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Suppose $a_n$ and $c_{j,n}$ are real numbers, that $\sum_{n}|a_n| < \infty$, that $\sup_{n}|c_{j,n}| < \infty$ for each $j$, and that $\lim_{j\to \infty} c_{j,n} = 0$ for each $n$.

Prove without using measure theory that if $\sup_{j,n}|c_{j,n}| < \infty$ then $$\lim_{j\to \infty}\sum_{n}a_n c_{j,n} =0$$ Prove the same thing using a theorem from measure theory. Can the hypothesis $\sup_{j,n}|c_{j,n}| < \infty$ be dropped?

I found this online and I am really confused how we are suppose to not use measure theory to prove this, any suggestions is greatly appreciated.

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    can you find the counterexample when the condition on $\sup$ is dropped? – 2017-01-07
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    for the counterexample take $c_{j,j}=j^2$ an $c_{i,j}=0$ for $i\neq j$. Let $a_i=\frac{1}{i^2}$ – 2017-01-07

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without measure theory:

Take $\epsilon > 0$.

Let $\alpha=\sup c_{j,n}$. We can take $N$ such that $\sum \limits_{n=N}^\infty |a_n| < \epsilon / 2\alpha$.

It follows by the comparison test that $|\sum\limits_{n=N}^\infty a_nc_{j,n}|<\sum\limits _{n=N}^\infty |a_nc_{j,n}|<\epsilon/2$ for all $j$.

Then we can take a $J$ such that $|c_{j,n}|<\epsilon/2N$ for all $j>J$ and $n

It follows that $|\sum\limits_{n=1}^N a_nc_{j,n}|\leq\sum\limits_{n=1}^N |a_nc_{j,n}|<\epsilon/2$ for all $j\geq J$.

We conclude that $|\sum\limits_{n=1}^\infty a_nc_{j,n}|<\epsilon$ for all $j>J$, as desired.

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    Do you know how to prove it with measure theory? – 2017-01-07
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    Just use Lebegue's dominated convergence thorem. – 2017-01-07
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    Can you show me? – 2017-01-07
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    it's direct, just consider the stepwise functions that vanish on the negative reals and satisfy: $f_j([n,n+1)) = c_{j,n}a_n$ and notice that they are dominated by the function $f([n,n+1))= |\alpha a_n|$. Notice that the series correspond to the integral of these functions, and just apply the lebesgue convergence theorem to the functions, which converge pointwise to the zero function. – 2017-01-07