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Let $f\colon\mathbb{R}^n\to\mathbb{R}$ be differentiable at $x^*$ and $f(x^*)=0$. Prove that if $n>1$, then $$\liminf_{x\to x^*}\frac{|f(x)|}{||x-x^*||}=0.$$ Is this true for $n=1$?

I know that because $f$ is differentiable in $x^*$, there is a linear transformation $L\colon \mathbb{R}^n\to\mathbb{R}$ such that $$0=\lim_{x\to x^*}\frac{|f(x)-f(x^*)-L(x-x^*)|}{||x-x^*||}=\lim_{x\to x^*}\frac{|f(x)-L(x-x^*)|}{||x-x^*||}.$$ How to proceed from there? And what is special about $n=1$?

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Why the difference between the cases $n=1$ and $n>1$?

For $n=1$, the kernel of a linear map can be reduced to the zero vector. This isn't the case for $n>1$.

The result is true for $n>1$

In that case $L$ is a linear form and its kernel is not reduced to the zero vector. Take $a \in \ker L \setminus \{0\}$. For $m \in \mathbb N$ and $x_m = \frac{a}{m} + x^*$, you have $L(x_m-x^*) = 0$ and $\lim\limits_{m \to \infty} \dfrac{|f(x_m)-L(x_m-x^*)|}{||x_m-x^*||}= \dfrac{|f(x_m)|}{||x_m-x^*||}=0$. As $\lim\limits_{m \to \infty} x_m = x^*$ you have $\lim \inf_{x\to x^*}\dfrac{|f(x)|}{||x-x*||}=0$.

The result is wrong for $n=1$

Just take $f(x)=x$ and $x^*=0$. You have $\dfrac{|f(x)|}{||x-x*||}=1$ for all $x \in \mathbb R \setminus \{0\}$.

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  • $\begingroup$ What is the role of $a$ in the proof? $\endgroup$ Aug 11, 2018 at 9:28
  • $\begingroup$ @GNUSupporter I updated a typo. $\endgroup$ Aug 11, 2018 at 9:30
  • $\begingroup$ $n$ is fixed as the dimension of $\mathbb{R}^n$ so you might want to choose another index for your sequence. ANd after you took the limes, there is still an $n$ apperaing, that is quite confusing. Where is the difference between a linear Transformation and a linear form? Which part of the proof doesn't work for $n=1$? Why can you interchange the limes and the absolut value? $\endgroup$
    – Algebra
    Aug 11, 2018 at 9:33
  • $\begingroup$ @mathstackuser I changed $n$ into $m$ for readability. $\endgroup$ Aug 11, 2018 at 9:35
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    $\begingroup$ A linear form is a linear map from a vector space ($\mathbb R^n$ in your case) to its field of scalars ( $\mathbb R$ in your case). For $n=1$ a linear form may have a kernel reduced to the zero vector. This is not possible for $n>1$. $\endgroup$ Aug 11, 2018 at 9:39

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