Let $f \in C^1(\mathbb{R})$ bounded and with bounded derivative. Is it possible to have an estimate like $$\left\vert\frac{f(x+y) - f(x)}{|y|}\right\vert \le K\Vert f' \Vert_\infty$$ or $$\left\vert\frac{f(x+y) - f(x)}{|y|}\right\vert \le K(\Vert f \Vert_\infty +\Vert f' \Vert_\infty)$$ where $K$ is a constant?

  • $\begingroup$ Is there some integration in a part you do not show or is there another reason for the $dx$ in the first formula? $\endgroup$ – LutzL Mar 15 '17 at 19:39
  • $\begingroup$ @LutzL That was just a typo. I edited it out. I originally meant to write $f'$ as $\frac{df}{dx}$ and then changed my mind. $\endgroup$ – user378822 Mar 15 '17 at 21:32

By the mean value theorem, you get $$ \frac{f(x+y)-f(x)}y=f'(x+\theta y),\qquad\theta\in(0,1) $$ so that your first inequality holds with $K=1$.

See also Lipschitz continuity and its connection to bounds on the first derivative.

For the general case use the fundamental theorem and write $$ f(x+y)-f(x)=\int_0^1f'(x+sy)y\,ds $$ which implies for the vector norms $$ \|f(x+y)-f(x)\|=\int_0^1\|f'(x+sy)\|\,\|y\|\,ds\le \|f'\|_\infty\,\|y\| $$

  • $\begingroup$ Thank you. I edited my question, the formula should have been $\frac{f(x+y) - f(x)}{|y|}$. Does this change anything in your answer? $\endgroup$ – user378822 Mar 15 '17 at 18:32
  • $\begingroup$ No, since the absolute value commutes with division. The vector valued version, where you really need the norm in the denominator, is not that different, the mean value equation gets replaced by a mean value inequality. $\endgroup$ – LutzL Mar 15 '17 at 18:41
  • $\begingroup$ Thanks. Would you mind adding to your answer some details on the following similar cases too?That is: $\frac{f(x+y)-f(x)}{\Vert y \Vert}$, with $x,y \in \mathbb{R}^N$ and $\frac{f(z,x+y)-f(z,x)}{\vert y \vert}$, with $x,y,z \in \mathbb{R}$. $\endgroup$ – user378822 Mar 15 '17 at 21:36
  • $\begingroup$ @Hideki If you Google the mean value inequality, you will find that precisely answers your first question (and the second is simply a special case). $\endgroup$ – Jason Mar 15 '17 at 22:10

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