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Given a smooth function $f : \Bbb R^n \to \Bbb R.$ A critical point $p$ of f is called non-degenerate if the Hessian of $f$ at $p$ is non-singular, i.e., the $n \times n$ matrix $$\left(\frac{\partial^2f}{\partial x_i\partial x_j}(p)\right)_{i,j=1}^n$$ is invertible. The map $f:\Bbb R^n\to \Bbb R$ is called a Morse function if all of its critical points are non-degenerate. Such functions are very important in geometry and topology, and any smooth manifold admits a Morse function. The first step to establish this is to prove it for $\Bbb R^n$, which is the goal of this problem.

Let $f: \Bbb R^n \to \Bbb R$ be a smooth function. Prove that for almost all $a = (a_1,\dots ,a_n) \in \Bbb R^n,$ the function $$f_a = f + \sum_{i=1}^n a_ix_i$$ is a Morse function on $\Bbb R^n$. Here “almost all” is in the sense of measure theory, i.e., the assertion holds except for a set of measure zero.

I started solving this problem by using the map $g:U\to \Bbb R^k$ defined as $$g=\left(\frac{\partial f}{\partial x_1},\dots , \frac{\partial f}{\partial x_k}\right),$$ the derivative of $f_a$ at point $p$: $$(df_a)_p=\left(\frac{\partial f_a}{\partial x_1}(p),\dots, \frac{\partial f_a}{\partial x_k}(p)\right)=g(p)+a.$$ If $g(p)=-a$ does that mean $p$ is critical point of $f_a$? I need some help at this point.

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    $\begingroup$ Welcome to MSE. We need to see serious effort here. In particular, we need to know what tools you have at your disposal. I personally would use the Transversality Theorem. It looks like you're just posting homework questions verbatim with no effort whatsoever. $\endgroup$ Apr 28, 2019 at 20:59
  • $\begingroup$ I try to use the following: $\endgroup$
    – Layan
    Apr 29, 2019 at 0:56
  • $\begingroup$ The Morse lemma:** The Morse lemma declares that about any nondegenerate critical point $p$ of a smooth function $f$ there is a coordinate neighborhood $x_i$ so that in those coordinates, $f(x_i) = x_1^2 + \cdots + x_i^2 - x_{i+1}^2 - \cdots - x_n^2 + f(p)$. $\endgroup$
    – Layan
    Apr 29, 2019 at 0:56
  • $\begingroup$ I doubt that will get you such a result. You must use at least Sard's Theorem. $\endgroup$ Apr 29, 2019 at 1:08

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A simpler expression is $f_a(x) = f(x) + \langle a,x\rangle$, so that $$Df_a(x)(v) = Df(x)(v) + \langle a,v\rangle = \langle \nabla f(x) + a, v\rangle.$$ Now, $x \in \Bbb R^n$ is a critical point of $f_a$ if and only if $\nabla f(x) = -a$. Under this assumption, we compute the second derivatives: $$\frac{\partial^2f_a}{\partial x^i\partial x^j}(x) = \frac{\partial^2f}{\partial x^i \partial x^j}(x).$$With this, $x$ is a non-degenerate critical point of $f_a$ if $\nabla f(x) = -a$ and the matrix of second partial derivatives of $f$ at $x$ has full rank. So, to show that $f_a$ is Morse for almost all $a$, we have to show that the values of $a$ for which this condition fails are the critical values of a smooth function, and then we apply Sard's Theorem. If I am not missing anything, the function that does this job is $E\colon \Bbb R^n \to \Bbb R^n$, $E(x) = -\nabla f(x)$.

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