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I got this question from my analysis book:

Let $f:U\to\mathbb{R}^n$ be of class $\mathcal{C}^1$ in an open convex $U\subseteq\mathbb{R}^m$, with $0\in U$ and $f(0)=0$. (a) If $|f'(x)|\le|x|$ for all $x\in U$ then $|f(x)|\le\frac{1}{2}|x|^2$ for all $x\in U$. (b) Conclude that if $f(0)=f'(0)=0$ with $f\in\mathcal{C}^2$ then $\left|\frac{\partial^2 f }{\partial u\partial v}\right|\le|u||v|$ implies $|f(x)|\le\frac{1}{2}|x|^2$.

I already done the (a) part and there should be only a few steps to conclude the (b) part but I'm struggling with it.

I know that $\frac{\partial^2 f }{\partial u\partial v}(x)=(f''(x))(u,v)$ if we see $f''(x)$ as a bilinear function $f''(x):\mathbb{R}^m\times\mathbb{R}^m\to\mathbb{R}^n$ but from here on I don't know what to do.

Can someone please shed some light on this?

For the (a) part see this question.

I also wonder how my question relates to this because it would imply that $|f(x)|\le\frac{1}{2}|x|^2$ for any $f$ satisfying the conditions in (b). I think that the author of that question misinterpreted the question statement given his comments on the answers, and the answers also not proving exactly what the author stated (and also he got the question from the same book that I'm reading).

AnalyticHarmony
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    Have you considered integrating $\frac{\partial^2 f}{\partial u\partial v}$ along two line segments parallel to the coordinate axis? Also, when you write $|f'(x)|$, what norm are you considering? This would only change thinks up to a multiplicative constant, but I don't know how important this is for your purposes. – Reveillark May 26 '20 at 22:12
  • I did not try integration since I have not yet studied integration in $R^n$, but if it involves only integration in one variable I can try. I'm considering the induced norm – AnalyticHarmony May 26 '20 at 22:18
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    It would actually be a pair of one dimensional integrals, since you’re integrating along lines. – Reveillark May 26 '20 at 22:19
  • Can you please give more details on this idea? – AnalyticHarmony May 26 '20 at 22:26

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Taylor's theorem with (Lagrange's?) remainder is helpful here. This post contains a statement, and outline of proof along with references.

Note that you assumption on the second directional derivatives implies that for all $x \in U$, the operator norm of the bilinear map $f''(x)$ is $\leq 1$; i.e $\lVert f''(x)\rVert \leq 1$. So, for any $x \in U$, we have \begin{align} \lVert f(x) - f(0) - f'(0)(x)\rVert \leq \left(\sup_{\xi \in U}\lVert f''(\xi)\rVert \right)\dfrac{\lVert x\rVert^2}{2!} = \dfrac{\lVert x\rVert^2}{2}. \end{align} But since $f(0) =0$ and $ f'(0) =0$, the claim follows.

peek-a-boo
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  • I have seen a variant of this question where one is asked to prove a bound on $$ \int_B (f(x)-f(0)) dx$$ and the solution is by first establishing the bound in @peek-a-boo 's answer. The nice twist is that you do not need to assume $Df(0)=0$ because by symmetry the integral of the linear part vanishes! I wonder though, if there is any better bound possible, because I remember that the question (which I do not completely recall:() required stringent bounds, better than estimate along rays from the origin. – Behnam Esmayli Feb 24 '21 at 19:52