ is a convex set, and $x_1, x_2, ... \in C$. Suppose further that $\sum_{i=1}^\infty a_i=1$, where $a_i\ge 0$. How do we show that $\sum_{i=1}^\infty a_ix_i \in C$? (Suppose the series converges.) Thank you!

syeh_106
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2 Answers2

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This is not true. For instance, if we work in the spare of square-summable sequences $\ell^2(\mathbb N)$, we note that it has as a subspace the set of sequences which have only finitely many non-zero terms. Being a subspace, this set is obviously convex. However, if we let $e_i$ be the sequence with a $1$ at the $i^{th}$ position and a $0$ everywhere else, we find that $\frac{1}2e_1+\frac{1}4e_2+\frac{1}8e_3+\ldots$ converges in $\ell^2$, but does not converge within the given subspace as it has infinitely many non-zero terms, despite every term $e_i$ being in the subspace.

If you restrict that $C$ is closed as well, this holds rather trivially, as one can see that the sequence $$x_n=\frac{\sum_{i=1}^na_ix_i}{\sum_{i=1}^na_i}$$ converges to $\sum_{i=1}^n a_ix_i$ whenever that exists and is a sequence of convex combinations.

Milo Brandt
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  • I'm not quite familiar with the "spare of the square-summable sequences". I'd appreciate it if you have an (counter-)example in $R^n$. Thanks a lot! – syeh_106 Oct 18 '16 at 03:05
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This does hold in $\mathbb R^n$, as we can show by induction (the base case $n=0$ being trivial). Let $x=\sum_{i=1}^\infty a_ix_i$, and let $$ V=\mathbb R_{>0}(C-x)=\{y\in\mathbb R^n\mid x+\alpha y\in C\text{ for some }\alpha>0\}. $$ Since $C$ is convex, $V$ is a convex cone. If $V=\mathbb R^n$ then $0\in V$ implies $x\in C$, and we are done. Suppose $V\neq\mathbb R^n$. As wikipedia notes, as a consequence of Farkas' lemma this implies $V$ is contained in a half-space $$ H=\{y\in\mathbb R^n\mid v\cdot y\leq0\} $$ where $v\in\mathbb R^n$ is nonzero. Hence $v\cdot c\leq\lambda$ for all $c\in C$ where $\lambda=v\cdot x$. In particular $$ \lambda=\sum_{i=1}^\infty a_i(v\cdot x_i)\leq\left(\sum_{i=1}^\infty a_i\right)\lambda=\lambda, $$ so we must have $v\cdot x_i=\lambda$ for each $i$ with $a_i>0$. Now the result follows from the inductive hypothesis applied to the hyperplane $$ \{y\in\mathbb R^n\mid v\cdot y=\lambda\}. $$

stewbasic
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  • By induction, do you mean successively reducing the affine dimension of the hyperplane where the $x_i$'s are on, until the dimension becomes 1? (At that point, the proof is easy.) Thank you! – syeh_106 Oct 18 '16 at 15:46
  • @syeh_106 Yes, I mean induction on the dimension of the space. – stewbasic Oct 18 '16 at 21:06
  • Thanks a lot, stewbasic! I think, alternatively, we may also show that $x=\sum_{i=1}^\infty a_ix_i$ is a limit point of $C$. Suppose $x \notin C$. Then $x$ must be on the boundary of $C$, and there must be a hyperplane supporting $C$ at $x$. So following your argument, the effective sequence (set) of ${x_i}$ must reside in this hyperplane, and by induction, the affine dimension of the effective ${x_i}$ set can be reduced to $0$, i.e. a singleton. This implies that $x$ equals the singleton and hence $x\in C$, contradicting our hypothesis. So $x \in C$. – syeh_106 Oct 19 '16 at 14:26