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.

  • $\begingroup$ I think you can just note that $q$ is a convex function, then take its first and second derivatives and see what happens $\endgroup$ – David M. Mar 2 at 15:09
  • $\begingroup$ 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. $\endgroup$ – John P Mar 2 at 15:17
  • $\begingroup$ 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. $\endgroup$ – David M. Mar 2 at 15:37

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