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)=limh→0f(x+h)−f(x)h=limh→0f(h)h=limh→0f(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(y−x)≥f(x).
Step III.
Since f is increasing, then the limit limx→0+f(x) exists. However
limx→0+f(x)=limn→∞f(1n)=limn→∞1nf(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(qn−x)⟶f(x),
since qn−x→0+, and thus limn→∞f(qn−x)=0.
Therefore, f(x)=xf(1), for all x∈[0,∞), and hence f′(x)=f(1).
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