# For almost all $a \in \Bbb R^n,$ the function $f_a$ is a Morse function on $\Bbb R^n$.

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.

• 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. – Ted Shifrin Apr 28 '19 at 20:59
• I try to use the following: – Layan Apr 29 '19 at 0:56
• 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, $f(x_i) = x_1^2 + \cdots + x_i^2 - x_{i+1}^2 - \cdots - x_n^2 + f(p)$. – Layan Apr 29 '19 at 0:56
• I doubt that will get you such a result. You must use at least Sard's Theorem. – Ted Shifrin Apr 29 '19 at 1:08

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 $$E\colon \Bbb R^n \to \Bbb R^n$$, $$E(x) = -\nabla f(x)$$.