Vtyjkyuk

Let $f:Osubset BbbR^ntoBbbR$ be a $C^3$ function. Let $x_0in O$ be a critical point of $f$. Let beginalignA=left(fracpartial ^2fpartial x_ipartial x_jright)_1leq i,jleq nendalign
If $A$ is positive, then $x_0$ is a local minimum.

Proof

By Taylor's formula
beginalignf(x)=f(x_0)+f'(x_0)(x-x_0)+fracf''(x_0)2!(x-x_0)^2+Vert x-x_0 Vert^2epsilon(x-x_0)endalign
But beginalignf'(x_0)=0endalignThis implies
beginalignf(x)-f(x_0)=fracf''(x_0)2!(x-x_0)^2+Vert x-x_0 Vert^2epsilon(x-x_0)endalign
Now,
beginalign
f''(x_0)(x-x_0)^2=langle A(x-x_0),(x-x_0)rangleendalign
beginalign
=sum^n_i=1sum^n_j=1fracpartial ^2f(x_0)partial x_ipartial x_j(x_j-x_0_j)^2endalign
beginalign
=sum^n_i=1sum^n_j=1(x_j-x_0_j)fracpartial ^2f(x_0)partial x_ipartial x_j(x_j-x_0_j)endalign
beginalign
=langle (x-x_0),A(x-x_0)rangleendalign
Thus,
beginalignf(x)-f(x_0)=fraclangle (x-x_0),A(x-x_0)rangle2!(x-x_0)^2+Vert x-x_0 Vert^2epsilon(x-x_0)endalign
beginalignf(x)-f(x_0)geqVert x-x_0Vert^2left[alpha+epsilon(x-x_0)right],;;textfor some;alpha>0endalign
Hence, $exists;delta>0$ such that $Vert x-x_0Vert<delta$ implies $left[alpha+epsilon(x-x_0)right]>0$. So, for $xin(x_0-delta,x_0+delta)$
beginalignf(x)geq f(x_0)endalign
Therefore, $x_0$ is a local minimum.

Can someone check if this proof is correct? Corrections will be highly welcome! Thanks

A few corrections/elaborations for you: First of all, you want $O$ to be an open subset of $Bbb R^n$ (or at least to contain a ball around $x_0$). Second, by $A$ you mean the Hessian matrix evaluated at $x_0$. Third, what do you mean by saying $A$ is positive? Usually we say positive definite. Fourth, your notation for the second-degree term in Taylor's formula really doesn't make sense. (Is your textbook using that?) It really needs to be what you wrote down after the "Now." :) Fifth, the $|$ at the end of the displayed equation doesn't belong there, but you do want to say that $epsilon(u)to 0$ as $uto 0$.

Now, I'm not sure what the point of your "Now" computation is. But you do need to explain that if the matrix $A$ is positive definite, then $langle Av,vrangle ge alpha|v|^2$ for some positive $alpha$. (This follows in one of two ways: (1) all the eigenvalues of $A$ are positive; or (2) the continuous function $Q(v)=langle Av,vrangle$ on the unit sphere has a minimum value, which must of course be positive. And there's your $alpha$.)

Last, I would suggest being more explicit in your last step. By the limit definition with $epsilon$, noting that $alpha>0$, you know that there's $delta>0$ so that whenever $|x-x_0|<delta$ we'll have $|epsilon(x-x_0)|<alpha$. That'll do it. (Why?)

Overall, with a bit of revision, good job! :)