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This is from wikipedia for the second derivative Hessian matrix test. From the determinant it seems to assume that $f_{xy} = f_{yx}.$ Why is this valid to assume? Is the test only valid for when $f_{xy} = f_{yx}?$

green frog
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2 Answers2

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If $f$ is twice-differentiable (not just $C^2$), $f(x+h) = f(x)+h\nabla f(x)+\frac{1}{2}(Hf(x)h,h)+o(|h|^2)$. If $x$ is a critical point, then $\nabla f (x) = 0$, so $f(x+h) = f(x)+\frac{1}{2}(Hf(x)h,h)$ plus a smaller order term. So, if $Hf(x)$ is positive definite, then $(Hf(x),h,h) > 0$ and thus $f$ has a local min at $x$. Analogously if $Hf(x)$ is negative definite.

mathworker21
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  • How does this address the question at all? The question is about why the author assumed that the mixed partials are equal, not about the sign of the determinant or whatever that implies about local extrema. – Allawonder Sep 06 '19 at 07:30
  • @Allawonder It is very clear that the last question is the main question of the post, i.e. whether the Hessian matrix second derivative test holds without the $C^2$ assumption. Would you prefer I ignore this last question and tell the OP that $-f_{xy}f_{yx} = -f_{xy}^2$ if and only if $f_{xy} = f_{yx}$? – mathworker21 Sep 06 '19 at 12:17
  • Thank you very much! – green frog Sep 06 '19 at 13:15
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This would be valid only for when $f_{xy}=f_{yx}.$ A sufficient condition for this equality to hold is for either of the sides to be continuous.

mathworker21
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Allawonder
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  • when you say "This would be valid", does "This" refer to the test itself or the determinant formula. – green frog Sep 02 '19 at 08:49
  • @伽罗瓦 The first "this" refers to their expression for $D(x,y).$ The second to the assumed equality that makes the expression for $D$ valid. – Allawonder Sep 02 '19 at 09:20
  • So why does the article just assume this equality? What if we don't have continuity and the second partials aren't the same... – green frog Sep 02 '19 at 09:22
  • @伽罗瓦 I don't know why they assume it (perhaps somewhere above they said they would assume it, or perhaps they assumed that the function is twice continuously differentiable). If we don't have continuity of at least one of the cross-partials, then there's no guarantee that they would be equal. Finally, if these partials are different, obviously the determinant isn't as given. In that case it should be $f_{xx}f_{yy}-f_{xy}f_{yx}.$ This is a more general expression for the determinant, and trying to write the last terms as a square is a detail which I see as being trivial here. – Allawonder Sep 02 '19 at 09:35
  • Dear Downvoter, could you please explain what's wrong here? – Allawonder Sep 06 '19 at 07:25
  • I didn't downvote, but I can guess why the downvoter downvoted, as I do partially agree with the downvote. The OP asked 2 questions. (1) Why is it valid to assume $-f_{xy}f_{yx} = f_{xy}^2$ and (2) Does the Hessian second derivative test work for any twice differentiable functions. You completely ignored question (2) and just rephrased question (1), not explaining why is it valid to assume $f_{xy} = f_{yx}$. – mathworker21 Sep 06 '19 at 12:28
  • @mathworker21 Well, there's not much more point to this now than just to clear the air: I'd treated the questions as related. I see that I might have misinterpreted what OP meant: I'd parsed the two questions as asking about why they took $f_{xy}$ equal to $f_{yx}$ in the article. To the question, Why is it valid to assume this? I stated the well-known sufficient condition -- I understood OP as asking for conditions that guarantee this. The second question I saw as an extension of the first, namely Is this valid only for when $f_{xy}=f_{yx}? was taken to refer to the equality. – Allawonder Sep 06 '19 at 21:39