Let f:[0,∞)→[0,∞) be a function such that f(x+y)=f(x)+f(y), for all x,y≥0. 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).
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