Scaling in utility maximisation If I have the wealth process $$dw_t=rw_tdt+n_tS_t(\sigma dB_t+(\mu-r)dt)-c_tdt,$$ where $n$ is number of $S_t$ and $B_t$ is Brownian motion. If we define the admissible set $A$ as follows: $(n_t,c_t)\in A(w)$ if $w_t\geq 0$ for all times.
My first question is why is $(n,c)\in A(w) \iff(\lambda n,\lambda c)\in A(\lambda w)$? My attempt is that if $n,c$ result in nonnegative weath then clearly the multiples of  $n,c$ will result in nonnegative weath. But how can I be sure. The reason should be the linearity of $w_t$. 
My second question is why is $$\sup_{(n,c)\in A(\lambda w)}E\left(\int_0^\infty u(c_t)dt\right)=\sup_{(n,c)\in A(w)}E\left(\int_0^\infty u(\lambda c_t)dt\right).$$ I am thinking that the statement is saying if one starts with wealth $\lambda w$ and consume at rate $c_t$ then their maximum utility is the same as on who starts with $w$ and consumes at rate $\lambda c_t$. This doesn't make sense. 
 A: (I assume $\lambda \geq 0$ was a requirement)
The first part of your request is to show that 
$$(n,c)\in A(w) \iff(\lambda n,\lambda c)\in A(\lambda w)$$
given 
$$dw_t=rw_tdt+n_tS_t(\sigma dB_t+(\mu-r)dt)-c_tdt$$
To do this we can multiply through by $\lambda$ to get
$$d(\lambda w_t)=r(\lambda w_t)dt+(\lambda n_t)S_t(\sigma dB_t+(\mu-r)dt)-(\lambda c_t)dt$$
If $(n,c)\in A(w)$ we know that $w \geq 0$ and the triple $n,c,w$ solves the sde.  From that last equation we also see that $(\lambda n, \lambda c) \in A(\lambda w)$ since $\lambda w$ solves the sde with $(\lambda n, \lambda c)$ and $\lambda w_t \geq 0$ if $\lambda \geq 0$.
For the second part we want to show:
$$\sup_{(n,c)\in A(\lambda w)}E\left(\int_0^\infty u(c_t)dt\right)=\sup_{(n,c)\in A(w)}E\left(\int_0^\infty u(\lambda c_t)dt\right)$$
From the first result it is true that $(n, c) \in A(\lambda w) \iff (n/\lambda, c/\lambda) \in A(w)$.  If we let $\tilde{n} = n/\lambda$ and $\tilde{c} = c/\lambda$ we can substitute and write the LHS above as
$$\sup_{(\tilde{n},\tilde{c})\in A(w)}E\left(\int_0^\infty u(\lambda \tilde{c_t})dt\right)$$
This is now the RHS where we have just renamed the $\sim$ variables with their original names (but there is now no relationship between them).  Confusing but valid.
