Suppose I have a function $\Lambda(t)$ for any $t>0$. This function has the following three properties:
It is stated that the function has the form $\Lambda(t) = \lambda t$, but how can I formally derive this from the above three properties. Thanks in advance.
Hint: Try proving these properties first, by giving $S$ and $T$ various values: $$\Lambda(0)=0$$ $$\Lambda(2t) = 2\Lambda(t)$$ $$\Lambda(nt) = n\Lambda(t)$$ $$\Lambda(m/n)=m/n\Lambda(1)$$
This should give linearity on $\mathbf{Q}_+$.
To expand linearity to $\mathbf{R}_+$, based on monotonicity only, see proofwiki (using density of the set of rationals in the set of real numbers and Peak Point lemma and Squeeze theorem).
See also Cauchy functional equation for more.
$\frac{d}{dt}\Lambda(t) > 0 \rightarrow \Lambda(t) = C t + u$
$\Lambda(S+T) = \Lambda(S) + \Lambda(T) \rightarrow u=0$
Hence
$\Lambda(t) = Ct$
Let $f (x)$ be such a function.From the second property we get, $$f(x+y)=f(x)+f(y)$$ Step 1 Differentiate partially w.r.t. $x$, $$f'(x+y)=f'(x)$$ Step 2 Put $x=0$, $$f'(y)=f'(0)=constant$$ Step 3 This is an Ordinary Differential Equation which gives, $$f(y)=f'(0)y+C$$ Since $C=0$ (as $f(0)=0$), we get, $$f(y)=\lambda y$$ Where $\lambda>0$ (from the first property)
