Saturday, 5 November 2016

calculus - If f:[0,infty)to[0,infty) and f(x+y)=f(x)+f(y) then prove that f(x)=ax




Let f:[0,)[0,) be a function such that f(x+y)=f(x)+f(y), for all x,y0. Prove that f(x)=ax, for some constant a.





My proof :



We have , f(0)=0. Then ,
f(x)=limh0f(x+h)f(x)h=limh0f(h)h=limh0f(h)f(0)h=f(0)=a(constant).



Then, f(x)=ax+b. As, f(0)=0 so b=0 and f(x)=ax.



Is my proof correct?


Answer




In your proof you assume that f is differentiable, which is not given.



Let me suggest how to obtain the formula of f:



Step I. Show that f(px)=pf(x), when p is a positive rational and x a non-negative real. (At first show this for p integer.) We obtain also that, f(0)=0.



Step II. Observe that f is increasing, since, for y>x, we have
f(y)=f(x)+f(yx)f(x).




Step III.
Since f is increasing, then the limit limx0+f(x) exists. However
limx0+f(x)=limnf(1n)=limn1nf(1)=0.



Step IV. Pick an arbitrary x(0,), and a decreasing sequence
{qn}Q tending to x. Then

f(qn)=qnf(1)


and
xf(1)qnf(1)=f(qn)=f(x)+f(qnx)f(x),

since qnx0+, and thus limnf(qnx)=0.



Therefore, f(x)=xf(1), for all x[0,), and hence f(x)=f(1).



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