# An estimate like $\left\vert\frac{f(x+y) - f(x)}{|y|}\right\vert \le K(\Vert f \Vert_\infty +\Vert f' \Vert_\infty)$

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?

• Is there some integration in a part you do not show or is there another reason for the $dx$ in the first formula? – LutzL Mar 15 '17 at 19:39
• @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. – 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 $K=1$.
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 $$\|f(x+y)-f(x)\|=\int_0^1\|f'(x+sy)\|\,\|y\|\,ds\le \|f'\|_\infty\,\|y\|$$
• 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? – user378822 Mar 15 '17 at 18:32
• 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 $x,y,z \in \mathbb{R}$. – user378822 Mar 15 '17 at 21:36
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