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Let $f= \begin{pmatrix} {f^1} \\{...} \\{f^m} \end{pmatrix} $ where $f^i : D \to \mathbb R$. Prove that $f$ is differentiable if and only if $f^i$ is differentiable for $i=1,...,m$.

I feel like this is straightforward, but I'm a little stumped with this.

I assume it would involve manipulating the formula

$\lim_{h\to 0}\frac{\|f(x + h) - f(x) - f'(x)h)\|}{\|h\|} = 0$

Could I make it

$\lim_{h\to 0}\frac{\|f^i(x + h) - f^i(x) - f^{i} {'}(x)h)\|}{\|h\|} = 0$ ?

Any help/suggestions would be appreciated!

supinf
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1 Answers1

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Note that for any vector $x\in\mathbb{R}^m$ we have $|x_i|\leq\|x\|\leq\sqrt{m}\max_{1\leq i\leq m}|x_i|$ for all $i=1,\dots, m$. You can use these simple inequalities to relate the two limits you have above. In particular, if $f$ is differentiable at some point $x$ then the first inequality implies that $f^i$ is also differentiable at $x$. To see this note that for any $i=1,\dots,m$ we have $$\lim_{h\to 0}\frac{|f^i(x+h)-f^i(x)-Df^i(x)h|}{\|h\|}\leq \lim_{h\to 0}\frac{\|f(x+h)-f(x)-Df(x)h\|}{\|h\|}=0.$$

Similarly, if all of the component functions are differentiable at some point $x$, then the second inequality implies that $f$ is differentiable at $x$.

Note that I wrote $Df$ instead of $f'$ since $f'$ is usually used to denote the derivative of a 1-dimensional function, instead of some function with domain $\mathbb{R}^m$.

Yousuf Soliman
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