# Convex subset of a linear space [duplicate]

I am dealing with the following exercise

A set $\mathcal M$ in a normed linear space $\mathcal R$ is said to be convex if $\mathcal M$ contains all elements of the form $\alpha x + \beta y$, where $\alpha, \beta \geq 0$, $\alpha+\beta=1$, provided that $\mathcal M$ contains x and y. Prove that the set of all elements $x \in \mathcal R$ satisfying the inequality $\|x-x_0 \| \leq c$, where $x_0$ is a fixed element of $\mathcal R$ and $c > 0$, is convex.

Assume $\|x-x_0 \|\leq c$, and $\|y-x_0 \| \leq c$, we have to proof that

$$\| \alpha x + (1-\alpha) y - x_0 \| \leq c.$$

By the triangular inequality,

\begin{align*}\| \alpha x + (1-\alpha) y - x_0 \| & \leq \| \alpha x \| + \|(1-\alpha) y - x_0 \| \\ & \leq \| \alpha x \| + \|(1-\alpha) y \| +\|- x_0 \| \\ & \leq |\alpha| \| x \| + |(1-\alpha)| \| y \| +\| x_0 \| \\ & = c.\end{align*}

Is it enough for proof the theorem? If not, what path should I take?

## marked as duplicate by Jason, gt6989b, Leucippus, user8795, JonMark PerryMay 19 '17 at 2:55

• Where does the last line come from? We know $\|x-x_0\|\le c$ and $\|y-x_0\|\le c$ - why does that imply your second to last line is $\le c$? – Jason May 18 '17 at 20:16
• I defined $c=|\alpha| \| x \| + |(1-\alpha)| \| y \| +\| x_0 \|$, but in fact it does not make many sense... – Dante May 18 '17 at 20:22
• You don't get to define what $c$ is - it is fixed! – Jason May 18 '17 at 20:23
• Yes, you are right. – Dante May 18 '17 at 20:55

Use triangle inequality differently: $$|ax + (1-a)y - x_0| = |a(x-x_0) + (1-a)(y-x_0)| \le a |x-x_0| + (1-a)|y-x_0|$$ Can you finish the proof?

• $a |x-x_0| + (1-a)|y-x_0| \leq a c + (1-a) c = c$. – Dante May 18 '17 at 20:54