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How can I prove: Let $f:\mathbb{R}^k\longrightarrow{\mathbb{R}}$, and, for each $a\in{\mathbb{R}^k}$ define $$f_a(x)=f(x)+a_1x_1+...+a_kx_k.$$ For almost all $a\in{\mathbb{R}^k}$, $f_a$ is a Morse function.

NextBear
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  • By definition a Morse function is a function whose critical points are non degenerate, that means that the matrix associated to the quadratic form at any critical point of f_a is invertible. So what you can do, firstly is compute the gradient of that function and see for which values of the x does vanish. Then compute the matrix associated to the quadratic form d^2 f and show that it does not vanish at that critical points you've found. – Salvatore Jun 04 '17 at 21:48
  • Are you sure that the function takes value in R^k and not in R ? – Salvatore Jun 04 '17 at 21:58
  • You're right, thank you. It has to be R. – NextBear Jun 05 '17 at 12:59
  • Word-by-word duplicate – Balarka Sen Jun 05 '17 at 15:04

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I assume that the codomain of $f$ is $\mathbb{R}$ not $\mathbb{R}^k$, otherwise, this is nonsense.

Let $g\colon\mathbb{R}^k\rightarrow\mathbb{R}^k$ defined by: $$g(x):=(\partial_{x_1}f(x),\ldots,\partial_{x_k}f(x)).$$ Then, notice that one has: $$\mathrm{d}_xf_a=g(x)+a.$$ Therefore, $x$ is a critical point of $f_a$ if and only if $g(x)=-a$.

Assume that that $-a$ is a regular value of $g$ and let $x$ be a critical point of $f$, then $x$ is a non-degenerated critical point of $f$, since ${\mathrm{d}^2}_xf_a=\mathrm{d}_xg$ is surjective and thus invertible.

Finally, using Sard's theorem for almost all $a$, $-a$ is a regular value of $g$, so that for almost all $a$, $f_a$ is a Morse function.

C. Falcon
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