Vtyjkyuk

If $displaystyle bigg|f(a+b)-f(b)bigg|leq fracab; forall; a,bin mathbbQ,bneq 0.$

Then show that $displaystyle sum^n_k=1bigg|f(2^n)-f(2^k)bigg|leq fracn(n-1)2$

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

So $displaystyle lim_hrightarrow 0bigg|fracf(b+h)-f(b)hbigg|leq lim_hrightarrow 0frac1bRightarrow |f'(b)|leq frac1b$

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

Recall the Absolute Value identity: $,left|a+bright|leleft|aright|+left|bright|,$:

$$
beginalign
left|f(a+b)-f(b)right|&lefracab=fraca+bb-1 \[2mm]
left|fleft(2^rright)-fleft(2^r-1right)right|&lefrac2^r2^r-1-1=2-1=colorred1 \[2mm]
sum_k=1^nleft|fleft(2^nright)-fleft(2^kright)right|&=sum_k=1^nbigg|fleft(2^nright)colorblue-fleft(2^n-1right)+fleft(2^n-1right)-dots \
&qquadcolorbluedots-fleft(2^k+1right)+fleft(2^k+1right)-fleft(2^kright)bigg| \[2mm]
&lesum_k=1^n,sum_r=k+1^n,left|fleft(2^rright)-fleft(2^r-1right)right| \[2mm]
&lesum_k=1^n,sum_r=k+1^ncolorred1=sum_k=1^n(n-k)=sum_k=0^n-1k=fracn(n-1)2
endalign
$$

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