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I have a Lagrangian of which I want to find the supremum in the primal variable $\boldsymbol{x}$:

$\mathscr{L}(\boldsymbol{x},\boldsymbol{\lambda})=f(\boldsymbol{x})^T\boldsymbol{a} + \boldsymbol{\lambda}^T(\boldsymbol{b}+cf(\boldsymbol{x})-f(\boldsymbol{x}))$

such that:

$\mathscr{L}:\mathbb{R}^M \times \mathbb{R}^N \to \mathbb{R}$

$f:\mathbb{R}^M \to \mathbb{R}^P$

In this way, the supremum will be at an $\boldsymbol{x}$ such that the gradient of the Lagrangian satisfies:

$\nabla_\boldsymbol{x}\mathscr{L}(\boldsymbol{x},\boldsymbol{\lambda})=\nabla_x f(\boldsymbol{x})^T(\boldsymbol{a} + c\boldsymbol{\lambda}-\boldsymbol{1})=0$

and it will be $\boldsymbol{\lambda}^T\boldsymbol{b}$. It is:

$$ sup \mathscr{L}(\boldsymbol{x},\boldsymbol{\lambda}) = \left\{ \begin{array}{ll} \boldsymbol{\lambda}^T\boldsymbol{b} & \nabla_x f(\boldsymbol{x})^T(\boldsymbol{a} + c\boldsymbol{\lambda}-\boldsymbol{1})=0 \\ \infty & otherwise \end{array} \right. $$

This statement is only truth if $\nabla_x f(\boldsymbol{x})=0$ implies $f(\boldsymbol{x})=0$

So, my questions are:

  1. Is there any family of functions $f(\boldsymbol{x})$ which satisfies $f(\boldsymbol{x})=0$ if $\nabla_x f(\boldsymbol{x})=0$ (appart from linear functions)?

  2. Is there a more general expression for the suppremum of this Lagrangian?

Thanks in advance!

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    $\begingroup$ what do you mean by $\nabla_w$ ? $\endgroup$
    – Carlo Beenakker
    Commented Apr 28, 2017 at 13:13
  • $\begingroup$ Sorry for the notation. I tried to fix it in the question. I mean $\boldsymbol{x}$ is an $\mathbb{R}^n$ input vector, and $\nabla f(x)$ the gradient of $f(\boldsymbol{x})$ w.r.t vector $\boldsymbol{x}$. If it is not clear yet, just let me know. $\endgroup$
    – Danilo Socovan
    Commented Apr 28, 2017 at 14:26
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    $\begingroup$ Like $f(x)=(g(x)-c)^2$, with $\nabla g\neq0$ everywhere and $c$ being in the range of $g$? $\endgroup$
    – Piero D'Ancona
    Commented Apr 28, 2017 at 14:35
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    $\begingroup$ apologies, I don't understand the question:$ \nabla f=0$ means $f={\rm constant}$, what else is there to say? $\endgroup$
    – Carlo Beenakker
    Commented Apr 28, 2017 at 14:35
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    $\begingroup$ @Neil: yes, as soon $n\ge2$. It's quite easy to construct a smooth function with any prescribed finite set $S$ of critical points, together with prescribed levels and Morse indices, plus, possibly, a set $S'$ of finitely many other critical points. To kill these additional critical points include them into an unbounded simple curve $\Gamma\sim [0,\infty)$, disjoint from $S$; consider a diffeo $\phi:\mathbb{R}^n\to\mathbb{R}^n\setminus\Gamma$ which is the identity outside a suitable nbd of $\Gamma$. Then $f\circ\phi$ has exactly $S$ as a critical set, and coincides with $f$ on a nbd of $S$. $\endgroup$
    – Pietro Majer
    Commented Apr 28, 2017 at 18:02

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