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The definition provided is that for a function $f:\mathbb{R}^n\to\mathbb{R}^m$, the derivative of $f$ at $\mathbf{x}_0\in\mathbb{R}^n$ is the matrix $L\in M_{m\times n}(\mathbb{R})$ satisfying the following property: \begin{equation*} \lim_{\mathbf{x}\to\mathbf{x}_0} \frac{\lVert{f(\mathbf{x}) - f\left(\mathbf{x}_0\right) - L \left(\mathbf{x}-\mathbf{x}_0\right)}\rVert_2}{\lVert{\mathbf{x}-\mathbf{x}_0}\rVert_2} = 0. \end{equation*}

I am then asked what happens if we (a) replace $\lVert\cdot\rVert_2$ with $\lVert\cdot\rVert_1$ in both the numerator and denominator, and (b) just in the numerator.

I am not entirely clear on what it is I should be seeing. I looked at this question, which (as far as I understood it) was stating that if the above limit exists (i.e., $f$ is differentiable with the Euclidean norm) then $f$ is also $L^1$ differentiable. But beyond that, I don't see what I should say 'happens'.

Any help would be appreciated.

Trisztan
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1 Answers1

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All norms in $\mathbb{R}^n$ are equivalent, that is, given any two norms $\|\cdot\|_a$ and $\|\cdot\|_b$ in $\mathbb{R}^n$, there exist positive numbers $\alpha,\beta$ (that may depend on $n$) such that for every $x\in\mathbb{R^n}$ $$\alpha\|x\|_b\leq \|x\|_a\leq \beta\|x\|_b$$ so if the limit is zero for one pair of norms (e.g. the $L_2$ norm both in the nominator and the denominator), then it is zero also with any other pair of norms.

uniquesolution
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  • So, for one, is that all that 'happens'? And second, I know that all norms on $\mathbb{R}^n$ are equivalent, but in question (b), we use two (distinct) norms at the same time (rather than the same norm throughout). Does that change anything? – Trisztan Feb 07 '23 at 00:45
  • So for one, nothing actually happens except the rate at which the limit becomes zero. And for question (b) it doesn't matter which norms you use, neither in the domain of $f$, nor in its range, provided both are finite dimensional. This is because the equivalence of the norms allows you to sandwich the expression between two multiples of the $L_2$ norm, for instance. – uniquesolution Feb 07 '23 at 00:46