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Let $f \in C^{1,1}(\mathbb{R}^d)$, i.e., $f \: \colon \mathbb{R}^d \rightarrow \mathbb{R}$ is a continuously differentiable function and its derivative $\nabla f$ is uniformly Lipschitz continuous on $\mathbb{R}^d$. I would like to prove the following claim:

There exists a constant $M > 0$ such that for all $x, y \in \mathbb{R}^d$ the following holds: \begin{equation} \lvert f(x+y) - f(x) - \langle \nabla f(x), y \rangle \rvert \leq M \lvert y \rvert^{2}. \end{equation}

So far, I have no idea how to tackle that problem, and I would be grateful for any suggestions.

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  • $\begingroup$ Is this inequality really supposed to hold everywhere in $\mathbb R^d$? I think I could come up with a proof for an a.e. version respectively with a proof, if $C^{1,1}$ is improved towards $C^2$, otherwise I don't see how one can tackle this. If you are interested, I can write this down, otherwise I'll be eager to see an answer by somebody else. $\endgroup$ – frog Nov 13 '15 at 9:39
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I don't know why one would need $C^2$ for this (as suggested in a comment). Unless I'm missing something, note that $$f(x+y)-f(x)=\int_0^1 \frac{d}{dt} f(x+ty)\,dt $$ so that $$f(x+y)-f(x)-\langle \nabla f(x), y\rangle = \int_0^1\langle \nabla f(x+ty) -\nabla f(x), y\rangle \,dt$$ So the norm of the lhs is bounded by $$\int_0^1 ||\nabla f(x+ty)- \nabla f(x)||\,||y||\,dt\le M \int_0^1 t||y||^2\,dt\le \frac{1}{2}M||y||^2$$ if $M$ is the (global) Lipschitz bound for $\nabla f$.

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  • $\begingroup$ I totally missed this integral representation of the difference. It's a cool answer. $\endgroup$ – frog Nov 13 '15 at 10:09

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