Problem with norms over $\mathbb{R}^2$

Question

Let $\|\cdot\|$ be a norm over $\mathbb{R}^2$ with $\|e_1\|=\|e_2\|=\|e_1+e_2\|=1$, where $e_1, e_2$ are the standard basis of $\mathbb{R}^2$. Show that $1≤\|e_1-e_2\|≤2.$

Now I can say that a norm with this conditions can be

$$ \|x\|=\sqrt{x_1^2-x_1x_2+x_2^2} \quad \text{or} \quad \|x\|=\sqrt{\left(\frac{x_1+x_2}{2}\right)^2+3\left(\frac{x_1-x_2}{2}\right)^2} $$

With both of these formulas I obtain $\|e_1-e_2\|=\sqrt3$ that indeed is $1≤\sqrt3≤2$.

Is this enough to be proved? What else can I consider?

Answer 1

This problem can be solved by using the triangle inequality (which holds for any norm),

$$ \| a + b \| \leq \|a\| + \|b\|.$$

To show that $\| e_1 - e_2 \| \leq 2$ is straightforward application of the triangle inequaltiy,

$$ \|e_1 - e_2\| \leq \|e_1\| + \|-e_2\| = 2. $$

Observe that $e_1 - e_2 = 2e_1 - (e_1 + e_2)$ and then apply the reverse triangle inequality,

$$\|a\| - \|b\| \leq \|a - b\|$$

so

$$ \|e_1 - e_2\| = \| 2e_1 - (e_1 + e_2)\| \geq \|2e_1\| - \|e_1 + e_2\| = 1. $$

Answer 2

Hint: Use $x_1-x_2=2x_1-(x_1+x_2)$ and the (reversed) triangular inequality.

Answer 3

No, that is not correct, since we are dealing with an arbitrary norm here.

Note that $\lVert e_1+e_2\rVert\leqslant\lVert e_1\rVert+\lVert e_2\rVert\leqslant1$ and that\begin{align}2&=\lVert 2e_1\rVert\\&=\bigl\lVert(e_1+e_2)+(e_1-e_2)\bigr\rVert\\&\leqslant\lVert e_1+e_2\rVert+\lVert e_1-e_2\rVert\end{align}and that therefore$$\lVert e_1-e_2\rVert\geqslant2-\lVert e_1+e_2\rVert=1.$$