Let matrix $A \in M_n(\mathbb{R})$ be positive definite. And let $f : \mathbb{R}^n \to \mathbb{R}$ be
$$f(x) := \frac12 \langle A x, x \rangle - \langle b, x \rangle$$
where $b \in \mathbb{R}^n$. So absolutely $H f(x)=A$. Why is then $f$ elliptic?
I think $f$ should only be strictly convex , but as it's written in my course elliptic! Need some help please and thanks.
Definition: $f$ being elliptic means that for all $x \in \mathbb{R^n}$, there is $a > 0$ such that $$\langle Hf(x) h, h \rangle \ge a \|h\|^2$$ where $h \in \mathbb{R^n}$.
My proof
If $A$ positive definite, then $\langle Au, u \rangle > 0$, for all $u \ne 0$. Let $u =\frac y{\|y\|}$, then:
$$\left\langle A\frac y{\|y\|},\frac y{\|y\|} \right\rangle > 0 \Rightarrow \langle Ay, y \rangle > a \|y\|^2$$
where $a > 0$. But $Hf(x)=A$, so $f$ is elliptic. Is that true?
Yes your proof seems okay. However, the first expression should be $\langle A\frac{y}{\|y\|}, \frac{y}{\|y\|} \rangle$. I see you have used two $A$. Also, $a$ is the smallest eigenvalue.