# inner product space generalizing to $n$ vectors $\|x+y\|=\|x\|+\| y\|$

Problem: Prove that if V is an inner product space, then $$\|x+y\|=\|x\|+\| y\|$$ if and only if one of the vectors $$x$$ or $$y$$ is a nonnegative scalar multiple of the other. Generalize it to the case of $$n$$ vectors.

Proof: $$\langle x+y,x+y\rangle=\langle x,x\rangle +2\|x\|\cdot\|y\|+\langle y,y\rangle$$ then, using linearity of the inner product I get

$$\langle x,x\rangle +\langle y,y\rangle+\langle x,y\rangle+\langle y,x\rangle=\langle x,x\rangle +2\|x\|\cdot\|y\|+\langle y,y\rangle$$

After all the cancellation,

$$\mathrm{Re}\langle x,y\rangle=\|x\|\cdot\|y|\$$

By Cauchy Schwarz, we prove that equation is equal.

Question is how am I supposed to generalize to $$n$$ vectors?

• The first statement in your question isn't true. Counterexample: $x=-y\ne0$. The correct statement is that $\|x+y\|=\|x\|+\|y\|$ if and only if one of $x$ or $y$ is a nonnegative scalar multiple of the other. Mar 9, 2020 at 4:30

Note that, if $$\|x + y + x\| = \|x\| + \|y\| + \|z\|$$, then $$\|x\| + \|y\| + \|z\| = \|x + y + x\| \le \|x + y\| + \|z\| \le \|x\| + \|y\| + \|z\|.$$ From this, we can conclude that $$\|x + y\| + \|z\| = \|x\| + \|y\| + \|z\| \implies \|x + y\| = \|x\| + \|y\|.$$ Similarly, we can deduce $$\|y + z\| = \|y\| + \|z\|$$ and $$\|x + z\| = \|x\| + \|z\|$$. Thus, $$x, y, z$$ are each positive multiples of each other.
• begining of $$\|x\| + \|y\| + \|z\| = \|x + y + x\| \le \|x + y\| + \|z\| \le \|x\| + \|y\| + \|z\|.$$. How do you generalize to three vectors, by triangle inequality? Mar 9, 2020 at 4:44
• The first equality is by assumption. The middle inequality is applying the triangle inequality $\|a + b\| \le \|a\| + \|b\|$ with $a = x + y$ and $b = z$. The last inequality is applying triangle inequality again with $a = x$ and $b = y$. Because we get an inequality beginning and ending with the same expression, the inequalities have to be equal. Mar 9, 2020 at 4:46