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)=lim



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)=p\,f(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(y-x)\ge f(x).




Step III.
Since f is increasing, then the limit \,\lim_{x\to 0^+}f(x)\, exists. However
\lim_{x\to 0^+}f(x)=\lim_{n\to\infty}f\Big(\frac{1}{n}\Big) =\lim_{n\to\infty}\frac{1}{n}\,f(1)=0.



Step IV. Pick an arbitrary x\in(0,\infty), and a decreasing sequence
\{q_n\}\subset\mathbb Q tending to x. Then

f(q_n)=q_n\,f(1)
and
x\,f(1)\longleftarrow q_n\,f(1)=f(q_n)=f(x)+f(q_n-x)\longrightarrow f(x),
since \,\,q_n-x\to 0^+, and thus \,\,\lim_{n\to\infty}f(q_n-x)=0.



Therefore, \,f(x)=x\,f(1),\, for all x\in\mathbb [0,\infty), and hence \,f'(x)=f(1).



No comments:

Post a Comment

real analysis - How to find lim_{hrightarrow 0}frac{sin(ha)}{h}

How to find \lim_{h\rightarrow 0}\frac{\sin(ha)}{h} without lhopital rule? I know when I use lhopital I easy get $$ \lim_{h\rightarrow 0}...