Inclusion of Lᵖ space

• categories: Theorem, Functional analysis

For a measure space $(X,\Sigma ,\mu )$, if $0<\mu (X)<\infty$ and $1\le p<q\le \infty$, then there is a continuous embedding of $L^p(X,\mu )$ into $L^q(X,\mu )$. This means that if $f\in L^p(X,\mu )$, then $f\in L^q(X,\mu )$, and $\Vert f{\Vert }_q\le C\Vert f{\Vert }_p$ for some constant $C$ depending on $p$, $q$, and $\mu (X)$.

Statement

Let $1\le p<q\le \infty$ and assume that $\mu (X)<\infty$. Then for any measurable function $f$,

$$\Vert f{\Vert }_q\le \Vert f{\Vert }_p\cdot \mu (X{)}^{\frac{1}{p}-\frac{1}{q}}.$$

Thus, if $f\in L^p(X,\mu )$, then $f\in L^q(X,\mu )$ and the inclusion map from $L^p(X,\mu )$ to $L^q(X,\mu )$ is continuous.

Intuition

When $p<q$, the $L^p$ norm penalizes larger values of $|f|$ less strongly than the $L^q$ norm does. Consequently, an $L^p$ function tends to have “more integrability” when viewed in the $L^q$ space, allowing for inclusion.

Proof Outline

1. Power Mean Inequality: Using Hölder’s Inequality, observe that for any $f\in L^p(X,\mu )$ and $\mu (X)<\infty$, $|f{|}^p$ is integrable, so $|f{|}^q$ is also integrable since $p<q$.
2. Estimate: Apply Hölder’s inequality with exponents $\frac{q}{p}$ and $\frac{q}{q-p}$ to obtain: $$\Vert f{\Vert }_q\le \Vert f{\Vert }_p\cdot \mu (X{)}^{\frac{1}{p}-\frac{1}{q}}.$$