Why does $sup {x_ky_k}lesup x_kcdot sup y_k$? Code Answer

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Let $x_n$ and $y_n$ both be bounded sequences. In the answers proving $limsup (x_n cdot y_n)lelimsup x_ncdot limsup y_n$, it is said that, in particular,$forall n in Bbb{N}$ $sup_{kge n} {x_ky_k}lesup_{kge n} x_kcdot sup_{kge n} y_k$. Why is that? Is this a factclaimcorollary? Something trivial or just something one remembers?


I think you probably mean this in the context of a positive sequence, since $x_{2k}=-1, x_{2k+1}=-2$ and $y_{2k}=-2, y_{2k+1}=-1$ gives a counterexample otherwise to the fact you quote that $2=sup_{kge n}x_ky_k> sup_{kge n}x_kcdotsup_{kge n}y_k=1$.

However, in the positive case we notice that we have an inclusion:

$$A={x_ky_k}_ksubseteq {x_my_n}_{m,n}=B$$

Now you can prove that $sup B=(sup x_m)(sup y_n)$ directly since clearly we just choose subsequences $n_j$ and $m_ell$ so that

$$begin{cases}x_{n_j}to sup x_n\y_{m_ell}to sup y_mend{cases}.$$

Then noting $x_ny_mle (sup x_n)y_mle sup x_ncdot sup y_m$–which is where we use positivity–you get the result on $B$.

Since $Asubseteq B$ we have that $sup Ale sup B$ by monotonicity of the sup function.

Note that even the case of $limsup$ requires the positivity, since my example from the start has that the limsup and sup are equal on $x_k$ and $y_k$.

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