Proving that $\lVert x\rVert=\sqrt{\langle x,x \rangle}$ is a norm (triangle inequality).

Question

To demonstrate that $\lVert x\rVert=\sqrt{\langle x,x \rangle}$ is a norm, I have to show that the properties of norms are coming from the properties of inner product. By example, let's demonstarte that $\lVert x\rVert = 0$ iff $x = 0$; on this purpose, I can say that because $\langle x,x \rangle = 0$ iff $x = 0$, then $\lVert x\rVert=\sqrt{\langle x,x \rangle} = 0$ iff $x = 0$. Now, I have to prove the triangle inequality… $\lVert x+y \rVert^2 = {\langle x,x \rangle} + 2\Re (\langle x,y \rangle) + \langle y,y \rangle$ and $(\lVert x \rVert + \lVert y \rVert)^2 = {\langle x,x \rangle} + 2\sqrt{\langle x,x \rangle \langle y,y \rangle} + \langle y,y \rangle$. The thing that cause me trouble is to demonstrate that $\Re (\langle x,y \rangle) \le \sqrt{\langle x,x \rangle \langle y,y \rangle}$. I know that I should use Chauchy-Schwartz inequality, but my teacher give us that information after he said that the inner product induces a norm.. and then he domostrated the triangle inequality with Dirac's Notation. But, what if I wanted to follow the consequentiality of the lessons? I would not be able to demonstarte $\lVert x\rVert=\sqrt{\langle x,x \rangle}$ due to the lack of Cauchy-Schwartz inequality? Or is there a method to demonstrate that $\Re (\langle x,y \rangle) \le \sqrt{\langle x,x \rangle \langle y,y \rangle}$?

Answer 1

Well, showing that $$ \Re{(\langle x,y\rangle)} \leq \sqrt{\langle x,x\rangle \langle y,y\rangle} $$ is precisely the Cauchy-Schwarz inequality (not accounting for absolute value), so of course you would have to show it to use it. However, the proof does not rely on any property other than the abstract definition of scalar products, so there is no issue in postponing it. Perhaps your lecturer just stated that $\lVert x \rVert = \sqrt{\langle x,x\rangle}$ defines a norm to anticipate what is to come.