buoyant kiln
f is a continuous function in [0,1] in which if x is in [0,1] f(x/2) + f((x+1)/2) = 2f(x) , i have already managed to prove that f is bounded and f(0) = m the maximum of the function, but i'm totally stuck for the second question asking us to show that f is constant, my idea was to show the fact that f(0) = m => f(x) = m for any x we chose in [0,1] by combining "/2" "(...+1)/2" operations but i don't know how to do write it properly
clever trenchBOT
sinful mural
f is a continuous function in [0,1] in which if x is in [0,1] f(x/2) + f((x+1)/2...
does it clarify that f is only continuous in [0,1]?
also no info on domain?
buoyant kiln
yes the function is defined and continuous in [0,1] and take values in R
buoyant kiln
so that means we would have f(1/2) = f(1) ?
sinful mural
so that means we would have f(1/2) = f(1) ?
for any x in [0,1] f(1/2) will be equal to that f(x)
buoyant kiln
why is it true for any x in [0,1] and not just 1 ?
sinful mural
why is it true for any x in [0,1] and not just 1 ?
you proved why it is bounded right?
buoyant kiln
yes
sinful mural
and you got the max on f(0)
buoyant kiln
we have the max m in [0,1] and f(0) = m
buoyant kiln
yes we have f(1/2) = m f(1) = m but how to show that it equals m for any x
no but it is like a density thing ? because we learned density
sinful mural
they are dense in [0,1]
buoyant kiln
ooooooooooo
so if we show that every for every diadic rationnal n we have f(n) = m then we prove that f is constant
wait tysm i think i have found something
sinful mural
good luck then