Show that $f$ is $C^{1}$, $Df(0) = I$ implies that $\lVert f(x) - f(y) \rVert \geq \frac{1}{2}\lVert x-y\rVert$

Show that given $f:\mathbb{R}^{n} \rightarrow \mathbb{R}^{n}$ is $C^{1}$, $Df(0) = I$, then $\lVert f(x) - f(y) \rVert \geq \frac{1}{2}\lVert x-y\rVert$ for $x,y$ in a sufficiently small $\delta$ neighborhood about $0$.

How can I show this explicitly?

My intuition:

As someone put it, the moral of all differential calculus is that where the derivative is defined for a function, the function behaves as the linearization given by the derivative. This particular case is no different, I believe; given that all partials $\frac{\delta f_{i}}{\delta x_{j}}$ are $1$, locally at $0$, $a = [x,y]^{T}$ maps (approximately) to $b = [x+y, x+y]^{T}$. Thus, the distance between $a$ and $b$ is at least $$(y^{2} + x^{2})^{1/2} = \lVert a\rVert$$. Working with another set of points and a similar equation as above should give something meaningful.

Note \begin{eqnarray} f(y)&=&f(x)+Df(x)(x-y)+R_2(\|y-x\|)\\ &=&f(x)+Df(0)(x-y)+R_1(\|x\|)(x-y)+R_2(\|y-x\|). \end{eqnarray} where $$\lim_{x\to 0}R_1(\|x\|)=0,\lim_{y\to x}\frac{R_2(\|y-x\|)}{\|y-x\|}=0.$$ So there is small $\delta>0$ such that, for $x,y\in B_\delta(0)$, $$\|R_1(\|x\|)\|<\frac14, \|R_1(\|x\|)\|<\frac14\|y-x\|.$$ Thus for $x,y\in B_\delta(0)$, $$\|f(y)-f(x)\|\ge\|y-x\|-\|R_1(\|x\|)\|\|y-x\|-\|R_2(\|y-x\|)\ge\frac12\|y-x\|.$$
• Sorry for the delay in responding. Implicit in your proof is the statement that $$D(f(x) - Df(0) = R_{1}(||x||)$$. Why is this true? I would infer from $f$ being $C^{1}$ that $$||D(f(x) - Df(0)|| = 0$$ as $x \rightarrow 0$ is only true, as opposed to the former one. The former one is much stronger. – Muno Mar 4 '17 at 2:31
• Since $f$ is $C^1$, $Df(x)$ is continuous and hence $Df(x)-Df(0)$ is $o(\|x\|)$. – xpaul Mar 4 '17 at 14:41
• Actually, it's not clear to me why the proof's last line is true: $$\|f(y)-f(x)\|\ge\|y-x\|-\|R_1(\|x\|)\|\|y-x\|-\|R_2(\|y-x\|)\ge\frac12\|y-x\|$$ In particular, I don't see why the inequality is in the direction that it is. – Muno Mar 6 '17 at 23:44
Note \begin{eqnarray} f(y)&=&f(x)+Df(x)(x-y)+R(\|y-x\|)\\ &=&f(x)+Df(0)(x-y)+O(\|x\|)(x-y)+R_2(\|y-x\|). \end{eqnarray} where $$\lim_{x\to 0}O(\|x\|)=0,\quad lim_{y\to x}\frac{R_2(\|y-x\|)}{\|y-x\|}=0.$$ So there is small $\delta>0$ such that, for $x,y\in B_\delta(0)$, $$\|R_1(\|x\|)\|<\frac14, \|R_1(\|x-y\|)\|<\frac14\|y-x\|.$$ Thus for $x,y\in B_\delta(0)$, $$\|f(y)-f(x)\|\ge\|y-x\|-\|R_1(\|x\|)\|\|y-x\|-\|R_2(\|y-x\|)\ge\frac12\|y-x\|.$$