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If $\displaystyle \bigg|f(a+b)-f(b)\bigg|\leq \frac{a}{b}\; \forall\; a,b\in \mathbb{Q},b\neq 0.$

Then show that $\displaystyle \sum^{n}_{k=1}\bigg|f(2^n)-f(2^k)\bigg|\leq \frac{n(n-1)}{2}$

Try: put $\displaystyle a=h>0$ Then $\displaystyle \bigg|f(b+h)-f(b)\bigg|\leq \frac{h}{b}$

So $\displaystyle \lim_{h\rightarrow 0}\bigg|\frac{f(b+h)-f(b)}{h}\bigg|\leq \lim_{h\rightarrow 0}\frac{1}{b}\Rightarrow |f'(b)|\leq \frac{1}{b}$

Could some help me how to solve it, please help me. Thanks

Community
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DXT
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2 Answers2

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You can show it by induction on $n$. The case $n=1$ is trivial. Suppose the result is true for an $n\geq 1$, we want to show that it holds for $n+1$. We have $$ \sum_{k=1}^{n+1}|f(2^{n+1})-f(2^k)|=\sum_{k=1}^{n}|f(2^{n+1})-f(2^k)|\leq\sum_{k=1}^{n}\left[|f(2^{n+1})-f(2^n)|+|f(2^{n})-f(2^k)|\right]\leq n|f(2^{n+1})-f(2^n)|+\frac{n(n-1)}{2} $$ where in the last step we used the induction hypothesis. Now estimate $|f(2^{n+1})-f(2^n)|$ using the condition for suitable $a,b$.

Redundant Aunt
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Recall the Absolute Value identity: $\,\left|a+b\right|\le\left|a\right|+\left|b\right|\,$:
$$ \begin{align} \left|f(a+b)-f(b)\right|&\le\frac{a}{b}=\frac{a+b}{b}-1 \\[2mm] \left|f\left(2^{r}\right)-f\left(2^{r-1}\right)\right|&\le\frac{2^{r}}{2^{r-1}}-1=2-1=\color{red}{1} \\[2mm] \sum_{k=1}^{n}\left|f\left(2^{n}\right)-f\left(2^{k}\right)\right|&=\sum_{k=1}^{n}\bigg|f\left(2^{n}\right)\color{blue}{-f\left(2^{n-1}\right)+f\left(2^{n-1}\right)-\dots} \\ &\qquad\color{blue}{\dots-f\left(2^{k+1}\right)+f\left(2^{k+1}\right)}-f\left(2^{k}\right)\bigg| \\[2mm] &\le\sum_{k=1}^{n}\,\sum_{r=k+1}^{n}\,\left|f\left(2^{r}\right)-f\left(2^{r-1}\right)\right| \\[2mm] &\le\sum_{k=1}^{n}\,\sum_{r=k+1}^{n}\color{red}{1}=\sum_{k=1}^{n}(n-k)=\sum_{k=0}^{n-1}k=\frac{n(n-1)}{2} \end{align} $$

Hazem Orabi
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