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If you have a differentiable function $f:\mathbb{R}^d\rightarrow\mathbb{R}$ that is $\beta$-smooth (for all $v$ and all $w$, you have $\|\nabla f(v)-\nabla f(w)\| \leq \beta \|v-w\|$), how can you show the equation below?

$f(v) \leq f(w) + \langle\nabla f(w), v-w \rangle + \frac{\beta}{2}\|v-w\|^2$

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Let $\gamma_t = w + t(v-w)$. We have $$f(v) - f(w) = f(\gamma_1) - f(\gamma_0) = \int_0^1 (f \circ \gamma)’(t)dt \\ = \int_0^1 \langle \nabla f(\gamma_t), \gamma’(t)\rangle dt = \int_0^1 \langle \nabla f(\gamma_t), v-w\rangle dt \\ = \langle\nabla f(w), v-w\rangle + \int_0^1 \langle \nabla f(\gamma_t) - \nabla f(w), v-w\rangle dt.$$ By Cauchy-Schwarz and your inequality $$\langle \nabla f(\gamma_t) - \nabla f(w), v-w\rangle \leq \lVert \nabla f(\gamma_t) - \nabla f(w) \rVert \lVert v-w\rVert \\ \leq \beta \lVert \gamma_t - w\rVert \lVert v-w\rVert = \beta \, t \, \lVert v-w\rVert^2 $$ so that $$ \int_0^1 \langle \nabla f(\gamma_t) - \nabla f(w), v-w\rangle dt \leq \beta \, \lVert v-w\rVert^2 \int_0^1 t dt = \frac \beta 2 \lVert v-w\rVert^2 $$ which proves your inequality.

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  • $\begingroup$ Thanks, this really is helpful! $\endgroup$ – learning machine Aug 28 '19 at 8:20
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Have you tried using the mean value theorem for integrals? You can relate $f(v)-f(w)$ to $\nabla f$ and use some simple estimates to get something like the above.

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  • $\begingroup$ Thanks for the advice! $\endgroup$ – learning machine Aug 28 '19 at 8:20
  • $\begingroup$ It's essentially is the same as the other answer, but in general when relating $f$ and $\nabla f$ it's best to always look for an integral relation. $\endgroup$ – bbbbbb Aug 28 '19 at 15:32

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