On Pg. 34 of this reference, I encountered Young's Convolution Inequality. The author states the inequality and manipulates it into various forms. I write this post to better understand the manipulations of Young's Convolution Inequality into different forms, and what the author is trying to achieve by doing so. I shall reproduce the text from the reference below (as quotes), and ask my questions inline.
It would be greatly appreciated even if you can help answer some of the following questions, if not all. Thank you!
If $f$ and $g:\mathbb R\to \mathbb R$ are bounded, integrable functions, the convolution $f * g$ of $f$ and $g$ is defined by $$f*g(x) = \int_\mathbb R f(y) g(x-y) dy$$
Young's Convolution Inequality: If $f \in L_p$, $g\in L_q$, and $$\frac 1p + \frac1q = 1 + \frac1s$$
then
$$\|f * g\|_s\le \|f\|_p \|g\|_q$$
Once we have Young’s inequality, we can give a meaning to convolutions of functions that are not both
integrable and bounded, provided that they lie in the correct $L_p$ spaces.
Q1. What is meant by "correct" $L^p$ spaces here?
Young’s inequality holds for convolution on any locally compact group, for example the circle. On compact groups it is sharp: there is equality for constant functions. But on $\mathbb R$, where constant functions are not integrable, the inequality can be improved (for most values of $p$ and $q$).
Q2. What is meant by compact groups and locally compact groups? Wikipedia says: (i) a compact (topological) group is a topological group whose topology is compact and (ii) a locally compact group is a topological group G for which the underlying topology is locally compact and Hausdorff. However, I am able to see which groups we are talking about in this context.
It was shown by Beckner [1975] and Brascamp and Lieb [1976a] that the correct constant in Young’s inequality is attained if $f$ and $g$ are appropriate Gaussian densities: that is, for some positive $a$ and $b$, $f(t) = e^{−at^2}$ and $g(t) = e^{−bt^2}$. (The appropriate choices of $a$ and $b$ and the value of the best constant for each $p$ and $q$ will not be stated here. Later we shall see that they can be avoided.)
Q3. What is meant by the correct constant in Young's inequality? First of all, where is the constant? Is it present in some other form of the inequality, that has not been stated here? Pretty confused about this.
How are convolutions related to convex bodies? To answer this question we need to rewrite Young’s inequality slightly. If $1/r+1/s = 1$, the $L_s$ norm $\|f ∗g\|_s$ can be realized as $$\int_\mathbb R (f*g)(x)h(x)$$ for some function with $\|h\|_r = 1$.
Q4. How was the inequality rewritten into the above format? It'd be great if someone could provide details because I'm unable to see how and what just happened.
So, the inequality says that if $1/p + 1/q + 1/r = 2$, then
$$\int \int f(y) g(x-y) h(x) dy dx\le \|f\|_p \|g\|_q \|h\|_r$$
We may rewrite the inequality again with $h(-x)$ in place of $h(x)$, since this doesn't affect $\|h\|_r$.
$$\int \int f(y) g(x-y) h(-x) dy dx\le \|f\|_p \|g\|_q \|h\|_r$$
Q5. What is the point of replacing $x$ by $-x$? What are we gaining?
This can be written in a yet more symmetric form, with the help of the map $\phi: \mathbb R^2 \to \mathbb R^3$ such that $\phi: (x,y) \to (y,x-y,-x) := (u,v,w)$. The range of $\phi$ is the subspace $H$ of $\mathbb R^3$, given by $$H = \{(u,v,w): u+v+w = 0\}$$ Besides the factor coming from the Jacobian, the integral can be written as $$\int_H f(u) g(v) h(w)$$ where the integral is with respect to the two-dimensional measure on the subspace $H$.
Q6. What are we gaining by rewriting the integral in the form (except that it looks nicer)? Also, does the Jacobian contribute to the constant factor that is being talked about in several places?
Q7. Which measure on $H$ are we talking about, explicitly?
So Young’s inequality and its sharp forms estimate the integral of a product function on $\mathbb R^3$ over a subspace. What is the simplest product function? If $f, g$, and $h$ are each the characteristic function of the interval $[−1, 1]$, the function $F$ given by $$F(u,v,w) = f(u)g(v)h(w)$$ is the characteristic function of the cube $[−1, 1]^3 ⊂ \mathbb R^3$. The integral of $F$ over a subspace of $\mathbb R^3$ is thus the area of a slice of the cube: the area of a certain convex body. So there is some hope that we might use a convolution inequality to estimate volumes.
Q8. What is the sharp form of Young's inequality that the author talks about here? How is it sharp?
Best Answer
Q2: I'm not an expert on this, so I would defer to someone else. Regarding the definition, "group" here is referring to a "topological group"; see Locally compact group for details and examples. I think Ball is implicitly referring to convolutions with respect to Haar measures on such spaces, but I'm not sure.
Q3, Q8: If you glance at the Brascamp and Lieb paper, you can infer that the constant Ball is referring to is basically the $c$ in $\|f * g\|_s \le c \|f\|_p \|g\|_q$. Presumably one is able to prove the inequality holds in general when $c=1$, but one might ask if you can prove the inequality for a smaller value of $c$. Ball notes that spaces where constant functions are integrable, equality is attained ($\|f*g\| = \|f\|_p \|g\|_q$), so you can't make $c$ smaller than $1$ in general. But in the specific case of $\mathbb{R}$, apparently the inequality holds for a smaller $c$.
Q4: For a function $u \in L^s$ and $h \in L^r$ with $1/s + 1/r=1$, Hölder's inequality $\int u(x)h(x) \, dx \le \|u\|_s \|h\|_r$ attains equality when $|u|^s$ and $|h|^r$ are equal almost surely up to a multiplicative constant. In particular, if $h$ is of the form $h = c u^{s/r}$ where $c = \|u^{s/r}\|_r^{-1}$ so that $\|h\|_r = 1$, then we have $\int u(x) h(x) \, dx = \|u\|_s$. Ball is applying this with $u=f*g$.
Response to comment: $\|h\|_r=1$ is only necessary for Ball's claim that "$\|f*g\|_s$ can be realized as $\int (f*g)(x) h(x) \, dx$ with $\|h\|_r=1$." It is not necessary for the subsequent inequalities that keep $\|h\|_r$ on the right-hand side. The purpose is to show the equivalence between the original convolution inequality and the modified form (you can see that the Brascamp and Lieb paper considers the second form).
Q5, Q6: The convolution integral is complicated because of the $g(x-y)$ term which involves both dummy variables $x$ and $y$ being integrated. I guess it is easier to to consider an integrand $f(u) g(v) h(w)$ that decomposes with respect to the dummy variables (each factor only involves one variable). Note that this isn't a "free lunch" scenario where we have magically untangled the dummy variables; the dummy variables are still intertwined by the subspace condition $u+v+w=0$. But this conversion is helpful for the purpose of getting intuition on why convolutions are relevant in estimating volumes (see the sentence describing the "area of a slice of a cube"), which is really the purpose of these few pages.
You are correct that the Jacobian factor would matter for determining the constant mentioned above, but I think Ball is just trying to give intuition here.
Q7: I think any measure that assigns measure to a set proportional to its surface area works. Exactly which measure this is (i.e., what is the multiplicative constant that converts from area to measure) would depend on an explicit parameterization of the plane, and I think the constant would just be the Jacobian constant (or could get absorbed into it). Since these measures are all proportional up to a constant, I don't think it matters for the sake of the discussion which only concerns intuition.
Response to comment: I did not really imply what you wrote. In the case of the plane $H$ defined in the text, I was referring to measures of the form $\mu(A) \propto \text{area}(A)$ for $A \subseteq H$. (No $\partial A$, and not $\mathbb{R}^n$.) In the context of the convolution inequality, this just comes out of the transformations. In terms of the broader context, I didn't read the rest of the text, but I guess there is some geometric context where you are interested in areas/volumes, so such measures (that are proportional to areas/volums) would be relevant.