Show that a subspace is proper dense in $l^1$ sequence space. L^1 space.

Question

Let Y = $L^1 $($\mu$) where $\mu$ is counting measure on N.

Let X = {$f$ $\in$ Y : $\sum_{n=1}^{\infty}$ n|$f(n)$|<$\infty$}, equipped with the $L^1$ norm.

Show that X is a proper dense subspace of Y.

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I don't know how to show that X is dense in Y.

I was thinking about constructing a sequence in Y so that every convergent sequence in X converges to the sequence in Y.

But then I think there might be another way to show denseness of X.

Thank you.

Answer 1

Hint: Think about sequences that are $0$ except for finitely many terms.

Answer 2

Let $e_i :\mathbb N\to\mathbb R$ defined as $$ e_i(j)=\left\{\begin{array}{lll} 1 &\text{if}&i=j,\\ 0 &\text{if}&i\ne j.\end{array}\right. $$ Then $e_i\in X$. If $f\in Y$, then setting $$ f_n=\sum_{i=1}^n f(i)\, e_i\in X, $$ we observe that $$ \|f-f_n\|_{L^1}=\sum_{i>n}\lvert\,f(i)\rvert. $$ Clearly, the right hand side of the above tends to zero, as $n\to\infty$, and hence $f_n\to f$ in $L^1(\mu)$.