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Question: Let $f\in L_{p_0}[0,1]$ for some $p_0>1.$ Show that $\lim_{p\to 1^+}\|f\|_{L_p}=\|f\|_{L_1}.$

Remark: This is not homework. The question above can be obtained from the book 'Banach Space Theory: The basis for Linear and nonlinear theory'.

I would like to solve the above question in as many ways as possible. So I tried googling to get solution online.

Unfortunately, I could not get any solution to the question. The closest that I get is this question posted by Parakee. However, OP asked $p\to\infty$ instead of $p\to 1+.$

I am sure that this question is very common. So if someone can provide a solution or a reference, I would be much appreciated.

Idonknow
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1 Answers1

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For any $p \in (1,p_0)$ $||f||_1\leq ||f||_p$. We have to show that $\limsup_{p \to 1} ||f||_p\leq ||f||_1$. Let $\epsilon>0$. There exists a continuous function $g$ such that $||f-g||_{p_0} <\epsilon$. This implies that $||f-g||_{p} <\epsilon$ for every $p \in [1,p_0]$. Hence $||f||_p \leq ||g||_p +\epsilon$ and if we show that $\limsup_{p \to 1} ||g||_p\leq ||g||_1$ we can complete the proof by noting that $||g||_1 <||f||_1 +\epsilon$ and $\epsilon$ is arbitrary. Now $\int |g|^{p} \leq M^{p-1} \int |g|$ where $M=\sup \{|g(x)|:0\leq x\leq 1\}$. Hence $||g||_p \leq M^{\frac {(p-1)} p} (\int |f|)^{1/p}$. Let $p \to1+$ to complete the proof.

Kavi Rama Murthy
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