Let $f(x): \mathbb{R}^n \to \mathbb{R}$ be a real-valued twice continuously differentiable function and $n>1$. I define the function $g(x) = f(x) + x^{\top} A x$ where $A$ is random matrix (say entries i.i.d from uniform distribution [-1,1]). Can we say that the Hessian of $g$ is invertible for all $x$ with probability one?
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$\begingroup$ The Hessian of $f'$, I guess you mean? (Can we call it something that looks less like a derivative?) $\endgroup$– Nate EldredgeCommented Feb 10, 2018 at 0:43
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$\begingroup$ Fixed, thank you! $\endgroup$– user31317Commented Feb 10, 2018 at 1:22
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2$\begingroup$ If $f$ is three times continuously differentiable, then the Hessian $Hf$ is a $C^1$ map from $\mathbb{R}^n$ into $\mathbb{R}^{n^2}$, so its image has Hausdorff dimension $n$ and hence Lebesgue measure zero. In particular $Hg$ is everywhere nonzero with probability one, which is a start. But if $Hf$ is merely continuous then it seems like we could get some bad behavior. $\endgroup$– Nate EldredgeCommented Feb 10, 2018 at 5:18
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$\begingroup$ Nice start Nate. $\endgroup$– user31317Commented Feb 10, 2018 at 5:38
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2 Answers
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Here is my counterexample: Take $f(x)=\|x\|^4$, then for any $x\neq 0$, by spherical symetry, $x$ is a eigenvector of $Hf|_x$. we have then $$Hf|_x(x)=a\|x\|^2 x $$ We can choose $x=v$ an eigenvector on $\frac{1}{2}(A+A^T)$. ie $\frac{1}{2}(A+A^T)(v)=\lambda v$ then with $t\in \mathbb{R}$ $$ (Hg)|_{tv}=(Hf+A)|_{tv}(v)=(a\|v\|^2 t^2 +\lambda) v $$ Therefore for $t^2=-\lambda/(a\|v \|^2)$. $Hg|_{tv}(v)=0$ and it is then not invertible.
For a more general counterexample, take $f$ such that there exist $x_1,x_2$ with $det(Hf(x_1))<0$ and $det(Hf(x_2))>0$, then for comparatively small random matrix $A$, $det(Hf(x_2)+A)>0$ and $det(Hf(x_1)+A)<0$ and by continuity. There exist $x_0$ such that $Hg(x_0)$ is not invertible.
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$\begingroup$ Hi, what I don't understand is that you assume $\lambda, \alpha$ have opposite signs with positive probability? $t$ is a real so $\frac{-\lambda}{\alpha ||v||^2}>0$. $\endgroup$– user31317Commented Feb 15, 2018 at 17:59
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$\begingroup$ What is $a$? (this comment was otherwise too short) $\endgroup$– lcvCommented Feb 15, 2018 at 21:42
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$\begingroup$ (a=4*3=12.) Yes, with positive probabily there exists $\lambda<0$, this particular example don't work otherwise. Insteed of $\|x\|^4$, one could have also taken $f(x)=F(\| x \|)$, with $F$ has a surjective seconde derivative. $\endgroup$ Commented Feb 16, 2018 at 8:47
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It's false for $n = 1,$ since then if $f$ is the double integral of your favorite surjective function $\mathbb{R}\to \mathbb{R},$ then the answer is NO.
answered Feb 10, 2018 at 4:54
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$\begingroup$ Nice example! This is not what I am looking for because it is a trivial case (when $n=1$), it is also true that $n^2 = n$ for the case $ n=1$. $\endgroup$– user31317Commented Feb 10, 2018 at 5:37
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$\begingroup$ Can we even show it for a neighborhood around a critical point of $f$? Because some union bound argument should work afterwards. Any ideas? $\endgroup$– user31317Commented Feb 11, 2018 at 0:33