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?

enter image description here

  • 1
    $\begingroup$ 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

1 Answer 1


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).


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.