How to prove that the sum of two convex closed cones is closed?

Question

Let $K_1, K_2$ be two convex closed cones in $\mathbb{R}^n$. If $(-K_1)\cap K_2=\{0\}$, show that $K_1+K_2=\{x+y \mid x\in K_1, y\in K_2\}$ is closed.

I wanted to say, let $(z_n)^{\infty}_1$ be a series in $K_1+K_2$, then there must exists a series $(x_n)^{\infty}_1$ in $K_1$ and one $(y_n)^{\infty}_1$ in $K_2$, such that $z_n=x_n+y_n$. Then we just need to prove that $\lim z_n=\lim (x_n+y_n)\in K_1+K_2$, suppose $\lim z_n$ exists. Is this the right approach? And if it is, how to prove that $\lim x_n$ and $\lim y_n$ exists?

Answer 1

Suppose that $(z_n)_{n\in\mathbb N}$ is a sequence in $K_1+K_2$ (so that there exists for each $n\in\mathbb N$ some $x_n\in K_1$ and $y_n\in K_2$ such that $z_n=x_n+y_n$) with $z\equiv\lim_{n\to\infty}z_n$. The claim is that $z\in K_1+K_2$.

I will show that $(\|x_n\|)_{n\in\mathbb N}$ must remain bounded. If not, one can find a subsequence with a strictly increasing sequence $(n_k)_{k\in\mathbb N}$ of index numbers such that

• $\lim_{k\to\infty}\|x_{n_k}\|=\infty$;
• $\|x_{n_k}\|>0$ for each $k\in\mathbb N$; and
• $\left(\frac{x_{n_k}}{\|x_{n_k}\|}\right)_{k\in\mathbb N}$ converges to some $v$ (by the compactness of the unit sphere) with $\|v\|=1$.

Note that the closedness and the conicality (is this a word?) of $K_1$ imply that $v\in K_1$. Define for each $k\in\mathbb N$ $$w_k\equiv\frac{y_{n_k}}{\|x_{n_k}\|}.$$ Observe also that $$\lim_{k\to\infty}\frac{z_{n_k}}{\|x_{n_k}\|}=0.$$ As a result, $$\lim_{k\to\infty}w_k=\lim_{k\to\infty}\frac{y_{n_k}}{\|x_{n_k}\|}=\lim_{k\to\infty}\frac{z_{n_k}}{\|x_{n_k}\|}-\lim_{k\to\infty}\frac{x_{n_k}}{\|x_{n_k}\|}=-v\in(-K_1).$$ But since $w_k\in K_2$ for each $k\in\mathbb N$ and $K_2$ is closed, it follows that $-v\in K_2$. Since $(-K_1)\cap K_2=\{0\}$, one has that $v=0$, which contradicts $\|v\|=1$.

The boundedness of $(\|x_n\|)_{n\in\mathbb N}$ easily implies that of $(\|y_n\|)_{n\in\mathbb N}$ as well, since $$\|y_n\|\leq\|y_n+x_n\|+\|(-x_n)\|=\|z_n\|+\|x_n\|$$ for every $n\in\mathbb N$. Therefore, there exists a subsequence to be denoted by a strictly increasing sequence $(n_k)_{k\in\mathbb N}$ of index numbers such that

• $(x_{n_k})_{k\in\mathbb N}$ converges to some $x$, which is in $K_1$; and
• $(y_{n_k})_{k\in\mathbb N}$ converges to some $y$, which is in $K_2$.

Consequently, $(x_{n_k}+y_{n_k})_{k\in\mathbb N}$ converges to $x+y$ but it also converges to $z$. Conclusion: $z=x+y\in K_1+K_2$, and we’re donezo.

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Note: nowhere in the proof has the convexity assumption been exploited. Therefore, the premise of the statement can be relaxed to merely closed cones.