Ryan Robertson

Answered

2022-07-08

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

My Attempt: Suppose $\stackrel{\to }{x},\stackrel{\to }{y}\in {N}_{r}\left(\stackrel{\to }{c}\right)$ and $t\in \left[0,1\right]$, we then have $‖\stackrel{\to }{x}-\stackrel{\to }{c}{‖}_{2} and $‖\stackrel{\to }{y}-\stackrel{\to }{c}{‖}_{2}. Now,
$‖t\stackrel{\to }{y}+\left(1-t\right)\stackrel{\to }{x}-\stackrel{\to }{c}{‖}_{2}=‖t\stackrel{\to }{y}-t\stackrel{\to }{x}+\stackrel{\to }{x}-\stackrel{\to }{c}{‖}_{2}\le ‖t\stackrel{\to }{y}-t\stackrel{\to }{x}{‖}_{2}+‖\stackrel{\to }{x}-\stackrel{\to }{c}{‖}_{2}<|t|‖\stackrel{\to }{y}-\stackrel{\to }{x}{‖}_{2}+r\le |t|\left(‖\stackrel{\to }{y}-\stackrel{\to }{c}{‖}_{2}+‖\stackrel{\to }{c}-\stackrel{\to }{x}{‖}_{2}\right)+r<|t|\left(r+r\right)+r=\left(1+2|t|\right)r$
Screeching halt.

My Question: How should I make $‖t\stackrel{\to }{y}+\left(1-t\right)\stackrel{\to }{x}-\stackrel{\to }{c}{‖}_{2}$ 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

$‖t\stackrel{\to }{y}+\left(1-t\right)\stackrel{\to }{x}-\stackrel{\to }{c}{‖}_{2}=‖t\stackrel{\to }{y}+\left(1-t\right)\stackrel{\to }{x}-t\stackrel{\to }{c}-\left(1-t\right)\stackrel{\to }{c}{‖}_{2}=‖t\left(\stackrel{\to }{y}-\stackrel{\to }{c}\right)+\left(1-t\right)\left(\stackrel{\to }{x}-\stackrel{\to }{c}\right){‖}_{2}\phantom{\rule{0ex}{0ex}}\le ‖t\left(\stackrel{\to }{y}-\stackrel{\to }{c}\right){‖}_{2}+‖\left(1-t\right)\left(\stackrel{\to }{x}-\stackrel{\to }{c}\right){‖}_{2}=t‖\stackrel{\to }{y}-\stackrel{\to }{c}{‖}_{2}+\left(1-t\right)‖\stackrel{\to }{x}-\stackrel{\to }{c}{‖}_{2}
Concerning $<$ note that not both $t$ and $1-t$ can be zero.

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