Let $ f: mathbb {R} to mathbb{R} $ be a right-continuous and such that for all $ p in

mathbb {N}^*, $ the sequence $ (f (frac {k} {2^p}))_{k inmathbb{N}} $ converges in $ mathbb {R}. $ Can we deduce that $ f $ admits a limit in $ + infty.$ I think no, it’s sufficient to take $displaystyle f(x)= sum _{n=0}^infty gBig( 4^n Big( x-Big(frac32Big)^n Big) Big)$ where g is defined by

$ g(x)= 1-x$ if $x in (0,1), g(x)= 1+x$ if $ x in (-1,0($ and $g(x)=0$ if $|x|>1 $. notice that $ f $ does not admit a limit in $ + infty.$ and for all $ p in

mathbb {N}^*, $ the sequence $ (f (frac {k} {2^p}))_{k inmathbb{N}} $ converges to 0

Finally, it seems to me that this proof is false