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Lets say I have a real valued function which is writable as a polynomial in terms of Frobenius norms of a pair of matrices as in it is of the form, $f_B(A) = f(||A||_F^2, ||AB||_F^2, ||A^TAB||_F^2)$ where $f$ is a polynomial. Then effectively its a polynomial in the entries of the $A$ (for a fixed $B$). Suppose that we know that for fixed $B$ when the Frobenius norm of $A$ is arbitrarily high the function is unbounded below.

  • Isn't the above property enough to guarantee that the function $f_B(A)$ is upperbounded?

In general,

  • Isnt it true (if yes then how can we prove) that a multivariable polynomial can only have a finite number of values at local maxima?

  • Do we have analytic tests which can decide if a multivariable polynomial (like say $f_B(A)$ above) is upperbounded?

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    $\begingroup$ I may be misinterpreting your question, but why would a polynomial function having no lower bound imply that it is bounded from above? The function $xy$ isn't bounded from either above or below. $\endgroup$
    – Alon Amit
    Jun 27, 2017 at 22:39
  • $\begingroup$ Here the thing is that its not lowerbounded for arbitrarily large values of Frobenius norm of $A$. Thats a very special kind of dependance. That means for every direction about the origin the function is asymptotically unbounded below in that direction. $\endgroup$
    – gradstudent
    Jun 27, 2017 at 22:43
  • $\begingroup$ @AlonAmit: I think what the OP may be after is a generalized version of the (trivial) fact that if $f(x^2)$ is unbounded below, then it is bounded above. $\endgroup$
    – Christian Remling
    Jun 28, 2017 at 0:14
  • $\begingroup$ Right. Somehow this is looking weirdly non-obvious in the general setting! $\endgroup$
    – gradstudent
    Jun 28, 2017 at 1:46
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    $\begingroup$ @Christian Again, I may be misunderstanding, but the polynomial looks like it has three independent inputs. Even if each one of them is a square or a quadratic form, there's no reason for that property to hold (e.g. $x^2-y^2$). $\endgroup$
    – Alon Amit
    Jun 28, 2017 at 1:55

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The set of critical points (in the domain) of a polynomial is the solution set of a system of polynomial equations viz the vanishing of the first derivatives. So it has finitely many irreducible components, hence finitely many connected components. And the polynomial is constant on each connected component of its critical locus. Therefore there are finitely many critical values.

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  • $\begingroup$ Thanks! I did not quite get the terminology you are using. It seems to be some algebraic geometry that I dont know. Could you please explain what you mean by "irreducible components" and what these "finitely many connected components"? Why is the original polynomial constant on these connected components? Can you please explain a little more? $\endgroup$
    – gradstudent
    Jun 28, 2017 at 14:05

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