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. $\newcommand\norm[2]{\left\|#1\right\|_{#2}}$

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.

(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.

*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:

- 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.
- 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.

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
\[ \Delta(y)^{-1/s} g(xy^{-1}) \in L_\infty' L_s .\]

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 \[ (\Delta^{1/p - 1/q}f) \star'_{\mu'} g = g \star_\mu (\Delta^{1/p - 1/q}f) \in L_q. \]