Let $x=(x_1,x_2,\ldots) \in l^4$, $x\ne 0$. For which of the following values of $p$ the series $\sum\limits_{i=1}^\infty x_iy_i$ converges for every $y=(y_1,y_2,\ldots) \in l^p$.
(A) $1$; (B) $2$; (C) $3$; (D) $4$
I have used holders inequality. I got $\frac{1}{p}=1-\frac{1}{4}$. But this option is not in the list. Please help me. Thank you in advance.
Hint. By Hölder's inequality $$\sum_{i} x_i y_i\leq \left(\sum_i |x_i|^4\right)^{1/4}\left(\sum_i |y_i|^q\right)^{1/q}\leq \left(\sum_i |x_i|^4\right)^{1/4}\left(\sum_i |y_i|^r\right)^{1/r}$$ with $1/4+1/q= 1$ and $1\leq r\leq q$.
P.S. Note that if $1 \leq r \leq q \lt \infty$ then $\|y\|_{q} \leq \|y\|_{r}$ (see Inequality between $\ell^p$-norms)
I think you made a mistake. The natural guess is $p=4/3.$
But you're right, that is not on the list. However, if $y\in l^1,$ then $y\in l^{4/3},$ and so Holder implies the sum in question will converge. Thus (A) is correct.
But why is $y\in l^{4/3}?$ Answer: If $\sum |y_n| < \infty,$ then $|y_n| \to 0,$ hence $|y_n|\le 1$ for large $n.$ For such $n,$ $|y_n|^{4/3} \le |y_n|.$ Thus $\sum |y_n|^{4/3} < \infty$ by the comparison test.
Other answers have given you fancier results to look at, but I always like to think about the $l^p$ spaces this way when possible.
