Semicircular function is convex and lower semicontinuous function

Question

I want to know how to show that $$f:\mathbb{R}\rightarrow \mathbb{\bar{R}},\; f(x)=\begin{cases} -\sqrt{1-x^2} & if\; |x| \leq 1 \\ +\infty & \, \text{ortherwise} \end{cases}$$ is convex and lower semicontinuous.

I know that in case $|x|\leq 1$, it can draw as the semicircular function which is obvious convex by the picture. But I still can not show through the definition. Do you have any suggestion for this?

score 1 · Answer 1 · answered May 31 '21 at 09:19

1

Hint: If $|x|, |y| \le 1$, you have $$ \left\|\begin{pmatrix}x\\ f(x)\end{pmatrix}\right\| = 1 ,\qquad \left\|\begin{pmatrix}y\\ f(y)\end{pmatrix}\right\| = 1. $$ For $z = \lambda x + (1-\lambda) y$ with $\lambda \in [0,1]$, you have $$ \left\|\begin{pmatrix}z\\ \lambda f(x) + (1-\lambda) f(y)\end{pmatrix}\right\| \le 1 $$ by the triangle inequality.

answered May 31 '21 at 09:19

gerw

31,359

Excuse me, I don't understand what is $\vert \left( \begin{array}{c} x\ f(x) \end{array}\right)\vert$ mean. – Math Brother May 31 '21 at 20:08
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This is the Euclidean norm of a vector with two elements. – gerw Jun 01 '21 at 05:46

Semicircular function is convex and lower semicontinuous function

1 Answers1