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We have a function $f(u)= u^{T}N^TNu$ where $u$ is a $n$-dimensional vector and $N$ is a $n\times n$ matrix. Is this a strictly convex function in $u$?

I know that if the hessian of $f(u)$ with respect to $u$, which is $N^TN$, is a positive definite matrix, then $f(u)$ is strictly convex in $u$. But, if $N^TN$ is a positive semi-definite matrix, then we cannot say it is not strictly convex (indeed it is convex), maybe it still is. Now, in the case that $N^TN$ is a positive semi-definite matrix, what can we say about whether it is strictly convex or not?

Thanks,

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    $\begingroup$ What if $N = 0$? $\endgroup$ – Robert Israel Sep 3 '15 at 3:47
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It is strictly convex if and only if $N$ is positive-definite.

Like you point out, $N^TN$ is always semidefinite. If it has an eigenvector $v$ with zero eigenvalue, then for any arbitrary $x$

$$f(x+tv) = f(x)$$ for all $t$ and $f$ is not strictly convex.

Put differently, if a function is convex yet has a singular Hessian, this must mean that convexity is rescued in the nullspace directions by the higher derivatives of $f$. This doesn't happen here because all of $f$'s higher-order derivatives are zero.

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  • $\begingroup$ Thanks, I think your first statement is not correct. It is not true to say if and only if. If N is positive-definite, then f is strictly convex. But, it is possible to have a N which is not positive-definite, but f is strictly convex. See here: math.stackexchange.com/questions/321090/… $\endgroup$ – m0_as Sep 3 '15 at 6:01
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    $\begingroup$ @m0_as Maybe you're thinking of examples like $ x^4$, which is strictly convex but has only a semidefinite Hessian at the origin. But for the particular function you gave, the if and only if statement is correct. $\endgroup$ – littleO Sep 3 '15 at 6:06
  • $\begingroup$ @user7530 Thanks. If instead of $N^T N$, we have a matrix $Q$ which is positive semi-definite, is it possible that f is strictly convex in u? $\endgroup$ – m0_as Sep 3 '15 at 18:32
  • $\begingroup$ @m0_as Yes, it's possible. littleO's example is one such case. $\endgroup$ – user7530 Sep 3 '15 at 19:19
  • $\begingroup$ @user7530 thanks, but the example of liitleO is not quadratic. I meant among quadratic form functions with positive semi-definite $Q$, is it possible that $f(u) = u^T Q u$ is strictly convex? $\endgroup$ – m0_as Sep 3 '15 at 19:33

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