Let $f$ be a smooth function on an open set $U\subset R^k$. For each $x \in U$ let $H(x)$ be the Hessian Matrix of $f$, whether $x$ is critical point or not. Prove that $f$ is Morse function if an only if
$$det(H)^2 + \sum_{i=1}^k (\frac {\partial f}{\partial x_i})^2>0$$
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Assume that $f$ is a Morse function, meaning every critical point $x$ is a non-degenration point. so at these point $x$ , $det(H) \not =0$, thus $det (H)^2 >0$.
If $x$ is a critical point then $\frac {\partial f}{\partial x}=0$ so $\sum_{i=1}^k (\frac {\partial f}{\partial x_i})^2=0$
if $x$ isn't a critical point, $\frac {\partial f}{\partial x} \not =0$ so $\sum_{i=1}^k (\frac {\partial f}{\partial x_i})^2>0$.
Since $det (H)^2 >0$, either $\sum_{i=1}^k (\frac {\partial f}{\partial x_i})^2=0$ or $\sum_{i=1}^k (\frac {\partial f}{\partial x_i})^2>0$. We still have
$$det(H)^2 + \sum_{i=1}^k (\frac {\partial f}{\partial x_i})^2>0$$
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I assume that $$det(H)^2 + \sum_{i=1}^k (\frac {\partial f}{\partial x_i})^2>0$$ I need to show that $det (H) \not =0$ for all critical point of $f$. For $x$ are critical point of $f$ then $\frac {\partial f}{\partial x}=0$ so $\sum_{i=1}^k (\frac {\partial f}{\partial x_i})^2=0$ . This implies that $det(H)^2 >0$, thus $det (H) \not =0$ for every critical point $x$ of $f$. Therefore, every critical point $x$ of $f$ is non-degeneration point. Hence $f$ is Morse.
I still have the feeling that I missed something important. I would be very appreciated if anyone could help me check this proof.