Is Function $f$ Convex if its Hessian Matrix is Positive Semidefinite?

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Here is this week's POTW:

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Let $f : \Bbb R^n \to \Bbb R$ be twice differentiable function whose Hessian matrix is everywhere positive semidefinite. Show that $f$ is convex.
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No one answered this week’s problem. You can read my solution below.

Spoiler

For fixed $\mathbf{x}, \mathbf{y}\in \Bbb R^n$, define $g(t;\mathbf{x}, \mathbf{y}) = f(\mathbf{x}+ t\mathbf{y})$, for all $t\in\Bbb R$. If $H_{\mathbf{z}}(f)$ denotes the Hessian matrix evaluated at $\mathbf{z}$, then $g’’(t;\mathbf{x}, \mathbf{y}) = \frac{1}{2}\mathbf{y}^TH_{\mathbf{x} + t\mathbf{y}}(f)\mathbf{y}$, which is nonnegative by the positive semidefiniteness of the Hessian matrix. Therefore $g(t;\mathbf{x},\mathbf{y})$ is convex. Since $\mathbf{x}$ and $\mathbf{y}$ are arbitrary, it follows that $f$ is convex. Indeed, for all $\mathbf{x},\mathbf{y}\in \Bbb R^n$ and $t\in [0,1]$, $$f((1-t)\mathbf{x} + t\mathbf{y}) = f(\mathbf{x} + t(\mathbf{y}-\mathbf{x})) = g(t; \mathbf{x},
\mathbf{y}-\mathbf{x}) \le (1-t) g(0;\mathbf{x}, \mathbf{y}-\mathbf{x}) + t g(1; \mathbf{y} - \mathbf{x}) = (1-t) f(\mathbf{x}) + t f(\mathbf{y})$$ as desired.