Can I interchange $\max$ and $\sup$? To be honest I do not know if this is true.
I am working with the following problem. I would like to prove that
$$\max \left\{\,\sup_{t>0}|f(t)|,\; t\sup_{t>0}|g(t)|\,\right\} \,=\, \sup_{t>0}\max\{|f(t)|,t|g(t)|\}.$$
My attempt:
For all $t>0$, we have that $|f(t)|\leq \sup_{t>0}|f(t)|$ and $t|g(t)|\leq t \sup_{t>0}|g(t)|$.
Then $|f(t)|\leq \max\{\sup_{t>0}|f(t)|, t \sup_{t>0}|g(t)|\}$, and similarly,
$$t|g(t)|\leq \max\{\sup_{t>0}|f(t)|, t \sup_{t>0}|g(t)|\}. $$
Therefore,
$$\max\{|f(t)|,t|g(t)|\}\leq \max\{\sup_{t>0}|f(t)|, t \sup_{t>0}|g(t)|\}$$ implies that
$$\sup_{t>0}\max\{|f(t)|,t|g(t)|\}\leq \max\{\sup_{t>0}|f(t)|, t \sup_{t>0}|g(t)|\}.$$
Am i right to here?
For the other inequality i can not yet. Some ideas are to use properties from $\sup$ as for all $\epsilon>0$ exists $t_*>0$: $\sup|f(t)|\leq |f(t_*)|+\epsilon$. A similar argument for $g$ perhaps help me.
$$\sup|g(t)|\leq |g(t_*)|+\epsilon, \\ t\sup|g(t)|\leq t|g(t_*)|+\epsilon t.$$
Perhaps somebody can help me please or give me hints or tell me that is no true my statement. Thank you.
I have a feeling that I'm misunderstanding your question. But if not, then the equality doesn't hold.
E.g., if $f(t)=1$, and $g(t)=0$ if $t\neq1$ and $g(1)=2$, then $\max(f(t),tg(t))=1$ whenever $t\neq1$ and is $2$ otherwise.
So, $\sup_{t>0}\max(f(t),tg(t))=2$. On the other hand, $t\sup_{t>0}g(t)=2t$, which is unbounded.