Can we relax the triangle inequality for $\| v \|$ = $\|v - v_0 + v_0\|$?

Question

Given some vector $v$ on vector space $X$ with a norm $\| \cdot \|$

Then $\| v \|$ = $\|v - v_0 + v_0\|$ where $v_0$ is some other vector

is it legal to then write $\| v - v_0 + v_0 \| = \|v -v_0\| + \|v_0\|$

as opposed to $\| v - v_0 + v_0 \| \leq \|v - v_0\| + \|v_0 \|$

Would the same conclusion hold for $\| v - v_0 + v_0 \|^2 = \|v -v_0\|^2 + \|v_0\|^2$

Answer 1

No, and a counter-example can come from the simple cases. Take $X=\mathbb{R}^2$, $v_0=\hat{i}$ and $v=\hat{i}+\hat{j}$.

In this case, $v=v-v_0+v_0=(1,1)$ and so $\|v-v_0+v_0\|=\sqrt{1+1}=\sqrt{2}$ while $\|v-v_0\|=\|\hat{j}\|=1$ and $\|v_0\|=1$.

In fact I think the equal sign holds only when $v-v_0$ is a non-negative multiple of $v_0$.

For the Pythagorean Theorem, it does not hold in general, either. When the space is equipped with an inner product and when $<v-v_0,v_0>=0$, the equation is correct.

Answer 2

Take $\mathbb R$ over $\mathbb R$ endowed with the absolute value norm, $v = 1$, and $v_0 = 2$. Then $$1 = |v| = |v - v_0 + v_0| \neq |v - v_0| + |v_0| = 1 + 2 = 3.$$ For inner product spaces in particular, $||x||^2 = \langle x, x \rangle.$ We have \begin{align*} ||x + y||^2 &= \langle x+y, x+y \rangle \\ &= \langle x,x \rangle + 2\langle x,y \rangle + \langle y,y \rangle \\ &\leq \langle x,x \rangle + 2\left|\langle x,y \rangle\right| + \langle y,y \rangle \\ &\leq ||x||^2 + 2||x||\,||y|| + ||y||^2 \\ &= (||x|| + ||y||)^2. \end{align*} Taking square roots yields the triangle inequality. Letting $x = v - v_0$ and $y = v_0$, we get $||v|| \leq ||v - v_0|| + ||v_0||$, with equality holding if and only if the Cauchy-Schwarz inequality is an equality and the inner product is non-negative if and only if $v - v_0$ is a positive multiple of $v_0$. Use the same counterexample for the second proposition and note that we need $\langle v - v_0, v\rangle = 0$ (from the second line).