A Weighted/Interpolated Young's Inequality

It turns out that for some applications the form of Young's inequality proven (for kernels) in my previous post is not quite sufficient. Here we will state and prove a more general version that allows interpolating between two different convolution kernels.

To keep the discussion general, we will work in the context of general measure spaces. $\newcommand\norm[2]{\left\|#1\right\|_{#2}}$

Notation  

Throughout we will let $(\Omega, \mathcal{E}, \mu)$ and $(\Omega', \mathcal{E}', \mu')$ be measure spaces, where $\Omega$ (also $\Omega'$) is the set, $\mathcal{E}$ the $\sigma$-algebra, and $\mu$ the measure. For convenience we will assume that $\mu$ is $\sigma$-finite; this in particular guarantees that the product measure on $\Omega\times \Omega'$ is well-defined and Fubini's theorem holds.

We we denote by $L_p$ the Lebesgue space with exponent $p$ on $(\Omega,\mu)$, and $L_p'$ the counterpart for $(\Omega', \mu')$. For functions $f:\Omega\times\Omega' \to \mathbb{R}$, the mixed norms are \[ \norm{f}{L_pL_q'} = \left( \int_\Omega \left( \int_{\Omega'} |f(x,y)|^q ~\mathrm{d}\mu'(y) \right)^\frac{p}{q} ~\mathrm{d}\mu(x) \right)^\frac{1}{p} \] and similarly $\norm{f}{L_q'L_p}$. So Fubini's theorem states that \begin{equation} \norm{f}{L_pL_p'} = \norm{f}{L_p'L_p}.\end{equation}

Furthermore, given a function $f:\Omega\times \Omega'\to\mathbb{R}$, we denote by $\norm{f}{L_p'}$ the function on $\Omega$ obtained by integrating over $\Omega'$ in the $p$ norm. In symbols: \[ \norm{f}{L_p'}(x) = \left( \int_{\Omega'} |f(x,y)|^p ~\mathrm{d}\mu'(y) \right)^{1/p}. \] The notation $\norm{f}{L_p}$ is defined similarly.

Preliminary computations

Let us take $1 \leq s_1 \leq r_1 \leq \infty$ and $1 \leq s_2\leq r_2\leq \infty$, and assume we are given two kernels $k_1, k_2: \Omega\times \Omega' \to \mathbb{R}$ satisfying \begin{equation} k_1 \in L_{r_1}' L_{s_1}, \quad k_2 \in L_{r_2} L_{s_2}'. \end{equation} Given $f:\Omega'\to\mathbb{R}$, we can define \begin{equation}\label{eq:def:tth} T^\theta f(x) = \int_{\Omega'} k_1^\theta(x,y) k_2^{1-\theta}(x,y) f(y) ~\mathrm{d}\mu'(y). \end{equation} We ask: for which $p, q\in [1,\infty]$ is $T^\theta$ a mapping from $L_p'$ to $L_q$?

First, we can perform a Hölder inequality in the $y\in \Omega'$ variable. This gives \[ |T^\theta f| \leq \norm{k_1^\theta f^\sigma}{L_\alpha'} \norm{f^{1-\sigma}}{L_\beta'} \norm{k_2^{1-\theta}}{L_\gamma'} \] where we split $f = f^\sigma f^{1-\sigma}$. Next we can integrate in $\Omega$ in the $L^q$ norm. The norm of $f^{1-\sigma}$ is a constant so comes out. The remaining two partial norms we can distribute among using Hölder again: \begin{equation} \norm{T^\theta f}{L_q} \leq \norm{k_1^\theta f^\sigma}{L_\alpha L_\alpha'} \norm{f^{1-\sigma}}{L_\beta'} \norm{k_2^{1-\theta}}{L_\delta L_\gamma'}. \end{equation} To the first term we can apply Fubini to evaluate the $L_\alpha$ integral first. As $f$ is independent of $x\in \Omega$, this integral only affects $k_1$. We can then apply Hölder again to the $L_\alpha'$ integral and obtain \begin{equation} \norm{T^\theta f}{L_q} \leq \norm{f^\sigma}{L_\epsilon'} \norm{f^{1-\sigma}}{L_\beta'} \norm{k_1^\theta}{L_\eta' L_\alpha} \norm{k_2^{1-\theta}}{L_\delta L_\gamma'} . \end{equation}

For this final expression to hold, there are some constraints we need to impose:

Variable ranges

The interpolation variables $\sigma,\theta \in [0,1]$.

The Lebesgue exponents, in order for Hölder to hold, requires

  • $\beta, \gamma \geq 1$
  • $\alpha, \delta \geq q$
  • $\epsilon, \eta \geq \alpha$

Hölder identities

\begin{gather} \label{eq:h1} \alpha^{-1} + \beta^{-1} + \gamma^{-1} = 1 \newline \label{eq:h2} \alpha^{-1} + \delta^{-1} = q^{-1} \newline \label{eq:h3} \epsilon^{-1} + \eta^{-1} = \alpha^{-1} \end{gather}

Matching exponents

To get the $k_1$, $k_2$, and $f$ in the desired spaces, we need

\begin{gather} \label{eq:m1} (1-\theta) \delta = r_2 \newline \label{eq:m2} (1-\theta) \gamma = s_2 \newline \label{eq:m3} \theta \eta = r_1 \newline \label{eq:m4} \theta\alpha = s_1 \newline \label{eq:m5} \sigma \epsilon = p \newline \label{eq:m6} (1-\sigma)\beta = p \end{gather}

Solving the system

Note that we have 8 unknowns $\alpha, \beta, \gamma, \delta, \epsilon,\eta$ and $\theta,\sigma$, with nine equations. So the system is overdetermined. This implies that we only expect the system to be solvable on a codimension 1 subspace of the $p, q, r_1, r_2, s_1, s_2$ variables; in other words, we expect a certain relation between those six variables to hold. This is as expected even in the usual case of Young's inequality.

We first note that \eqref{eq:h2}, \eqref{eq:m1}, and \eqref{eq:m4} forms a closed system for $\theta, \alpha, \delta$. Performing the algebra shows that \begin{gather} \delta = q \cdot \frac{r_2 - s_1}{q - s_1} \newline \alpha = q \cdot \frac{r_2 - s_1}{r_2 - q} \newline 1 - \theta = \frac{r_2}{q} \cdot \frac{q - s_1}{r_2 - s_1}\newline \theta = \frac{s_1}{q} \cdot \frac{r_2 - q}{r_2 - s_1} \end{gather} Note that in view of the constraint that both $\alpha$ and $\delta$ are at least $q$, we need $q$ to be between $r_2$ and $s_1$. (But so far the system is compatible with either $r_2 \leq q \leq s_1$ or $s_1 \leq q \leq r_2$.) (In both cases the constraint on $\theta\in [0,1]$ is also satisfied.)

Next, $\eta$ is also uniquely determined by \eqref{eq:m3}, since $\theta$ is now known: \begin{equation} \eta = q \cdot \frac{r_1}{s_1} \cdot \frac{r_2 - s_1}{r_2 - q}. \end{equation} This forces, via \eqref{eq:h3}, the value of $\epsilon$ to be \begin{equation} \epsilon = q \cdot \frac{r_1}{r_1 - s_1} \cdot \frac{r_2 - s_1}{r_2 - q} = \left( \frac{\theta}{s_1} - \frac{\theta}{r_1}\right)^{-1}. \end{equation}

Similarly we can solve for $\gamma$ using \eqref{eq:m2}, which gives \begin{equation} \gamma = q \cdot \frac{s_2}{r_2} \cdot \frac{r_2 - s_1}{q - s_1}. \end{equation} (Note that $\gamma > 0$ as long as $q$ is between $s_1$ and $r_2$.) This implies that \begin{equation} \gamma^{-1} + \alpha^{-1} = \frac{1}{s_2} \left( 1 + \frac{r_2 - q}{q} \frac{s_2 - s_1}{r_2 - s_1} \right) = \frac{1-\theta}{s_2} + \frac{\theta}{s_1} \end{equation} and as long as both $s_2, s_1 \geq 1$ this sum is $\leq 1$. And thus by \eqref{eq:h1} we can solve for \begin{equation} \beta = \left( 1 - \frac{1-\theta}{s_2} - \frac{\theta}{s_1}\right)^{-1}. \end{equation}

Finally, combining \eqref{eq:m5} and \eqref{eq:m6} we obtain \[ \frac{1}{p} = \frac{1}{\epsilon} + \frac{1}{\beta}.\] Expanding using our results above we have that \begin{equation} p^{-1} = 1 - \frac{1 - \theta}{s_2} - \frac{\theta}{r_1} \end{equation} which is the compatibility relation required for the system to be solvable.

The theorem

Based on the computations above, we have the following theorem.

Theorem  
Given $1 \leq s_1 \leq r_1 \leq \infty$ and $1 \leq s_2 \leq r_s \leq \infty$ and kernels $k_1, k_2:\Omega\times\Omega' \to \mathbb{R}$ satisfying \[ k_1 \in L_{r_1}'L_{s_1}, \quad k_2 \in L_{r_2} L_{s_2}', \] then, for any $\theta\in [0,1]$, the operator $T^\theta$ defined by \eqref{eq:def:tth} is a mapping from $L_p'(\Omega')$ to $L_q(\Omega)$ provided \begin{gather} \frac{1}{q} = \frac{1-\theta}{r_2} + \frac{\theta}{s_1} \newline 1 - \frac{1}{p} = \frac{1-\theta}{s_2} + \frac{\theta}{r_1} \end{gather}

(One can also prove this by complex interpolation in the obvious way.)

Application to Lie groups

One application of this theorem is to obtain a Young's convolution inequality on general Lie groups. The reader should refer to this previous post for basic material concerning Lie groups.

Let us first define the notion of a convolution.

Definition    [Convolution on Lie groups]
Let $\mu$ denote a left-invariant Haar measure on a Lie group $G$. The convolution of two functions $f, g:G \to \mathbb{R}$ is defined to be \[ (f \star_\mu g )(x) = \int_G f(y) g(y^{-1} x) ~\mathrm{d}\mu(y) .\]

We first note that commutation on general Lie groups are not commutative. Since the measure is left-invariant, for any $z\in G$ the integral \[ \int_G f(zy) ~\mathrm{d}\mu(y) = \int_G f(y) ~\mathrm{d}\mu(y). \] Applying this to the definition with $z = x$, we see that \[ (f \star_\mu g)(x) = \int_G f(xy) g(y^{-1}) ~\mathrm{d}\mu(y). \] This expression differs from $(g\star_\mu f)(x)$ in two ways:

  1. First, as explained in the previous post, the group involution generally don't leave the Haar measure invariant: it converts left Haar measures to right Haar measures and vice versa. Denote by $\mu'$ the pushforward of $\mu$ under this involution. We then have \[ (f \star_\mu g)(x) = \int_G f(xy^{-1}) g(y) ~\mathrm{d}\mu'(y). \] So the two expressions differ as they integrate against different measures.
  2. Secondly, the group operation on a general Lie group is not commutative, and so $xy^{-1} \neq y^{-1} x$ in general.

In the case where $G$ is Euclidean space (which is a unimodular abelian group), both of these obstructions vanish, and hence we have the convolution on Euclidean spaces is commutative.

The discussion above also suggests that for a right-invariant Haar measure $\mu'$ on $G$, the convolution should be defined as \[ (f \star_{\mu'}' g)(x) = \int_G f(y) g(x y^{-1}) ~\mathrm{d}\mu'(y).\] So if $\mu'$ is the pushforward of $\mu$ under the group involution, we in fact have the relation \begin{equation} f \star_{\mu} g = g \star_{\mu'}' f. \end{equation}

The two measures $\mu'$ and $\mu$ turns out to have a simple relation.

Lemma  
If $\mu$ is a left Haar measure and $\mu'$ its right invariant counterpart obtained by the group involution, then \[ \mathrm{d}\mu(x) = \Delta(x) ~\mathrm{d}\mu'(x) \] where $\Delta$ is the modular function (see Definition 9 in the previous post on Lie groups).
Proof [Sketch]:
From the definition of the modular function it is clear that $\Delta$ is a group homomorphism from $G$ to $(0,\infty)$ as a multiplicative group, with $\Delta(xy) = \Delta(x)\Delta(y)$. Next we verify that $\Delta(x) ~\mathrm{d}\mu'(x)$ is left invariant. \[ \int_G f(xy) \Delta(y) ~\mathrm{d}\mu'(y) = \int_G f(xy) \Delta(xy) \Delta(x^{-1}) ~\mathrm{d}\mu'(y) = \int_G f(xy) \Delta(xy) ~\mathrm{d} (\lambda_{x^{-1}})_\star \mu'(xy) = \int_G f(y) \Delta(y) ~\mathrm{d}\mu'(y). \] This shows that $\mathrm{d}\mu$ and $\Delta \mathrm{d}\mu'$ differ by at most a constant multiple. Continuity near the identity element shows that the multiple is exactly 1.
Remark  

And now we see why it is important to have Theorem 2. Let $\mu$ be a left-invariant Haar measure on $G$, and $\mu'$ its right-invariant counterpart. Set $k(x,y) = g(x y^{-1})$ for some $g \in L_s$. We see immediately that \[ \norm{k}{L_s'} = \int_G |g(xy^{-1})|^s ~\mathrm{d}\mu'(y) = \int_G |g(xy)|^s ~\mathrm{d}\mu(y) = \int_G |g(y)|^s ~\mathrm{d}\mu(y) \] where we used the involutive relationship between $\mu$ and $\mu'$, and the fact that $\mu$ is left-invariant. This means that $k(x,y) \in L_\infty L_s'$. On the other hand, we don't have $k(x,y) \in L_\infty' L_s$ in general. While \[ \norm{k}{L_s} = \int_G |g(x y^{-1})|^s ~\mathrm{d}\mu(x) \] since $\mathrm{d}\mu(x)$ is not right-invariant, this expression is in general dependent on $y$! In fact, we have \[ \int_G |g(x y^{-1})|^s ~\mathrm{d}\mu(x) = \int_G |g(x y^{-1})|^s \Delta(x) ~\mathrm{d}\mu'(x) = \int_G |g(x)|^s \Delta(xy) ~\mathrm{d}\mu'(x) \] where in the last step we used the right-invariance of $\mu'$. This implies that \[ \int_G |g(x y^{-1})|^s ~\mathrm{d}\mu(x) = \Delta(y) \int_G |g(x)|^s ~\mathrm{d}\mu(x).\] Here we used that $\Delta$ is a group homomorphism. Note that since $\Delta$ is a group homomorphism, in the case where $\Delta$ is not identically 1, it must be unbounded, and this shows that $k \not\in L_\infty' L_s$.

On the other hand, the above computation also shows that we do have \[ \Delta(y)^{-1/s} g(xy^{-1}) \in L_\infty' L_s .\]

Corollary  
Let $\Omega = \Omega' = G$, and $\mu$ be a left-invariant Haar measure, and $\mu'$ its right-invariant counterpart. Given $f \in L_p'$, $g\in L_s$. Denote by $\Delta^{\alpha}f$ the function $y\mapsto \Delta^{\alpha}(y) f(y)$. Then the convolution \[ (\Delta^{-1/q} f) \star'_{\mu'} g = g \star_\mu (\Delta^{-1/q}f) \in L_q \] provided \[ 1 - \frac1p = \frac1s - \frac1q.\]
Proof:
We wish to apply Theorem 2, with $r_1 = r_2 = \infty$ and $s_1 = s_2 = s$. Adding the conditions on $p^{-1}$ and $q^{-1}$ we immediately obtain the identity relating $p, q, s$. Set $\theta = s/q \in [0,1]$, note that $1/q = \theta/s$. Writing \[ (\Delta^{-1/q} f) \star'_{\mu'} g = \int_G \Delta^{-\theta/s}(y) f(y) g(xy^{-1}) ~\mathrm{d}\mu'(y), \] we can set $k_1(x,y) = \Delta^{-1/s}(y) g(xy^{-1})$ and $k_2(x,y) = g(xy^{-1})$, after comparing against \eqref{eq:def:tth}. The discussion in Remark 5 shows that $k_2 \in L_\infty L_s'$, while $k_1\in L_\infty' L_s$, and hence Theorem 2 can be applied to obtain the desired conclusion.

If one doesn't want to work with $f\in L_p'$, and instead want $f$ belonging to some $L_p$ space, we can note that if $f\in L_p$, we have \[ \int_G |f(y)|^p ~\mathrm{d}\mu(y) = \int_G |f(y)|^p \Delta(y) ~\mathrm{d}\mu'(y) \] so that $\Delta^{1/p} f\in L_p'$. And hence we have that whenever \[ 1 - \frac1p = \frac1s - \frac1q \] and $f\in L_p$ and $g\in L_s$, then \[ (\Delta^{1/p - 1/q}f) \star'_{\mu'} g = g \star_\mu (\Delta^{1/p - 1/q}f) \in L_q. \]

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Willie WY Wong
Assistant Professor

My research interests include partial differential equations, geometric analysis, fluid dynamics, and general relativity.

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