Ryan Robertson

Answered

2022-07-08

Definition: Here a neighborhood of the point $\overrightarrow{c}\in {\mathbb{R}}^{n}$ is the set ${N}_{r}(\overrightarrow{c})=\{\overrightarrow{x}\in {\mathbb{R}}^{n}:\Vert \overrightarrow{x}-\overrightarrow{c}{\Vert}_{2}<r\}$ for some $r>0$, and a subset $U\subset {\mathbb{R}}^{n}$ is said to be convex if $\overrightarrow{x},\overrightarrow{y}\in U,t\in [0,1]$ implies $t\overrightarrow{y}+(1-t)\overrightarrow{x}\in U$.

My Attempt: Suppose $\overrightarrow{x},\overrightarrow{y}\in {N}_{r}(\overrightarrow{c})$ and $t\in [0,1]$, we then have $\Vert \overrightarrow{x}-\overrightarrow{c}{\Vert}_{2}<r$ and $\Vert \overrightarrow{y}-\overrightarrow{c}{\Vert}_{2}<r$. Now,

$\Vert t\overrightarrow{y}+(1-t)\overrightarrow{x}-\overrightarrow{c}{\Vert}_{2}=\Vert t\overrightarrow{y}-t\overrightarrow{x}+\overrightarrow{x}-\overrightarrow{c}{\Vert}_{2}\le \Vert t\overrightarrow{y}-t\overrightarrow{x}{\Vert}_{2}+\Vert \overrightarrow{x}-\overrightarrow{c}{\Vert}_{2}<|t|\Vert \overrightarrow{y}-\overrightarrow{x}{\Vert}_{2}+r\le |t|(\Vert \overrightarrow{y}-\overrightarrow{c}{\Vert}_{2}+\Vert \overrightarrow{c}-\overrightarrow{x}{\Vert}_{2})+r<|t|(r+r)+r=(1+2|t|)r$

Screeching halt.

My Question: How should I make $\Vert t\overrightarrow{y}+(1-t)\overrightarrow{x}-\overrightarrow{c}{\Vert}_{2}$ less than $r$? My intuition is that I need the Cauchy-Schwarz inequality; but I am unsure where to apply it. Any hint would be greatly appreciated.

My Attempt: Suppose $\overrightarrow{x},\overrightarrow{y}\in {N}_{r}(\overrightarrow{c})$ and $t\in [0,1]$, we then have $\Vert \overrightarrow{x}-\overrightarrow{c}{\Vert}_{2}<r$ and $\Vert \overrightarrow{y}-\overrightarrow{c}{\Vert}_{2}<r$. Now,

$\Vert t\overrightarrow{y}+(1-t)\overrightarrow{x}-\overrightarrow{c}{\Vert}_{2}=\Vert t\overrightarrow{y}-t\overrightarrow{x}+\overrightarrow{x}-\overrightarrow{c}{\Vert}_{2}\le \Vert t\overrightarrow{y}-t\overrightarrow{x}{\Vert}_{2}+\Vert \overrightarrow{x}-\overrightarrow{c}{\Vert}_{2}<|t|\Vert \overrightarrow{y}-\overrightarrow{x}{\Vert}_{2}+r\le |t|(\Vert \overrightarrow{y}-\overrightarrow{c}{\Vert}_{2}+\Vert \overrightarrow{c}-\overrightarrow{x}{\Vert}_{2})+r<|t|(r+r)+r=(1+2|t|)r$

Screeching halt.

My Question: How should I make $\Vert t\overrightarrow{y}+(1-t)\overrightarrow{x}-\overrightarrow{c}{\Vert}_{2}$ less than $r$? My intuition is that I need the Cauchy-Schwarz inequality; but I am unsure where to apply it. Any hint would be greatly appreciated.

Answer & Explanation

Miles Mueller

Expert

2022-07-09Added 11 answers

$\Vert t\overrightarrow{y}+(1-t)\overrightarrow{x}-\overrightarrow{c}{\Vert}_{2}=\Vert t\overrightarrow{y}+(1-t)\overrightarrow{x}-t\overrightarrow{c}-(1-t)\overrightarrow{c}{\Vert}_{2}=\Vert t(\overrightarrow{y}-\overrightarrow{c})+(1-t)(\overrightarrow{x}-\overrightarrow{c}){\Vert}_{2}\phantom{\rule{0ex}{0ex}}\le \Vert t(\overrightarrow{y}-\overrightarrow{c}){\Vert}_{2}+\Vert (1-t)(\overrightarrow{x}-\overrightarrow{c}){\Vert}_{2}=t\Vert \overrightarrow{y}-\overrightarrow{c}{\Vert}_{2}+(1-t)\Vert \overrightarrow{x}-\overrightarrow{c}{\Vert}_{2}<tr+(1-t)r=r.$

Concerning $<$ note that not both $t$ and $1-t$ can be zero.

Concerning $<$ note that not both $t$ and $1-t$ can be zero.

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