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
$$det(H)^2 + \sum_{i=1}^k (\frac {\partial f}{\partial x_i})^2>0$$
=>
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
$$det(H)^2 + \sum_{i=1}^k (\frac {\partial f}{\partial x_i})^2>0$$
<=
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.
You aren't doing anything specifically wrong, it just can be written out more succinctly than how you're doing it.
If you have $n$ numbers $a_1,...,a_n$, then $a_1^2 + ... + a_n^2 > 0$ if and only if at least one $a_i$ is nonzero. So you're being asked to show the function is Morse if and only if at each point $x$ either $det(H) \neq 0$ or the vector $\nabla f$ is nonzero. This is exactly the same as the requirement that at each $x$, if $\nabla f$ is zero then $det(H)$ is nonzero. This is the definition of a Morse function.