I can prove that if $Q$ is a refinement of $P$, ie $P\subseteq Q$ then $L(P,f)≤ L(Q,f)$ and $U(Q,f)≤ U(P,f)$.
An interesting thing to note here is that $P \subseteq Q \implies ||Q||<||P||$ but the converse is not true.
However, I'm wondering whether the following implication holds:
If $||P_1|| < ||P_2||$ then $L(P_2,f)≤L(P_1,f)$ and $U(P_1,f)≤U(P_2,f)$.
Intuitively, I think it should hold as $P_1$ is in some sense finer than $P_2$ but I'm not sure how to prove it rigorously because the definition of norm as maximum of the lengths of the subintervals doesn't seem easy to use. Any help?
The implication is false. The trick is that if your function is constant on some interval then any partition of the interval where the function is constant will give the same Riemann sum and we can therefore "trick" the norm. So for instance let $$f(x)=\begin{cases}x,&0 \leq x\leq 1 \\ 1,& 1\leq x\leq 2.\end{cases}$$ Take $P_{2} = \{0, 1/4, 1/2, 3/4, 1, 2\}$ and $P_{1}=\{0, 1/2, 1, 3/2, 2\}$ then $\|P_{1}\|=1/2<1=\|P_{2}\|$ but $L(P_{1}, f)< L(P_{2}, f)$ since $P_{2}$ is a refinement of $P_{1}$ on $[0,1]$.