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I am reading the following proof of second derivative test for higher dimension, but get stuck with one key step.

In the following, $H(x)$ stands for the Hessian matrix of $f:\Bbb R^n \to \Bbb R$ at $x$ with continuous second-order mixed derivatives near and at $x$. It looks like an easy piece that "when $v$ is small enough" then the inequality follows, but I cannot figure out how this is true following the author's reasoning.

Note in the proof, $t$ is a fixed value in $(0,1)$ from the Lagrange remainder, but we can let $\bf v$ go to $\bf 0$. So ${\bf H}({\bf x}+t{\bf v})$ is actually a function of $\bf v$. Also every element of ${\bf H}({\bf x}+t{\bf v})$ goes to the corresponding element of $\bf H(x)$ as $\bf v \to 0$ since the second order derivatives of $f$ are assumed to be continuous.

PS: I did some investigation and find that the eigenvalues of ${\bf H}({\bf x}+t{\bf v})$ should be continuous with respect to $\bf v$ and goes to the eigenvalues of $\bf H(x)$, but this requires a lot of external knowledge to prove this theorem first, and it looks to me that the following proof does not use this.

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where 7.19 is derived from Taylor's formula.

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$f \in C^2(U)$, so $A(\mathbf{v}) = H(\mathbf{x}+t\mathbf{v})-H(\mathbf{x})$ is a continuous function of $\mathbf{v}$ with $A(0)=0$. Hence, given $\delta^2/2$, it is possible to find $\varepsilon>0$ so that $\lVert A(\mathbf{v}) \rVert < \delta^2/2$ whenever $\lVert v \rVert < \varepsilon$, where $\lVert \cdot \rVert$ is a sensible norm on the space of matrices.

I don't know what the book uses, but the sensible one to take here is given by the absolute value of the eigenvalue of $A$ with the largest modulus, from which we immediately find that for the set of $\mathbf{v}$ we found above, $$ \mathbf{v}^T A(\mathbf{v}) \mathbf{v} \geq -\lVert A(\mathbf{v}) \rVert \lVert \mathbf{v} \rVert^2 > -\frac{1}{2}\delta^2 \lVert \mathbf{v} \rVert^2. $$