In case p≥1, it is a classical result on convolution (discrete Young inequality)
Dear Gerard
I suppose there is a typo in your question, namely that you meant to define w_j = \sum_{m\ge 0} u_m v_{j+m} (whereby "v" has been inserted), so as to have something like a convolution of u and v.
It seems to me that the answer is affirmative: w does belong to l_p. But this is only so because you are dealing with the sequence spaces, so that
l_p \subset l_r \subset l_\infty whenever 0 < p < r < \infty.
To show it one may first obtain that \sum_{m\ge 0} |u_m v_{j+m}|^p < \infty. This follows from the embeddings above, as u is a bounded sequence.
Now we obtain \sum_j |w_j|^p \le \sum_j ( \sum_m |u_m||v_{j+m}| )^p, where the sum over m can be estimated by the corresponding norm in l_p, by the above embeddings---and afterwards an interchange in the summation order gives an estimate by the norm in l_1 of u times the p'th power of the norm of v in l_p.
Hopefully this will work for you.
Best regards, Jon
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