The Parallelogram Identity for an inner product space

For a given inner product space $V$, I would like to prove the inequality $$\|\mathbf{x}+\mathbf{y}\|^2 + \|\mathbf{x}-\mathbf{y}\|^2 = 2(\|\mathbf{x}\|^2 + \|\mathbf{y}\|^2).$$

We see that $$\|\mathbf{x}+\mathbf{y}\|^2 + \|\mathbf{x}-\mathbf{y}\|^2 = \langle\mathbf{x}+\mathbf{y},\mathbf{x}+\mathbf{y}\rangle + \langle\mathbf{x}-\mathbf{y},\mathbf{x}-\mathbf{y}\rangle,$$

which by linearity $$=\langle\mathbf{x},\mathbf{x}+\mathbf{y}\rangle + \langle\mathbf{y},\mathbf{x}+\mathbf{y}\rangle + \langle\mathbf{x},\mathbf{x}-\mathbf{y}\rangle - \langle\mathbf{y},\mathbf{x}-\mathbf{y}\rangle.$$

I've been trying to figure out how to go about this proof using linearity in the second argument of an inner product, but my textbook does not say that linearity necessarily holds in the second component. If it does, how would I go about the rest of this proof? If not, should I be potentially using some aspect of conjugate symmetry to prove this statement?

Solution 1:

$\langle x,x+y\rangle=\langle x+y,x\rangle^{*}=\langle x,x\rangle^{*}+\langle y,x\rangle^{*}=\langle x,x\rangle+\langle x,y\rangle$ $\langle y,x+y\rangle=\langle x+y,y\rangle^{*}=\langle x,y\rangle^{*}+\langle y,y\rangle^{*}=\langle y,x\rangle+<y,y\rangle$ $\langle x,x-y\rangle=\langle x-y,x\rangle^{*}=\langle x,x\rangle^{*}-\langle y,x\rangle^{*}=\langle x,x\rangle-\langle x,y\rangle$ $\langle y,x-y\rangle=\langle x-y,y\rangle^{*}=\langle x,y\rangle^{*}-\langle y,y\rangle^{*}=\langle y,x\rangle-\langle y,y\rangle$

Plug this back into the expression that you've gotten so far using linearity and you should end up with the identity.