Limit of operator norm

Question

Let $n$ be an integer. Let $T : \mathbb{R}^n \to \mathbb{R}^n$ be a linear map.

How to show that $$ \lim_{p \to 1} \|T: \ell^1_n \to \ell^p_n\| =\|T: \ell^1_n \to \ell^1_n\| ? $$

Any reference is welcome.

Answer 1

Indeed the key point is that $p$-norm is a continuous function as stated in the comments. Although, a small argument is needed:

Fix $x=(x_1,...,x_n)\in \mathbb{R^n}$ (i assume that your $p$ is $>1$). Then, by Hölder's Inequality used in the vectors $x,y\in \mathbb{R^n}$ where $y=(1,...,1)$ we have

\begin{align} \sum_{k=1}^{n}|x_k|&=\sum_{k=1}^{n}|x_k|\cdot 1\\ &\leq \biggl(\sum_{k=1}^{n}|x_k|^p\biggr)^{1/p}\biggl(\sum_{k=1}^n1\biggr)^{1-1/p}\\ &=n^{1-1/p}\cdot ||x||_p \end{align}

Hence, $$\tag{1} ||x||_1\leq n^{1-1/p}||x||_p$$

Now, denote by $||T||_{1,1}$ the norm of the operator $T:l_1\to l_1$ and by $||T||_{1,p}$ the norm of the operator $T:l_1\to l_p$. Using $(1)$ we have $$||Tx||_1\leq n^{1-1/p}\cdot ||Tx||_p\leq n^{1-1/p}\cdot ||T||_{1,p}\cdot ||x||_1$$ Hence, $||T||_{1,1}\leq n^{1-1/p}||T||_{1,p}$ so, taking $\liminf_{p\to 1}$ in the last inequality we get $$||T||_{1,1}\leq \liminf_{p\to 1}||T||_{1,p}$$

For the other direction we use the continuity of the $p-$norm. Let $\epsilon>0$, by the definition of $||T||_{1,p}$ there is an $||x||_1\leq 1$ such that $$\tag{2} ||T||_{1,p}-\epsilon<||Tx||_p\leq ||T||_{1,p}$$

Then, using $\lim_{p\to 1}||Tx||_p=||Tx||_1$ in $(2)$ we get $$\tag{3}\limsup_{p\to 1}||T||_{1,p}-\epsilon\leq ||Tx||_1\leq ||T||_{1,1}\cdot ||x||_1=||T||_{1,1}$$

Hence, $\limsup_{p\to 1}||T||_{1,p}\leq ||T||_{1,1}$. Putting all together,

$$\limsup_{p\to 1}||T||_{1,p}\leq ||T||_{1,1}\leq \liminf_{p\to 1}||T||_{1,p}$$ which is equivalent to $\lim_{p\to 1}||T||_{1,p}=||T||_{1,1}$.