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Prove that triangle's inequality in the definition of a norm is equivalent to the convexity of the closed unit ball.

(I proved triangle inequality implies convexity, but I could not prove converse. For converse I tried like this: I took any $x$, $y$ in the normed space $X$, then I derived two new unit vectors $x/||x||$, $y/||y||$ so that the new vectors lie in the closed unit ball and so I can use the convexity of the ball for the vectors $x$ & $y$. But further I could not get the result this way.)

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  • $\begingroup$ Do you have any attempts to share? Perhaps what you have tried, you could post what you know and people will be able to help you from there. $\endgroup$ – smokeypeat Aug 8 '17 at 10:55
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Triangle's inequality in the definition of a norm is as follows:

$$\forall x,y\in X,\ \|x+y\|\le \|x\|+\|y\| $$

Now, the above inequality follows the convexity of the unit closed ball. Let $\lambda\in [0,1]$ and $\mathbf{x},\mathbf{y}\in \mathbb{B}[0,1]\implies \|\mathbf{x}\|\le 1 $ and $\|\mathbf{y}\|\le 1$. Now consider, \begin{align*} \|\lambda \mathbf{x}+(1-\lambda) \mathbf{y}\| & \le \|\lambda \mathbf{x}\|+\|(1-\lambda)\mathbf{y}\|\\ & = \lambda \|\mathbf{x}\|+(1-\lambda)\|\mathbf{y}\|\\ & \le \lambda + (1-\lambda)=1 \end{align*} Hence, $ \lambda \mathbf{x}+(1-\lambda) \mathbf{y}\in \mathbb{B}[0,1]. $

For the converse see, $$ \frac{\mathbf{x}+\mathbf{y}}{\|\mathbf{x}\|+\|\mathbf{y}\|}=\frac{\|\mathbf{x}\|}{\|\mathbf{x}\|+\|\mathbf{y}\|}\frac{\mathbf{x}}{\|\mathbf{x}\|}+\frac{\|\mathbf{y}\|}{\|\mathbf{x}\|+\|\mathbf{y}\|}\frac{\mathbf{y}}{\|\mathbf{y}\|}\in \mathbb{B}[0,1] \implies \|\mathbf{x}+\mathbf{y}\|\le \|\mathbf{x}\|+\|\mathbf{y}\|$$

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