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I am currently reading a paper that does the following simplification. I have broken it down and worked it out by some examples, but can anyone show me how they made this simplification using the notation they used?

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    $\begingroup$ The only tricky part is $\nabla \phi(y) = 2y$. The rest is just algebra (and the fact that $\|a\|^2 = \langle a, a \rangle$). $\endgroup$
    – copper.hat
    May 20, 2015 at 20:54

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You need to know the properties of the inner product and of the norm induced by an inner product. In the following, $x,y,z$ are vectors and $a$ is a scalar. The space is also assumed to be real (the situation is slightly different in a complex space).

$$\langle ax+z,y \rangle = a \langle x,y \rangle + \langle z,y \rangle \\ \langle x,x \rangle = \| x \|^2 \geq 0 \\ \| x \|^2 = 0 \Leftrightarrow x = 0 \\ \langle x,y \rangle = \langle y,x \rangle.$$

It's not hard to calculate that $\nabla \phi(y) = 2y$. The rest of it is a calculation with the bilinear property (the first and fourth lines together), followed by identifying $\langle x-y,x-y \rangle$ as $\| x-y \|^2$.

What's going on is very much like the derivation of the parallelogram law (but in reverse).

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