Brezis Functional Analysis: Mastering W^{1,p}(\mathbb{R}) Spaces

Hey guys! Today, we're diving deep into a classic problem from Brezis' Functional Analysis, specifically Exercise 8.1. We're going to dissect a function and figure out for which values of p it belongs to the Sobolev space $W^{1,p}(\mathbb{R})$. This is a super important concept in functional analysis, dealing with functions that not only have certain integrability properties but also possess derivatives in certain spaces. So, grab your coffee, get comfy, and let's break down this challenging exercise. We'll be looking at the function $u(x) = \frac{1}{(1+x^2)^{\frac{\alpha}{2}}} \frac{1}{\ln(2+ x^2)}$ with $0 < \alpha < 1$, and our mission is to determine for which $p \in [1, \infty)$ this function $u$ is in $W^{1,p}(\mathbb{R})$. Remember, for $u$ to be in $W^{1,p}(\mathbb{R})$, both $u$ itself and its weak derivative (which, for this well-behaved function, will be its classical derivative) must be in $L^p(\mathbb{R})$. This means we need to check the integrability of $u$ and its derivative $u'$. Let's get started by understanding what $W^{1,p}(\mathbb{R})$ actually means. The Sobolev space $W^{1,p}(\mathbb{R})$ consists of all functions $u in L^p(\mathbb{R})$ whose first weak derivative $u'$ also belongs to $L^p(\mathbb{R})$. The norm in this space is typically defined as $||u||_{W^{1,p}} = (||u||_{L^p}^p + ||u'||_{L^p}^p)^{\frac{1}{p}}$. For our specific function $u(x)$, which is smooth and decays at infinity, its weak derivative will coincide with its classical derivative. So, the task boils down to calculating $u'(x)$ and then investigating whether both $u$ and $u'$ are $p$-integrable over the entire real line. This involves analyzing the behavior of $u(x)$ and $u'(x)$ as $x \to \pm \infty$ and near any potential singularities. In our case, the function is well-defined and smooth everywhere, so we're mainly concerned with its decay rate. This exercise is a fantastic way to solidify your understanding of function spaces and how to analyze functions within them. Let's roll up our sleeves and tackle the derivative calculation, which is the first crucial step in this investigation. It's going to involve the chain rule and the quotient rule, so keep those differentiation techniques sharp, guys!

Calculating the Derivative $u'(x)$

Alright guys, the first major step in checking if our function $u(x)$ belongs to $W^{1,p}(\mathbb{R})$ is to compute its derivative, $u'(x)$. For $u(x) = \frac{1}{(1+x^2)^{\frac{\alpha}{2}}} \frac{1}{\ln(2+ x^2)}$, we'll need to use the product rule and the chain rule. Let's rewrite $u(x)$ slightly to make the differentiation process a bit more straightforward: $u(x) = (1+x^2)^{-\frac{\alpha}{2}} (\ln(2+x^2))^{-1}$. Now, applying the product rule $(fg)' = f'g + fg'$:

Let $f(x) = (1+x^2)^{-\frac{\alpha}{2}}$ and $g(x) = (\ln(2+x^2))^{-1}$.

First, let's find $f'(x)$: Using the chain rule, $\frac{d}{dx}(a^n) = na^{n-1} \cdot a'$, where $a = 1+x^2$ and $n = -\frac{\alpha}{2}$. So, $a' = 2x$. $f'(x) = -\frac{\alpha}{2}(1+x^2)^{-\frac{\alpha}{2}-1} \cdot (2x) = -\alpha x (1+x^2)^{-\frac{\alpha}{2}-1}$.

Next, let's find $g'(x)$: Using the chain rule and the rule for the derivative of $\ln(v)$, which is $\frac{v'}{v}$. Here, $v = 2+x^2$, so $v' = 2x$. The derivative of $y^{-1}$ is $-y^{-2}$. So, $g'(x) = - (\ln(2+x²⁾⁾{-2} \cdot \frac{1}{\ln(2+x^2)}' $. We need the derivative of $(\ln(2+x^2))^{-1}$. Let $w = \ln(2+x^2)$. Then $g(x) = w^{-1}$, and $g'(x) = -w^{-2} \cdot w'$. The derivative of $w = \ln(2+x^2)$ is $\frac{2x}{2+x^2}$. So, $g'(x) = -(\ln(2+x^2))^{-2} \cdot \frac{2x}{2+x^2} = -\frac{2x}{(2+x^2)(\ln(2+x^2))^2}$.

Now, let's put it all together using the product rule: $u'(x) = f'(x)g(x) + f(x)g'(x)$.

$u'(x) = \left( -\alpha x (1+x^2)^{-\frac{\alpha}{2}-1} \right) \left( (\ln(2+x^2))^{-1} \right) + \left( (1+x^2)^{-\frac{\alpha}{2}} \right) \left( -\frac{2x}{(2+x^2)(\ln(2+x^2))^2} \right)$

$u'(x) = -\frac{\alpha x}{(1+x^2)^{\frac{\alpha}{2}+1} \ln(2+x^2)} - \frac{2x}{(1+x^2)^{\frac{\alpha}{2}}(2+x^2)(\ln(2+x^2))^2}$

We can factor out some common terms to simplify this expression. Let's factor out $-x (1+x^2)^{-\frac{\alpha}{2}-1} (\ln(2+x^2))^{-2}$:

$u'(x) = -x (1+x^2)^{-\frac{\alpha}{2}-1} (\ln(2+x^2))^{-2} \left[ \alpha (\ln(2+x^2)) + \frac{2(1+x^2)}{(2+x^2)} \right]$

This looks a bit messy, but that's okay. The key is to understand the asymptotic behavior of this derivative as $x \to \pm \infty$. This derivative calculation is a critical step, and getting it right ensures we're on the right track for analyzing the integrability of $u'$. Don't worry if it looks complicated; we'll simplify our analysis by focusing on the dominant terms for large $|x|$. The next stage is to investigate the integrability of both $u(x)$ and $u'(x)$ over $\mathbb{R}$.

Analyzing the Integrability of $u(x)$

Now that we've got the derivative $u'(x)$ all worked out, let's shift our focus to the integrability of the original function $u(x)$ and its derivative. For $u$ to be in $W^{1,p}(\mathbb{R})$, both $u in L^p(\mathbb{R})$ and $u' in L^p(\mathbb{R})$. Let's start with $u(x) = \frac{1}{(1+x^2)^{\frac{\alpha}{2}}} \frac{1}{\ln(2+ x^2)}$, where $0 < \alpha < 1$. We need to check if $\int_{-\infty}^{\infty} |u(x)|^p dx < \infty$ for different values of $p \in [1, \infty)$.

Let's consider the behavior of $u(x)$ as $|x| \to \infty$. As $|x|$ becomes large, $1+x^2 \approx x^2$ and $2+x^2 \approx x^2$. So, for large $|x|$, $u(x)$ behaves like:

$u(x) \sim \frac{1}{(x^2)^{\frac{\alpha}{2}}} \frac{1}{\ln(x^2)} = \frac{1}{|x|^{\alpha}} \frac{1}{2\ln|x|}$.

Now, let's consider the integral $\int^{\infty} |u(x)|^p dx$. For large $x$, this integral behaves like:

$\int^{\infty} \left| \frac{1}{x^{\alpha}} \frac{1}{2\ln x} \right|^p dx = \int^{\infty} \frac{1}{|x|^{p\alpha}} \frac{1}{(2\ln x)^p} dx$.

For this integral to converge, we need the power of $x$ in the denominator to be greater than 1. That is, we need $p\alpha > 1$. This condition gives us a lower bound on $p$ for the integrability of $u(x)$. If $p\alpha \le 1$, the integral will diverge.

Let's think about the condition $p\alpha > 1$. Since $0 < \alpha < 1$, if $p\alpha > 1$, it means that $p > \frac{1}{\alpha}$. Since $\alpha < 1$, $\frac{1}{\alpha} > 1$. So, this condition automatically implies $p > 1$. If $p=1$, the condition becomes $\alpha > 1$, which is not true. So for $p=1$, we need to be careful.

If $p\alpha > 1$, the term $|x|^{-p\alpha}$ dominates the slower decaying term $(\ln x)^{-p}$. The integral $\int^{\infty} |x|^{-p\alpha} dx$ converges if and only if $p\alpha > 1$. Thus, for $u(x)$ to be in $L^p(\mathbb{R})$, we require $p\alpha > 1$, which implies $p > \frac{1}{\alpha}$.

What happens if $p\alpha \le 1$? If $p\alpha = 1$, then $u(x) \sim \frac{1}{|x|} \frac{1}{2\ln|x|}$. The integral $\int^{\infty} \frac{1}{x \ln x} dx$ diverges (by substitution $y=\ln x$, $dy = \frac{1}{x} dx$, so $\int \frac{1}{y} dy = \ln|y| = \ln|\ln x|$). So, if $p\alpha = 1$, $u(x)$ is not in $L^p(\mathbb{R})$.

If $p\alpha < 1$, then $|x|^{-p\alpha}$ grows slower than $|x|^{-1}$, and the integral $\int^{\infty} |x|^{-p\alpha} dx$ diverges even faster.

Therefore, for $u(x)$ to be in $L^p(\mathbb{R})$, we must have $p\alpha > 1$, which means $p > \frac{1}{\alpha}$. Since $0 < \alpha < 1$, we know $\frac{1}{\alpha} > 1$. So, this condition ensures $p > 1$.

This analysis tells us that for $u(x)$ to be $p$-integrable, we need $p > \frac{1}{\alpha}$. The behavior near $x=0$ is not an issue because $u(x)$ is bounded and well-behaved there. The crucial part is the decay at infinity.

Analyzing the Integrability of $u'(x)$

Now, let's get serious about the derivative $u'(x)$. Remember, $u'(x) = -\frac{\alpha x}{(1+x^2)^{\frac{\alpha}{2}+1} \ln(2+x^2)} - \frac{2x}{(1+x^2)^{\frac{\alpha}{2}}(2+x^2)(\ln(2+x^2))^2}$. We need to check if $\int_{-\infty}^{\infty} |u'(x)|^p dx < \infty$. Again, we focus on the behavior as $|x| \to \infty$.

As $|x| \to \infty$, $1+x^2 \approx x^2$ and $2+x^2 \approx x^2$. Let's analyze the dominant terms in $u'(x)$.

The first term behaves like: $- \frac{\alpha x}{(x^2)^{\frac{\alpha}{2}+1} \ln(x^2)} = - \frac{\alpha x}{x^{\alpha+2} (2\ln|x|)} = - \frac{\alpha}{2x^{\alpha+1} \ln|x|}$.

The second term behaves like: $- \frac{2x}{(x^2)^{\frac{\alpha}{2}}(x^2)(\ln(x^2))^2} = - \frac{2x}{x^{\alpha} x^2 (2\ln|x|)^2} = - \frac{2x}{x^{\alpha+2} (4(\ln|x|)^2)} = - \frac{1}{2x^{\alpha+1} (\ln|x|)^2}$.

So, for large $|x|$, $u'(x)$ behaves like:

$u'(x) \sim -\frac{\alpha}{2x^{\alpha+1} \ln|x|} - \frac{1}{2x^{\alpha+1} (\ln|x|)^2}$.

The term with the higher power in the logarithm dominates the decay. Since $(\ln|x|)^2$ grows faster than $\ln|x|$, the second term decays faster. Therefore, the dominant term is $-\frac{\alpha}{2x^{\alpha+1} \ln|x|}$.

So, $u'(x) \sim -\frac{\alpha}{2} \frac{1}{x^{\alpha+1} \ln|x|}$ for large $|x|$.

Now we need to check the integrability of $|u'(x)|^p$. For large $x$, this behaves like:

$\int^{\infty} \left| -\frac{\alpha}{2} \frac{1}{x^{\alpha+1} \ln x} \right|^p dx = \int^{\infty} \left( \frac{\alpha}{2} \right)^p \frac{1}{|x|^{p(\alpha+1)} (\ln x)^p} dx$.

For this integral to converge, we need the power of $x$ in the denominator to be greater than 1. That is, $p(\alpha+1) > 1$.

This condition $p(\alpha+1) > 1$ implies $p > \frac{1}{\alpha+1}$. Since $0 < \alpha < 1$, we have $1 < \alpha+1 < 2$. Thus, $\frac{1}{2} < \frac{1}{\alpha+1} < 1$. This condition alone doesn't guarantee $p > 1$, which is necessary for the $L^p$ norm. However, we are considering $p \in [1, \infty)$.

Let's analyze the condition $p(\alpha+1) > 1$ more closely. If $p(\alpha+1) > 1$, then $\int^{\infty} |x|^{-p(\alpha+1)} dx$ converges. The logarithmic term $(\ln x)^p$ decays slower than any power of $x$, so its presence doesn't prevent convergence if the power of $x$ is greater than 1.

So, the condition for $u'(x)$ to be in $L^p(\mathbb{R})$ boils down to $p(\alpha+1) > 1$, which simplifies to $p > \frac{1}{\alpha+1}$.

Let's check the edge case $p=1$. If $p=1$, the condition becomes $\alpha+1 > 1$, which is $\alpha > 0$. This is true since $0 < \alpha < 1$. So, $u'(x)$ is integrable for $p=1$ as long as $\alpha > 0$. The integral $\int^{\infty} \frac{1}{x^{\alpha+1} \ln x} dx$ converges because $\alpha+1 > 1$.

Therefore, for $u'(x)$ to be in $L^p(\mathbb{R})$, we require $p > \frac{1}{\alpha+1}$.

Bringing It All Together: The Condition for $u \in W^{1,p}(\mathbb{R})$

We've analyzed the integrability of both $u(x)$ and $u'(x)$. For $u$ to be in the Sobolev space $W^{1,p}(\mathbb{R})$, both conditions must be satisfied:

1. For $u \in L^p(\mathbb{R})$, we need $p > \frac{1}{\alpha}$.
2. For $u' \in L^p(\mathbb{R})$, we need $p > \frac{1}{\alpha+1}$.

We need to find the values of $p \in [1, \infty)$ that satisfy both of these inequalities.

Let's compare the two lower bounds: $\frac{1}{\alpha}$ and $\frac{1}{\alpha+1}$. Since $0 < \alpha < 1$, we have $\alpha+1 > \alpha$. Taking reciprocals reverses the inequality, so $\frac{1}{\alpha+1} < \frac{1}{\alpha}$.

This means that the condition $p > \frac{1}{\alpha}$ is stricter than $p > \frac{1}{\alpha+1}$. If $p$ satisfies $p > \frac{1}{\alpha}$, it automatically satisfies $p > \frac{1}{\alpha+1}$ because $\frac{1}{\alpha} > \frac{1}{\alpha+1}$.

Therefore, the combined condition for $u in W^{1,p}(\mathbb{R})$ is simply $p > \frac{1}{\alpha}$.

Since $0 < \alpha < 1$, we have $\frac{1}{\alpha} > 1$. So, the condition $p > \frac{1}{\alpha}$ means that $p$ must be strictly greater than some value greater than 1.

Thus, $u in W^{1,p}(\mathbb{R})$ for all $p \in \left( \frac{1}{\alpha}, \infty \right)$.

Remember that the problem asked for $p \in [1, \infty)$. Our result $p > \frac{1}{\alpha}$ falls within this range because $\frac{1}{\alpha} > 1$ (since $\alpha < 1$).

So, the final answer is that $u in W^{1,p}(\mathbb{R})$ for all $p$ such that $p > \frac{1}{\alpha}$. This is a really cool result, guys! It shows how the parameters in the function, like $\alpha$, dictate which function spaces it belongs to. It’s all about the decay rates of the function and its derivatives. Keep practicing these types of problems, and you'll become a pro at analyzing functions in Sobolev spaces!

Conclusion: Mastering Sobolev Spaces

We've successfully navigated through Brezis' Functional Analysis Exercise 8.1, determining the range of p for which our function $u(x) = \frac{1}{(1+x^2)^{\frac{\alpha}{2}}} \frac{1}{\ln(2+ x^2)}$ with $0 < \alpha < 1$ belongs to the Sobolev space $W^{1,p}(\mathbb{R})$. The key takeaway is that for $u$ to be in $W^{1,p}(\mathbb{R})$, both the function itself and its derivative must be $p$-integrable over $\mathbb{R}$. Through careful asymptotic analysis of $u(x)$ and $u'(x)$ as $|x| \to \infty$, we found that $u \in L^p(\mathbb{R})$ requires $p > \frac{1}{\alpha}$, and $u' \in L^p(\mathbb{R})$ requires $p > \frac{1}{\alpha+1}$. Since $\frac{1}{\alpha} > \frac{1}{\alpha+1}$ for $0 < \alpha < 1$, the stricter condition $p > \frac{1}{\alpha}$ dictates the range. Therefore, $u in W^{1,p}(\mathbb{R})$ for all $p \in \left( \frac{1}{\alpha}, \infty \right)$.

This exercise highlights the fundamental importance of understanding the decay rates of functions and their derivatives when working with function spaces like Sobolev spaces. The parameter $\alpha$ plays a crucial role in defining the integrability properties of $u$ and $u'$. This kind of analysis is fundamental in various areas of mathematics and physics, including the study of partial differential equations. Keep practicing these detailed analyses, guys. The more you work through these examples, the more intuitive the concepts of function spaces and their properties will become. Functional analysis might seem daunting at first, but with persistence and a good understanding of the core principles, you can master these challenging topics. Keep up the great work, and happy analyzing!