From Boyd & Vandenberghe's Convex Optimization, example 3.5:
Suppose the quadratic function:
$$f(x, y) = x'Ax + 2x'By + y'Cy$$
(where $A$ and $C$ are symmetric) is convex in $(x, y)$,
We can express
$$g(x) = inf[f(x, y)]$$ as $$g(x) = x'(A − B*inv(C)*B' )x$$
My questions is how to derive the infimum of $f(x,y)$ on $y$ to get $$g(x) = x'(A − B*inv(C)*B' )x$$ Please provide details steps of derivation.
I guess I would have to accept that infimum of f(x, y) on y is to take derivative over y. I don't know why but if it is by definition, it means treat x as constant pointwise (for each x in dom of f(x,y)), then $$inf(f(x,y))$$ over y becomes a function of x, after taking derivative on y and set it to zero, plugin y as represent by x into f(x,y) then the infimum can be represent by a function of x, g(x).
Rest is trivial. Take the derivative of f(x, y) on y and set to 0, then $$ 2B'X + Cy + C'y = 0$$ since C is symmetric, $$ y = -inv(C)*B'X $$ plugin to f(x, y),
$$′+2′+′ = x'Ax - 2x'B*inv(C)*B'x + X'B*inv(C)*C*inv(C)*B'X =x'Ax - x'B*inv(C)*B'x = x'(A-B*inv(C)*B')x $$