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Let $f:\mathbb{R}^n \rightarrow \mathbb{R}$ be a $C^2 $ function and $x^*$ be a point such that $\bigtriangledown^2f(x^*)$ is positive definite.Is it always true that,there exists a neighborhood around $x^*$ such that for all points $x$ in that neighborhood ,we have $\bigtriangledown ^2f(x)$ is positive definite?

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Yes: since $\nabla^2f(x^*)>0$, all the determinants of the up-left square submatrices are positive. By continuity of the partial derivatives of order $2$, the map $g_k\colon x\mapsto \det (\nabla^2f(x))_k$, where $A_k$ means the submatrix of $A=(a_{i,j})_{1\leq i,j\leq n}$, namely $A_k=(a_{i,j})_{1\leq i,j\leq k}$, is continuous.

Therefore, we can find a neighborhood $V$ of $x^*$ such that $g_k(x)>0$ for all $x$ in $V$ and each $k\in \{1,\dots,n\}$. We conclude by Sylvester's criterion.

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Alternatively, you can argue that since $h: \mathbb{R}^n \to \mathbb{R}, h(x) := \lambda_{min}(\nabla^2(f(x)))$ is continuous (see, e.g., Continuity of eigenvalues and spectral radius for a general matrix), and since $h(x^*) > 0$, you can always find a neighborhood $\mathcal{N}(x^*)$ around $x^*$ such that $h(x)>0$ for all $x\in \mathcal{N}(x^*)$.