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I'm working on a problem related to the Kullback-Leibler divergence but I'm stuck on one part. $$f(u) - f(v) - \nabla f(v)^\intercal (u-v)$$ The function $f(v) = \sum_{i=1}^n v_i \log v_i$ and $u,v \in \mathbb{R}_{++}^n$. I have already proved that $f$ is differentiable and convex. Could someone explain how to prove that $f(u) \geq f(v) + \nabla f(v)^\intercal (u-v)$? Thank you very much.

Cindy Philip
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    This inequality is true for all differentiable convex functions, by the way. It's nothing specific to your function or to the Kullback-Liebler divergence. – littleO Mar 28 '21 at 04:10
  • If a function is convex, then its graph lies above any tangent plane to the graph. The inequality says exactly that. – Deane Mar 28 '21 at 05:28

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I suppose that your understanding of "convexity" means that $\nabla^2 f \succeq {\bf 0}$ in the sense of symmetric positive semi-definite matrices. By Taylor expansion, there exists some $\xi$ such that $$f(u) = f(v) + \nabla f(v)^\intercal (u-v) + (u-v)^\intercal\,\nabla^2 f(\xi)\,(u-v),$$ do you see how to achieve the advertised conclusion ?

Fei Cao
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