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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?

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    $\begingroup$ 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. $\endgroup$
    – user1551
    Mar 9, 2020 at 4:30

1 Answer 1

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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.

Now extend via induction.

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  • $\begingroup$ Thanks, can you specify the method of approach? I don't really follow. $\endgroup$
    – shine
    Mar 9, 2020 at 4:40
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    $\begingroup$ Which part are you not following? $\endgroup$
    – user744868
    Mar 9, 2020 at 4:42
  • $\begingroup$ 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? $\endgroup$
    – shine
    Mar 9, 2020 at 4:44
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    $\begingroup$ 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. $\endgroup$
    – user744868
    Mar 9, 2020 at 4:46
  • $\begingroup$ thanks for the explanation! $\endgroup$
    – shine
    Mar 9, 2020 at 4:47

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