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Proof: extremum has a semi definitie Hessian matrix

  1. Apr 14, 2015 #1
    Consider a function ##f : U \subseteq \mathbb{R}^{n} -> \mathbb{R}## that is an element of ##C^{2}## which has an minimum in ##p \in U##.

    According to Taylor's theorem for multiple variable functions, for each ##h \in U## there exists a ##t \in ]0,1[## such that :

    ##f(p+h)-f(p) = h^{T}.D^{2}(p+th).h##

    Where ##D^{2}## is the Hessian or the matrix with the mixed partial derivatives, ##D^{2}(p+th)## means it's not in point p. I already assume it's a minimum so the Jacobian matrix that should be there is 0.

    Now I should take a limit and somehow show that in both cases where p is a strict or not strict minimum that is ##f(p)<f(x)## of ##f(p)\leq f(x)## that in both cases ##D^{2}## will be negative semi definite. Can someone help me finish/understand this final step formally? Because I'm not sure about how a limit will work in the equation above, the left term just goes to zero.
     
  2. jcsd
  3. Apr 15, 2015 #2

    wabbit

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    You could divide both sides by ## ||h||^2 ##.
     
  4. Apr 15, 2015 #3
    Not sure what that would give me. Left I'd get a similar expression to the derivative but since it's in multiple variables I'm not sure. It would be equal to the limit of ##\frac{D*h}{||h||}## I guess?
     
  5. Apr 15, 2015 #4

    wabbit

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    Well you can then have ##h\rightarrow 0## but ##h/||h||## remains finite, and you could set this to be any vector ##u=h/||h||## of your choice and see what this tells you about D when h approaches 0 but u is fixed.

    Or if you prefer that formulation, set ##h=\theta u## and take the limit as ##\theta\rightarrow 0##

    I dont know what your "*" is but if it is just usual matrix-vector multiplication, no this isn't what you get.
     
  6. Apr 15, 2015 #5
    I reasoned that that's what's it supposed to be according to ##lim \frac{f(p+h)-f(p)-D*h}{||h||}=0## where ##D## is the derivative matrix.
     
  7. Apr 15, 2015 #6

    wabbit

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    But as you stated in your op, the first derivative is 0 and D is the Hessian, so suddenly using D as the jacobian instead is truly bizarre reasonning.

    Oh sorry I see you were using ##D^2## to denote the hessian - please read my previous statement as referring to that.
     
  8. Apr 15, 2015 #7
    Oh right, should have thought a bit longer before replying.
     
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