# quadratic form corresponding to function at critical point is positive definite implies local minimum

Let $f: \mathbb{R}^n \to \mathbb{R}$ be a $C^3$ function. Have $x_0$ be a critical point of $f$. How would I go about proving that if the quadratic form $q(h)$ corresponding to $f$ at $x_0$ is positive-definite, then $f(x_0)$ is a local minimum? I think I will probably have to push some epsilons and utilize some form of Taylor's theorem. Any help would be appreciated. Thanks!

You answered the question yourself. If you agree that $f(x_0+h)=f(x_0)+q(h)+\text{remainder}$ where the remainder is $o(\|h\|^2)$, then $q$ has a positive minimum $m$ on the unit sphere $S^{n-1}$ by compactness of the sphere, and for small enough $\epsilon$, $\|h\|<\epsilon$ implies that the $|\text{remainder}|$ is smaller than $m\|h\|^2$ (while $q(h)\geq m\|h\|^2$).
Let $R_2(h) = f(x_0 + h) - T_2(h)$ where $T_2(h) + f(x_0) + h\nabla f(x_0) + q(h)$ (observe here that $q(h)$ includes the ${1\over2}$ coefficient. We know that$$\lim_{\|h\| \to 0} {{R_2(h)}\over{\|h\|}} = 0$$by Taylor's theorem.
Because $\nabla f(x_0) = 0$, we have that $f(x_0 + h) - f(x_0) = q(h) + R_2(H)$. To show that $f(x_0)$ is a local minimum, we want to demonstrate that there exists $\delta > 0$ such that $f(x_0 + h) - f(x_0)>0$ for all $\|h\| < \delta$. So it is enough to show that $q(h) + R_2(h)$ is positive for small enough $h \neq 0$.
Evidently the expression $${{q(h) + R_2(h)}\over{\|h\|^2}} > 0$$ if and only if $q(h) + r_2(h) > 0$. And because $q$ is a quadratic form $${{q(h) + R_2(h)}\over{\|h\|^2}} = q\left({h\over{\|h\|^2}}\right) + {{R_2(h)}\over{\|h\|^2}}.$$ We interpret $$q\left({h\over{\|h\|^2}}\right): \mathbb{R}^n - \{0\} \to \mathbb{R}$$as a function $\tilde{q}(x): S^{n-1} \to \mathbb{R}$ since each $${h\over{\|h\|^2}} \in S^{n-1}.$$ Furthermore since $S^{n-1}$ is closed and bounded, it is compact, and since the function $q$ and by extension $\tilde{q}$ is continuous, its image in $\mathbb{R}$ attains its maximum value. Hence $\tilde{q}(x) \ge m$ for all $x \in S^{n-1}$, and because $q$ is positive-definite, $m > 0$. Thus $${{q(h) + R_2(h)}\over{\|h\|^2}} \ge m + {{R_2(h)}\over{\|h\|^2}}.$$ Now because $$\lim_{\|h\| \to 0} {{R_2(h)}\over{\|h\|^2}} = 0$$ we know there exists some $\delta > 0$ such that $$\left|{{R_2(h)}\over{\|h\|^2}}\right| < {m\over2}$$ for all $\|h\| < \delta$. Thus for $h \in B_\delta(0)$, $${{q(h) + R_2(h)}\over{\|h\|^2}} \ge m - {m\over2} = {m\over2} > 0.$$ So $f(x_0 + h) - f(x_0) > 0$ in this $\delta$ neighborhood, and $f(x_0 + h) - f(x_0) > 0$ in this $\delta$ neighborhood, and $f(x_0)$ is a local minimum as we wanted to show.