# Minimum of $\langle Ax,x \rangle - 2 \langle b,x \rangle$.

In exercise 5 of section 0 of Fundamentals of convex analysis by Hiriart-Urrut, Lemaréchal, we're supposed to prove that if a self-adjoint linear operator $$A:\mathbb{R}^n \rightarrow \mathbb{R}^n$$ is positive semi-definite and $$b \in \textrm{Im}A$$, then $$q(x)=\langle Ax,x \rangle - 2 \langle b,x \rangle$$ has a finite infimum, and this infimum is a minimum.

I'm not sure how to prove this. An outline of a possible approach:

We will show that $$q$$ grows large as we move away far away from the origin. If that is true, then for a properly chosen $$U=\overline{B}(0,K)$$ we have that $$q$$ is continuous and so it achieves a minimum on $$U$$. Because $$q$$ only "grows" outside of $$U$$, this minimum is global.

Denote the symmetric bilinear form $$(x,y) \mapsto \langle Ax,y \rangle$$ by $$L$$. Then it's easy to see that if $$L(v,v)>0$$, then $$L(kv,kv)=k^2 L(v,v)$$, and so $$q(kv)$$ goes to infinity as we raise $$k>0$$. By the spectral theorem there exists a basis $$\{ b_1,...,b_n \}$$ such that $$L$$ with respect to this basis is diagonal and the elements on the diagonal are non-negative.

If the elements on the diagonal are all positive we are done by the previous part. If the $$i$$th element on the diagonal is zero, then $$q(b_i)=L(b_i,b_i)-2 \langle b,x \rangle = L(b_i,b_i) - 2 \langle Ay, b_i \rangle = L(b_i,b_i) - 2L(b_i,y)=0-0=0$$ (using the fact that $$b \in \textrm{Im} A$$).

However this seems excessive and I doubt I'm supposed to use the spectral theorem, so I think I must be missing something. EDIT: spectral theorem might not be necessary - symmetric elementary operations are enough to transform $$L$$ into a diagonal bilinear form with respect to some basis.

• I think you can just note that $q$ is a convex function, then take its first and second derivatives and see what happens – David M. Mar 2 at 15:09
• I'm doing an exercise from a book called "Fundamentals of convex analysis" and I completely overlooked this option haha. Thank you, that sounds like a reasonable approach. – John P Mar 2 at 15:17
• Sometimes it’s the simplest thing. Your observation about $q$ being unbounded at infinity (i.e. radially unbounded) is useful for general convex analysis, though it doesn’t hold here unless $A$ is strictly positive definite. – David M. Mar 2 at 15:37