Let r<p. Prove that if the series sum from m=1
Let $r<p$. Prove that if the series $\displaystyle\sum_{m=1}^{\infty}\left(\sum_{n=1}^{\infty}|u(n,m)|^r\right)^{1/r}$ converges, then the series $\displaystyle\sum_{m=1}^{\infty}\left(\sum_{n=1}^{\infty}|u(n,m)|^p\right)^{1/p}$ also converges. Can the sums $\sum_{m=1}^{\infty}$ and $\sum_{n=1}^{\infty}$ be interchanged and the statement still hold?
Explanation of Method
The main statement is Problem 151's norm-monotonicity lemma applied to each "column" $(u(n,m))_{n=1}^{\infty}$ of the infinite matrix, for fixed $m$. Whether the roles of $m$ and $n$ can be swapped — i.e. whether a hypothesis on row $r$-norms (indexed by $m$) forces a conclusion about column $p$-norms (indexed by $n$) — is a genuinely different question, since it asks about the matrix's other dimension entirely; a separable (product) counterexample shows the answer is no.
- Apply the Problem 151 lemma to each fixed-$m$ column to prove the main statement.
- Construct a separable matrix $u(n,m)=x_ny_m$ to test the interchanged claim.
- Choose $x_n,y_m$ so the hypothesis (on $m$) holds but the interchanged conclusion (on $n$) fails.
Step 1 · Main statement
For each fixed $m$, apply the Problem 151 lemma (norm-monotonicity) to the sequence $(u(n,m))_{n=1}^{\infty}$: since $r<p$, $$ A_m^{(p)}=\left(\sum_{n=1}^{\infty}|u(n,m)|^p\right)^{1/p} \le \left(\sum_{n=1}^{\infty}|u(n,m)|^r\right)^{1/r}=A_m^{(r)}. $$ Summing over $m$ and comparing: $\sum_mA_m^{(p)}\le\sum_mA_m^{(r)}<\infty$, so $\sum_m A_m^{(p)}$ converges.
Step 2 · Testing the interchange with a separable matrix
Let $u(n,m)=x_ny_m$. Then $A_m^{(r)}=|y_m|\cdot\|x\|_r$, so $\sum_mA_m^{(r)}=\|x\|_r\sum_m|y_m|$ — finite iff $x\in\ell^r$ and $y\in\ell^1$. The interchanged conclusion asks about $\sum_n\left(\sum_m|u(n,m)|^p\right)^{1/p}=\sum_nB_n^{(p)}$ where $B_n^{(p)}=|x_n|\cdot\|y\|_p$, so $\sum_nB_n^{(p)}=\|y\|_p\sum_n|x_n|$ — finite iff (given $y\in\ell^1\subset\ell^p$ automatically) $x\in\ell^1$.
Step 3 · Choose a counterexample ($r=2$, $p=3$)
Take $x_n=n^{-3/4}$ and $y_m=m^{-2}$. Since $\frac32\cdot\frac34\cdot2=$ well: $x\in\ell^2$ because $\sum n^{-3/2}<\infty$; $y\in\ell^1$ because $\sum m^{-2}<\infty$. So the hypothesis holds: $\sum_mA_m^{(2)}=\|x\|_2\sum_m m^{-2}<\infty$. But $x\notin\ell^1$ since $\sum n^{-3/4}=\infty$ (exponent $\tfrac34<1$), so $\sum_nB_n^{(3)}=\|y\|_3\sum_n n^{-3/4}=\infty$ — the interchanged sum diverges.
Summary Table
| Statement | Verdict |
|---|---|
| Main claim (fixed order) | True, by the Problem 151 lemma |
| Interchanged claim | False, counterexample: $x_n=n^{-3/4}$, $y_m=m^{-2}$ |
Additional Notes
- The main statement is a direct "row-by-row" application of Problem 151's lemma; the difficulty only appears once the question of interchange is posed, since row-$r$-summability and column-$p$-summability are, in general, unrelated properties of a matrix.
- The separable (rank-one) matrix $u(n,m)=x_ny_m$ is the natural test case for any question about interchanging double-series operations, since it reduces every double sum to a product of two single sums, making the two hypotheses ($x\in\ell^r$, $y\in\ell^1$) and the two conclusions ($x\in\ell^1$, $y\in\ell^p$) fully explicit and independently tunable.
- This failure of interchange is conceptually related to Problem 150's demonstration that even summing the same double series in two different orders can give different (or here, differently-behaved) results — both problems illustrate that structure along one index of a matrix says nothing in general about structure along the other.
Result
The main statement holds; the interchanged version does not: