Limit of operator norm

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.

• Isn't it trivially true? $p$-norm is a continuous function right? If limit of the image equals, operator norm can be written into norm of the image, and it's a matter of working the definition? Am I missing something? Commented Nov 21, 2020 at 13:29

1 Answer

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 $$\lim_{p\to 1}||T||_{1,p}=||T||_{1,1}$$.