Limit $\ell^p$ norms Let $x=(x_n)\in \ell^1$. Then one can show that $x\in\ell^p,\,\forall p\geq1$. I must prove that $\lim_{p\to 1^+}||x||_p = ||x||_1$, where $||\cdot||_p$ and $||\cdot||_1$ denote the usual $\ell^p$ and $\ell^1$ norms. I've managed to show that $\limsup_{p\to1^+}||x||_p\leq||x||_1$, but I'm having difficulty proving $\liminf_{p\to1^+}||x||_p\geq||x||_1$. My teacher told me to try using the fact that given $\epsilon>0$ there exists $N\in\mathbb{N}$ such that $\sum_{n=N+1}^\infty|x_n|^p<\epsilon$, and then that $\lim_{p\to1^+}\sum_{n=1}^N|x_n|^p = \sum_{n=1}^N|x_n|$, but that doesn't seem to help, since you end up with the same inequality. Any tips?
 A: Fix $\epsilon > 0$. We can find $N$ sufficiently large so that
$$ \sum_{n=N+1}^\infty |x_n|^p < \epsilon.$$
Now we can write
$$\|x\|_p^p = \sum_{n=1}^\infty |x_n|^p = \sum_{n=1}^N |x_n|^p + \sum_{n=N+1}^\infty |x_n|^p < \sum_{n=1}^N |x_n|^p + \epsilon.$$
Taking the $p$th root of both sides,
$$\|x\|_p < \sum_{n=1}^N |x_n| + \epsilon^{1/p} \leq \sum_{n=1}^\infty |x_n| + \epsilon^{1/p} = \|x\|_1 + \epsilon^{1/p}.$$
Take $p \rightarrow 1^+$ to get $\limsup_{p \rightarrow 1^+} \|x\|_p \leq \|x\|_1 + \epsilon$ for all $\epsilon > 0$, hence $\limsup_{p \rightarrow 1^+} \|x\|_p \leq \|x\|_1$.
For the other direction, fix $\epsilon> 0$ and observe that there is an $N$ large enough so that
$$ \sum_{n=N+1}^\infty |x_n| < \epsilon.$$
Now we can write
$$\|x\|_1 = \sum_{n=1}^N |x_n| + \sum_{n=N+1}^\infty |x_n| < \sum_{n=1}^N |x_n| + \epsilon.$$
Since there are only finitely many terms we need to worry about, we can choose a constant $C$ so that $|Cx_n| \geq 1$ (or $|Cx_n| = 0$ if $x_n = 0$) for $1 \leq n \leq N$. For $p \geq 1$ this gives us
$$\|Cx\|_1 < \sum_{n=1}^N |Cx_n| + |C| \epsilon \leq \sum_{n=1}^n |Cx_n|^p + |C| \epsilon = |C|^p \sum_{n=1}^N |x_n|^p + |C| \epsilon.$$
Divide both sides by $|C|$ to get
$$ \|x\|_1 \leq |C|^{p-1}\sum_{n=1}^N |x_n|^p + \epsilon \leq |C|^{p-1} \|x\|_p^p + \epsilon.$$
Now take $p$th roots to get
$$\|x\|_1^{1/p} \leq  |C|^{(p-1)/p} \|x\|_p + \epsilon^{1/p}.$$
Let $p \rightarrow 1^+$ to get
$$ \|x\|_1 \leq \liminf_{p \rightarrow 1^+} \|x\|_p + \epsilon.$$
This holds for all $\epsilon > 0$, so let $\epsilon \rightarrow 0$ to get our desired answer.
