# Interchange of $\ell^r$ and $L^p$-norm

Let $(f_i)_{i\in\mathbb{N}}$ be a sequence of $L^p$-functions. What is the relation between $\Vert \Vert (f_i)_{i\in\mathbb{N}}\Vert_{\ell^r}\Vert_{L^p}$ and $\Vert \left(\Vert f_i\Vert_{L^p}\right)_{i\in\mathbb{N}}\Vert_{\ell^r}$, where $\Vert (f_i)_{i\in\mathbb{N}}\Vert_{\ell^r}$ means the pointwise $\ell^r$-norm.

• I'm not sure what to make of $\|(f_i)_{i\in\mathbb{N}}\|_{\ell^r}$ when each $f_i$ is an $L^p$ function? Perhaps for clarity you could write in full your definitions of everything? E.g., what is the domain of $f_i$, are you using usual Lebesgue measure for $L^p$, etc.? – charlestoncrabb Dec 4 '15 at 19:25
• Can you explain better what you mean with $\|(f_i)_{i\in\mathbb{N}}\|_{l^r}$? – mrprottolo Dec 4 '15 at 19:26

What you are looking for is Minkowski's inequality. If $p \leq r \leq \infty$ you have

$$\| (\|f_i\|_{L^p}) \|_{\ell^r} \leq \| \|(f_i)\|_{\ell^r} \|_{L^{p}}$$

if on the other hand $r \leq p \leq \infty$ you have

$$\| \|(f_i)\|_{\ell^r} \|_{L^{p}} \leq \| (\|f_i\|_{L^p}) \|_{\ell^r}$$

The Wikipedia discussion linked to above is brief. Better to consult Inequalities by Hardy, Littlewood, and Polya.

• I'm sort of lost about the notations (both in the OP's question and this answer). I can't seem to understand the notation $\|(f_i)\|_{\ell^r}$, and how to map Minkowski's inequality to the statement above... – Clement C. Dec 5 '15 at 14:10
• The $\ell^r$ norm of a sequence $(a_i)$ is the standard definition $$\| (a_i)\|_{\ell^r} = \left( \sum_{i \in \mathbb{N}} |a_i|^r \right)^{\frac1r}$$ Now that $(f_i)$ is a sequence of functions, the pointwise sum $\ell^r$ is the function $$f = \|(f_i)\|_{\ell^r}$$ given by $$f(x) = \left( \sum_{i \in \mathbb{N}} |f_i(x)|^r \right)^{\frac1r}$$ – Willie Wong Dec 5 '15 at 15:33
• Thanks! ${}{}{}{}{}$ – Clement C. Dec 5 '15 at 15:35
• For Minkowski's inequality: the usual version is stated for $p = 1$ and $r \geq 1$ (or vice versa). To get the version above replace $f_i$ by $|f_i|^p$ and $p$ by 1 and $r$ by $r/p$ etc. Lastly, Minkowski's integral inequality is stated for arbitrary measures: if you replace one of the measures by the counting measure on $\mathbb{N}$ you get the infinite sum. – Willie Wong Dec 5 '15 at 15:36