# Can I assume utility functions strictly increasing?

I am required to show that if: $$f:R^L \rightarrow R$$ is a strictly monotonic function and $$u:R^L \rightarrow R$$ is a utility function representing a preference relation $$\succsim$$, then the function $$v:R^L \rightarrow R$$ defined by $$v(x) = u(f(x))$$ does not always represent the preference $$\succsim$$.

For such a proposition, can I assume that if $$x>y$$, then $$u(x) > u(y)$$. The context is that I am trying to prove this by contradiction and my contradiction hinges on $$x >y \implies u(x) > u(y)$$, where x,y are elements of $$R^L$$.

• If the domain of $u$ is $R^L$, and the codomain of $f$ is $R$, then $u(f(x))$ doesn't make sense. Also, if $x$ and $y$ are in $R^L$, then $x>y$ doesn't make sense unless you've defined an ordering on $R^L$. Oct 17 at 4:55

• Yes, and you can even find one explicit utility function and one explicit increasing function $f$ that gives you a counterexample. Oct 16 at 23:55