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See the book OPTIMIZATION: Algorithms and Applications by Rajesh Kumar Arora, Page-$56$.

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The necessary conditions for $x^*$ to be a minimum are that
$$∇f(x*) = 0 qquad (3.1)$$

and $x^T Hx $ is positive definite $(x^T H x > 0)$. To ensure this,
eigenvalues of $H$ are to be positive. Consider a two-variable
function $$f_x = {x_1}^2 + {x_2}^2 - 2x_1 qquad (3.2)$$

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I have three questions:

1. What is $Hx$?


2. What does it mean by the term positive-definite?


3. Why are
   (a) $displaystyle ∇f(x*) = 0$, and
   (b) $displaystyle x^T H x > 0$
   necessary conditions for $x^*$ to be minimum?

$1$. What is $H$?

The matrix $H$ is referred to as the Hessian matrix, i.e.

$$H = begin{bmatrix}
dfrac{partial^2 f}{partial x_1^2} & dfrac{partial^2 f}{partial x_1,partial x_2} & cdots & dfrac{partial^2 f}{partial x_1,partial x_n} \[2.2ex]
dfrac{partial^2 f}{partial x_2,partial x_1} & dfrac{partial^2 f}{partial x_2^2} & cdots & dfrac{partial^2 f}{partial x_2,partial x_n} \[2.2ex]
vdots & vdots & ddots & vdots \[2.2ex]
dfrac{partial^2 f}{partial x_n,partial x_1} & dfrac{partial^2 f}{partial x_n,partial x_2} & cdots & dfrac{partial^2 f}{partial x_n^2}
end{bmatrix}$$

$2$. What does it mean by the term positive-definite?

Positive definite means that if you grab any vector $z$, the term $z^T H z$ is always positive for any non-zero $z$. Informally speaking, it is a way of saying that your matrix is "positive". To see that, take a scalar and say it is positive-definite, it just turns out that the scalar is positive.

$3$. Why are
(a) $∇f(x*) = 0$, and
(b) $x^T H x > 0$
necessary conditions for $x^*$ to be minimum?

In 1-dimensional functions, so that you have a local minimum around $x^*$, wiggling around $x^*$ should make your function larger, right? This means that we must have
$$f(x^*+epsilon )gt f(x^*)Rightarrow f(x^*+epsilon )- f(x^*)gt 0$$
Taylor series tells us that
$$f(x^*+epsilon)approx f(x^*)+frac{df}{dx}|_{x^*}epsilon+frac 12frac{d^2f}{dx^2}|_{x^*}epsilon^2+O(epsilon^3)$$
We can substitute the approximation of $f(x^*+epsilon)$ to the definition of minimum
$$f(x^*+epsilon )- f(x^*)=f(x^*)+frac{df}{dx}|_{x^*}epsilon+frac 12frac{d^2f}{dx^2}|_{x^*}epsilon^2-f(x^*)gt 0$$
$$frac{df}{dx}|_{x^*}epsilon+frac 12frac{d^2f}{dx^2}|_{x^*}epsilon^2gt 0$$

Now in $n-$dimensional, which is your case, (note that $epsilon$ becomes a vector now)
$$nabla f(x^*)epsilon+ epsilon^T frac 12nabla^2 f(x^*)epsilongt 0$$
We now use the assumption that $nabla f(x^*)=0$, which gives
$$epsilon^T nabla^2 f(x^*)epsilongt 0$$
Since the term $epsilon^T A epsilongt 0$ is always positive definite for every $epsilon$ the condition for local minimum becomes
$$nabla^2 f(x^*)gt 0$$