Proving The Stein-Weiss Inequality: A Detailed Guide

Hey everyone! Today, we're diving deep into a fascinating topic from harmonic analysis: the Stein-Weiss inequality. This inequality provides a powerful tool for understanding the behavior of sums of functions in the context of weak $L^p$ spaces. If you're scratching your head already, don't worry! We'll break it down step by step, making it super easy to grasp.

Understanding the Stein-Weiss Inequality

So, what exactly is the Stein-Weiss inequality? In essence, it gives us a bound on the weak $L^{1,\infty}$ norm of the sum of several functions. Specifically, if we have functions $f_1, f_2, ..., f_N$, the inequality states that:

$$\| f_1 + f_2 + ... + f_N \|_{L^{1,\infty}} \lesssim \| f_1 \|_{L^{1,\infty}} + \| f_2 \|_{L^{1,\infty}} + ... + \| f_N \|_{L^{1,\infty}}$$

Where the symbol $\lesssim$ means "less than or equal to, up to a constant factor". This constant factor is often independent of the functions $f_i$, which is a crucial part of the inequality's usefulness.

Before we jump into the proof, let's quickly recap what weak $L^p$ spaces are all about. These spaces, denoted as $L^{p,\infty}$, are "bigger" than the usual $L^p$ spaces. A function $f$ belongs to $L^{p,\infty}$ if its distribution function decays sufficiently fast. The distribution function, $d_f(\alpha)$, measures the size of the set where $|f|$ exceeds a certain threshold $\alpha$:

$$d_f(\alpha) = m(\{x : |f(x)| > \alpha\})$$

Where $m$ denotes the Lebesgue measure. For $f$ to be in $L^{p,\infty}$, we need:

$$d_f(\alpha) \lesssim \frac{\|f\|_{L^{p,\infty}}^p}{\alpha^p}$$

In other words, the set where $|f|$ is large must have a small measure. The $L^{p,\infty}$ norm is then defined as the infimum of all constants for which this inequality holds. For the special case of $p=1$, we have the weak $L^1$ space, $L^{1,\infty}$, which consists of functions whose distribution function satisfies:

$$d_f(\alpha) \lesssim \frac{\|f\|_{L^{1,\infty}}}{\alpha}$$

Okay, with these definitions in hand, we're ready to tackle the proof!

Proof of the Stein-Weiss Inequality

Let's break down the proof of the Stein-Weiss inequality into manageable steps. Our goal is to show that:

$$\| f_1 + ... + f_N \|_{L^{1,\infty}} \lesssim \sum_{i=1}^N \| f_i \|_{L^{1,\infty}}$$

Step 1: Understanding the Distribution Function

First, we need to analyze the distribution function of the sum $f_1 + ... + f_N$. We want to estimate the measure of the set where the absolute value of this sum is greater than some threshold $\alpha$:

$$m(\{x : |f_1(x) + ... + f_N(x)| > \alpha\})$$

The key idea here is to use the triangle inequality. The triangle inequality tells us that the absolute value of a sum is less than or equal to the sum of the absolute values:

$$|f_1(x) + ... + f_N(x)| \leq |f_1(x)| + ... + |f_N(x)|$$

Therefore, if the sum $|f_1(x)| + ... + |f_N(x)|$ is less than or equal to $\alpha$, then certainly $|f_1(x) + ... + f_N(x)|$ must also be less than or equal to $\alpha$. This observation allows us to write:

$$\{x : |f_1(x) + ... + f_N(x)| > \alpha\} \subseteq \{x : |f_1(x)| + ... + |f_N(x)| > \alpha\}$$

This inclusion is crucial because it allows us to bound the measure of the set we're interested in by the measure of a related set.

Step 2: Applying the Subadditivity of Measure

Now, let's introduce a clever trick. For each $i$, consider the set where $|f_i(x)|$ is greater than $\alpha/N$. Then, we can write:

$$\{x : |f_1(x)| + ... + |f_N(x)| > \alpha\} \subseteq \bigcup_{i=1}^N \{x : |f_i(x)| > \alpha/N\}$$

Why is this true? Well, if the sum of the absolute values of the $f_i$'s is greater than $\alpha$, then at least one of the $|f_i(x)|$ must be greater than $\alpha/N$. If none of them were greater than $\alpha/N$, then their sum could be at most $\alpha$, contradicting our assumption.

Now we use a fundamental property of measures called subadditivity. Subadditivity states that the measure of a union of sets is less than or equal to the sum of the measures of the individual sets:

$$m(\bigcup_{i=1}^N \{x : |f_i(x)| > \alpha/N\}) \leq \sum_{i=1}^N m(\{x : |f_i(x)| > \alpha/N\})$$

Combining this with our previous inclusion, we get:

$$m(\{x : |f_1(x) + ... + f_N(x)| > \alpha\}) \leq \sum_{i=1}^N m(\{x : |f_i(x)| > \alpha/N\})$$

Step 3: Using the Weak $L^1$ Property

This is where the weak $L^1$ property comes into play. Since each $f_i$ belongs to $L^{1,\infty}$, we know that:

$$m(\{x : |f_i(x)| > \beta\}) \leq \frac{\|f_i\|_{L^{1,\infty}}}{\beta}$$

For any $\beta > 0$. Applying this to each term in our sum, with $\beta = \alpha/N$, we obtain:

$$m(\{x : |f_i(x)| > \alpha/N\}) \leq \frac{\|f_i\|_{L^{1,\infty}}}{\alpha/N} = N \frac{\|f_i\|_{L^{1,\infty}}}{\alpha}$$

Plugging this back into our inequality, we get:

$$m(\{x : |f_1(x) + ... + f_N(x)| > \alpha\}) \leq \sum_{i=1}^N N \frac{\|f_i\|_{L^{1,\infty}}}{\alpha} = \frac{N}{\alpha} \sum_{i=1}^N \|f_i\|_{L^{1,\infty}}$$

Step 4: Final Conclusion

Finally, we can rewrite this inequality as:

$$\alpha m(\{x : |f_1(x) + ... + f_N(x)| > \alpha\}) \leq N \sum_{i=1}^N \|f_i\|_{L^{1,\infty}}$$

This tells us that the $L^{1,\infty}$ norm of the sum $f_1 + ... + f_N$ is bounded by $N$ times the sum of the $L^{1,\infty}$ norms of the individual functions. In other words:

$$\| f_1 + ... + f_N \|_{L^{1,\infty}} \lesssim \sum_{i=1}^N \| f_i \|_{L^{1,\infty}}$$

We've successfully proven the Stein-Weiss inequality! The constant factor here is $N$, which depends on the number of functions we are summing. Remember that the $\lesssim$ notation hides this constant, which is why we say the inequality holds "up to a constant factor."

Importance and Applications

The Stein-Weiss inequality might seem abstract, but it has numerous applications in harmonic analysis and related fields. It is particularly useful when dealing with operators that map functions into weak $L^p$ spaces. Here are a few examples:

• Singular Integrals: Many singular integral operators, such as the Hilbert transform and the Riesz transforms, are bounded from $L^1$ to $L^{1,\infty}$. The Stein-Weiss inequality can then be used to analyze the behavior of sums of functions transformed by these operators.
• Maximal Functions: Maximal functions, like the Hardy-Littlewood maximal function, also exhibit boundedness properties involving weak $L^p$ spaces. The Stein-Weiss inequality can be helpful in understanding the behavior of these maximal functions applied to sums of functions.
• PDEs: In the study of partial differential equations (PDEs), weak $L^p$ spaces often arise as natural spaces for solutions or their derivatives. The Stein-Weiss inequality can provide valuable estimates in this context.

By providing a bound on the weak $L^1$ norm of sums of functions, the Stein-Weiss inequality allows us to control the size of the set where the sum is large. This is particularly useful when we only have information about the individual functions, but not about their sum directly.

Conclusion

Alright, guys, we've journeyed through the proof of the Stein-Weiss inequality, unpacking each step to make it crystal clear. Remember, this inequality is a cornerstone in harmonic analysis, offering insights into how sums of functions behave in weak $L^p$ spaces. Whether you're wrestling with singular integrals, maximal functions, or PDEs, the Stein-Weiss inequality is a valuable tool to have in your arsenal. Keep practicing and exploring, and you'll master these concepts in no time! Now you know how to prove Stein-Weiss inequality.