In one of the Gateaux derivative examples mentioned here on page 5, it is mentioned that the function $f(x) = \lvert x\rvert $ has the Gateaux derivative defined at $x=0.$ In particular, it says that $$ \lim_{\epsilon\to 0}\frac {\left\lvert {\epsilon h} \right\rvert} {h} = \left\lvert h \right\rvert $$

I feel that the above limit is not unique. It can be either $ +h \text{ or} -h $. The Gateaux derivative is not defined at $x = 0$

  • 2
    $\begingroup$ I would not trust your cited source. $\endgroup$ Dec 29, 2017 at 16:39

1 Answer 1


By definition $$ d_hf(0)=\lim_{\epsilon\to0}\frac{f(\epsilon h)-f(0)}{\epsilon}=\lim_{\epsilon\to0}\frac{|\epsilon h|}{\epsilon}=\lim_{\epsilon\to0}\frac{|\epsilon| }{\epsilon}|h|. $$ You wrote $h$ in the denominater instead of $\epsilon$. And here you see that the limit doesn't exists because $$ \lim_{\epsilon\to0^+}\frac{|\epsilon| }{\epsilon}|h|=\lim_{\epsilon\to0^+}|h|=|h|\neq-|h|=\lim_{\epsilon\to0^-}-|h|=\lim_{\epsilon\to0^-}\frac{|\epsilon| }{\epsilon}|h|. $$ So there is a mistake in your reference.


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