Question

If \(f\) is absolutely continuous on \([\epsilon, 1]\) for \(0<\epsilon<1\) and \(\int_{0}^{1} x\left|f^{\prime}(x)\right|^{p} d x<\infty\), then \(\lim _{x \rightarrow 0} f(x)\) exists (and is finite) if \(p>2,|f(x)| /|\log x|^{1 / 2} \rightarrow 0\) as \(x \rightarrow 0\) if \(p=2\), and \(|f(x)| / x^{1-(2 / p)} \rightarrow 0\) as \(x \rightarrow 0\) if \(p<2\).

Short answer

The limit of \( f(x) \) as \( x \to 0 \) exists and is finite in each case based on conditions given for \( p \).

Step-by-step solution

Understand the Conditions

The problem gives conditions that help determine the behavior of a function \( f \) as \( x \) approaches 0. It assumes \( f \) is absolutely continuous on \( [\epsilon, 1] \) and involves the integral \( \int_{0}^{1} x |f'(x)|^{p} dx < \infty \). The exercise asks about the limit of \( f(x) \) as \( x \rightarrow 0 \) under different scenarios based on \( p \).

Analyze the Case \( p > 2 \)

If \( p > 2 \), the fact that \( f \) is absolutely continuous and \( \int_{0}^{1} x |f'(x)|^{p} dx < \infty \) ensures that \( f(x) \) doesn't oscillate too much. The finiteness of this integral implies control over the behavior of \( f \) near 0, suggesting that the limit exists and is finite as \( x \rightarrow 0 \).

Consider the Case \( p = 2 \)

When \( p = 2 \), an additional condition is given: \( |f(x)| / |\log x|^{1/2} \rightarrow 0 \) as \( x \rightarrow 0 \). This condition ensures that any growth in the function \( f(x) \) is slower than \( |\log x|^{1/2} \), restricting it from diverging as \( x \rightarrow 0 \) and ensuring a finite limit exists.

Examine the Case \( p < 2 \)

In this scenario, \( |f(x)| / x^{1-(2/p)} \rightarrow 0 \) as \( x \rightarrow 0 \). Here, the condition means that the rate of growth of \( f(x) \) must be slower than \( x^{1-(2/p)} \), serving as a stricter requirement to control \( f(x) \) as it approaches 0, ensuring the limit exists.

Key concepts

Absolutely Continuous

Absolutely continuous functions have a special property that makes them easy to work with in calculus, especially when it comes to integration. A function is called absolutely continuous on an interval if, for any tiny positive number (no matter how small), we can divide the interval into even smaller segments such that the sum of changes of the function on these segments is as small as we want.
This concept helps ensure that the function behaves predictably over its entire domain, particularly when integrating. If a function is absolutely continuous, we can guarantee that it has a well-behaved derivative and that we can recover the original function from knowing its derivative and an initial value, like piecing a puzzle back together.
For the given problem, assuming the absolute continuity of function \( f \) over the interval \([\epsilon, 1] \) holds a significant implication. It means that the integral of the derivative of \( f \) weighted by \( x \) won't be too large or undefined, helping ensure we can talk sensibly about the function's limit as \( x \) approaches zero.

Limit of a Function

The limit of a function at a point is core to understanding how the function behaves as it gets closer to that point. In real analysis, we often look at what happens to \( f(x) \) as \( x \rightarrow 0 \). Essentially, we are asking what value \( f(x) \) is inching towards as \( x \) gets smaller and smaller.In this exercise, we explore different cases of \( p \), influencing how the limit exists and behaves. When \( p > 2 \), the function tends not to have wild fluctuations near zero, which often leads to a tidy, finite limit. When \( p = 2 \), we have to ensure that \( f(x) \) grows slower than \( \log(x)^{1/2} \) to keep things under control, yet still allowing a finite limit. With \( p < 2 \), we impose stricter growth conditions such that \( f(x) \) softens as it approaches zero, avoiding infinity. These cases help us predict what happens to \( f(x) \) in a variety of scenarios as \( x \) heads towards zero.

Integral Convergence

Understanding when an integral converges is a cornerstone of real analysis. Convergence means that the integral has a finite value. This is important as it reflects on how the function behaves over an interval or as a parameter approaches a certain value. In this problem, we're specifically interested in the integral \[\int_{0}^{1} x |f'(x)|^{p} dx < \infty \]and its behavior.
The convergence of this integral over \( [0, 1] \) suggests that the impact of the function's derivative, when multiplied by \( x \), isn't explosive or unsustainably large near zero - it is set to a manageable size. When such an integral converges, it implies that the average behavior of \( f'(x) \) is under control, and thus \( f \) isn't diverging wildly as \( x \rightarrow 0\). It provides a necessary condition that helps in concluding about the limit of \( f(x) \) as \( x \rightarrow 0 \).

Behavior Near Zero

As \( x \) approaches zero, understanding the behavior of a function is crucial for many real analysis problems. Functions can behave erratically, grow infinitely large, or stabilize as \( x \rightarrow 0 \). The challenge is to make predictions about their behavior without using direct computation.The exercise provides a clear lens into this by giving conditions under which \( f(x) \) can stabilize or converge as \( x \rightarrow 0 \). By testing different scenarios of \( p \), it teaches us how control conditions (like slower growth rates) help reach conclusions about the function's behavior. When \( p \) is greater than 2, the function's behavior near zero is quite simple as the conditions automatically suggest convergent behavior. As \( p \) equals or goes below 2, we start needing additional restrictions on \( f(x) \)'s growth to ensure it stabilizes. This exploration helps us wrap intuition around why certain conditions are necessary and how they affect functions we handle in analysis.