Defining Nonnegative Function With Supremum Condition Real Analysis Exploration

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This article explores the fascinating question of defining a nonnegative function $f$ with specific properties on a particular domain. We delve into the realm of real analysis, sequences, series, and functions to determine if such a function can exist. Specifically, we investigate whether we can define a function $f: X o [0, extbf{∞})$ for $X = {1/n | n ∈ extbf{N}} ∪ {0}$ such that the supremum of a certain ratio involving $f$ is less than 1. This problem combines elements of sequence behavior and functional inequalities, making it a compelling topic in mathematical analysis. Understanding the nuances of function definition and supremum properties is crucial in tackling this question. The exploration involves carefully analyzing the given condition and constructing a function that satisfies it, or proving that no such function can exist. The domain $X$, consisting of reciprocals of natural numbers and zero, adds a discrete flavor to the problem, requiring us to consider the behavior of $f$ at these specific points. The condition involving the supremum introduces an inequality constraint that must hold for all natural numbers $n$, making the problem both intriguing and challenging.

Problem Statement: Defining a Nonnegative Function

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Let's formally state the problem we aim to address: Can we define a function $f: X o [0, ∞)$ where $X = {1/n | n ∈ N} ∪ {0}$, such that $\sup_{n∈N} \frac{\frac{1}{n+1} + f(\frac{1}{n+1})}{\frac{1}{n} + f(\frac{1}{n})} < 1$? This question lies at the intersection of real analysis and function theory. We're tasked with constructing a function that satisfies a particular inequality condition involving its values at reciprocal integers. The key challenge is ensuring the supremum of the ratio remains strictly less than 1. This constraint puts a significant restriction on how $f$ can behave as $n$ varies. Our approach will involve analyzing the implications of this supremum condition and attempting to construct such a function or proving its non-existence. The domain $X$ includes the reciprocals of all natural numbers, which form a sequence converging to 0, along with 0 itself. This suggests that the behavior of $f$ near 0 might play a crucial role in determining whether the supremum condition can be satisfied. The non-negativity of $f$, i.e., $f(x) ≥ 0$ for all $x ∈ X$, is another essential constraint that guides our search for a solution. We need to find a balance such that the ratio decreases sufficiently fast as $n$ increases, ensuring the supremum is less than 1. This exploration will shed light on the interplay between function values and the given supremum inequality.

Analyzing the Supremum Condition

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The heart of the problem lies in the supremum condition: $\sup_{n∈N} \frac{\frac{1}{n+1} + f(\frac{1}{n+1})}{\frac{1}{n} + f(\frac{1}{n})} < 1$. To understand this condition, let's denote $x_n = \frac{1}{n}$. Then, the condition can be rewritten as $\sup_{n∈N} \frac{x_{n+1} + f(x_{n+1})}{x_n + f(x_n)} < 1$. This means that for all $n ∈ N$, we must have $\frac{x_{n+1} + f(x_{n+1})}{x_n + f(x_n)} < 1$. This inequality implies that $x_{n+1} + f(x_{n+1}) < x_n + f(x_n)$ for all $n ∈ N$. Let's define a sequence $a_n = x_n + f(x_n) = \frac{1}{n} + f(\frac{1}{n})$. The inequality then becomes $a_{n+1} < a_n$ for all $n ∈ N$. This tells us that the sequence $(a_n)$ must be strictly decreasing. Since $f$ is a nonnegative function, i.e., $f(x) ≥ 0$ for all $x ∈ X$, we have $a_n = \frac{1}{n} + f(\frac{1}{n}) > \frac{1}{n} > 0$. Thus, $(a_n)$ is a strictly decreasing sequence bounded below by 0. This means that $(a_n)$ converges to some limit $L ≥ 0$. The question now is whether we can construct a function $f$ such that this convergence occurs while maintaining the strict decrease of the sequence $(a_n)$. The supremum condition places a strong constraint on the rate at which $a_n$ decreases. If $a_n$ decreases too slowly, the supremum might be equal to 1, violating the condition. Conversely, if $a_n$ decreases too rapidly, it might force $f$ to take on specific values that are difficult to manage. Understanding this balance is key to solving the problem. We need to carefully consider how the choice of $f(\frac{1}{n})$ affects the sequence $(a_n)$ and whether we can fine-tune $f$ to satisfy both the strict decrease and the supremum condition. This involves delving deeper into the properties of sequences, functions, and inequalities.

Constructing a Candidate Function

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To construct a candidate function, we need to carefully consider the implications of the condition $a_{n+1} < a_n$, where $a_n = \frac{1}{n} + f(\frac{1}{n})$. We also need to ensure that $f$ is nonnegative. A natural starting point is to try a linear function or a function that scales with $\frac{1}{n}$. Let's explore the possibility of $f(\frac{1}{n}) = c \frac{1}{n}$ for some constant $c ≥ 0$. In this case, $a_n = \frac{1}{n} + c \frac{1}{n} = (1+c) \frac{1}{n}$. Then, the ratio in the supremum condition becomes:

$\frac{\frac{1}{n+1} + f(\frac{1}{n+1})}{\frac{1}{n} + f(\frac{1}{n})} = \frac{\frac{1}{n+1} + c \frac{1}{n+1}}{\frac{1}{n} + c \frac{1}{n}} = \frac{(1+c) \frac{1}{n+1}}{(1+c) \frac{1}{n}} = \frac{n}{n+1}$

The supremum of this ratio over all $n ∈ N$ is:

$\sup_{n∈N} \frac{n}{n+1} = 1$

This shows that a linear function of the form $f(\frac{1}{n}) = c \frac{1}{n}$ does not satisfy the condition $\sup_{n∈N} \frac{\frac{1}{n+1} + f(\frac{1}{n+1})}{\frac{1}{n} + f(\frac{1}{n})} < 1$. This is because the ratio approaches 1 as $n$ tends to infinity. We need a function that decreases more rapidly than $\frac{1}{n}$ to ensure the ratio stays strictly below 1. Let's consider a function of the form $f(\frac{1}{n}) = \frac{c}{n^p}$ where $p > 1$ and $c ≥ 0$. This function decreases faster than $\frac{1}{n}$ as $n$ increases. Substituting this into the ratio, we get:

$\frac{\frac{1}{n+1} + \frac{c}{(n+1)^p}}{\frac{1}{n} + \frac{c}{n^p}} = \frac{\frac{1}{n+1}(1 + \frac{c}{(n+1)^{p-1}})}{\frac{1}{n}(1 + \frac{c}{n^{p-1}})} = \frac{n}{n+1} \cdot \frac{1 + \frac{c}{(n+1)^{p-1}}}{1 + \frac{c}{n^{p-1}}}$

Now, we need to analyze the behavior of this expression as $n$ becomes large and see if we can choose $c$ and $p$ such that the supremum is less than 1. The term $\frac{n}{n+1}$ still approaches 1 as $n$ approaches infinity, so we need to ensure that the second term $\frac{1 + \frac{c}{(n+1)^{p-1}}}{1 + \frac{c}{n^{p-1}}}$ is strictly less than 1 and decreases sufficiently fast to compensate for the $\frac{n}{n+1}$ term. This requires a more detailed analysis of the second term's behavior.

Further Analysis and Potential Solutions

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Let's further analyze the term $\frac{1 + \frac{c}{(n+1)^{p-1}}}{1 + \frac{c}{n^{p-1}}}$ from the previous section. We can rewrite it as follows:

$\frac{1 + \frac{c}{(n+1)^{p-1}}}{1 + \frac{c}{n^{p-1}}} = \frac{(n+1)^{p-1} + c}{n^{p-1} + c} \cdot \frac{n^{p-1}}{(n+1)^{p-1}} = \frac{n^{p-1}}{(n+1)^{p-1}} \cdot \frac{(n+1)^{p-1} + c}{n^{p-1} + c}$

For the supremum condition to hold, we need this expression to be less than 1 for all $n ∈ N$. This expression involves powers of $n$ and $n+1$, making it challenging to directly compute the supremum. However, we can analyze its behavior as $n$ becomes large. Let's consider the case when $p = 2$. The expression becomes:

$\frac{1 + \frac{c}{n+1}}{1 + \frac{c}{n}} = \frac{n(n+1+c)}{(n+1)(n+c)} = \frac{n^2 + n + cn}{n^2 + n + cn + c} = 1 - \frac{c}{n^2 + n + cn + c}$

In this case, the ratio becomes:

$\frac{n}{n+1} \cdot \left(1 - \frac{c}{n^2 + n + cn + c}\right) = \frac{n}{n+1} - \frac{cn}{(n+1)(n^2 + n + cn + c)}$

As $n$ tends to infinity, the first term approaches 1, while the second term approaches 0. This suggests that for large enough $n$, the ratio will be less than 1. However, we need to ensure this holds for all $n ∈ N$ and that the supremum is strictly less than 1. To make the analysis more rigorous, we can consider the function $g(x) = \frac{x}{x+1} \left(1 - \frac{c}{x^2 + x + cx + c}\right)$ for $x ≥ 1$. We need to find the supremum of $g(x)$ on the interval $[1, ∞)$. Taking the derivative of $g(x)$ and setting it to 0 would give us the critical points, which could help us determine the supremum. This approach involves calculus and may lead to a more precise understanding of the conditions on $c$ for which the supremum is less than 1. Another approach is to choose a specific value for $f(\frac{1}{n})$ that decreases rapidly enough to offset the $\frac{n}{n+1}$ term. For instance, we could try $f(\frac{1}{n}) = e^{-n}$. This function decreases very quickly as $n$ increases. In this case, the ratio becomes:

$\frac{\frac{1}{n+1} + e^{-(n+1)}}{\frac{1}{n} + e^{-n}}$

Analyzing the supremum of this ratio directly might be challenging, but we can consider the limit as $n$ approaches infinity. The terms $e^{-(n+1)}$ and $e^{-n}$ will approach 0 much faster than $\frac{1}{n+1}$ and $\frac{1}{n}$, respectively. This suggests that for large enough $n$, the ratio will behave like $\frac{\frac{1}{n+1}}{\frac{1}{n}} = \frac{n}{n+1}$, which approaches 1. However, the exponential terms might help keep the ratio strictly less than 1 for all $n$, but this requires further investigation. We could also consider a recursive definition for $f(\frac{1}{n})$. We could set a value for $f(1)$ and then define $f(\frac{1}{n+1})$ in terms of $f(\frac{1}{n})$ such that the ratio $\frac{\frac{1}{n+1} + f(\frac{1}{n+1})}{\frac{1}{n} + f(\frac{1}{n})}$ is always less than some constant $k < 1$. This approach might lead to a function that satisfies the supremum condition, but it requires careful selection of the recursion formula.

Conclusion: Existence of such Nonnegative function

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In conclusion, the problem of defining a nonnegative function $f: X o [0, ∞)$ for $X = {1/n | n ∈ N} ∪ {0}$ such that $\sup_{n∈N} \frac{\frac{1}{n+1} + f(\frac{1}{n+1})}{\frac{1}{n} + f(\frac{1}{n})} < 1$ is a complex one that requires careful analysis. We have explored several candidate functions and approaches, including linear functions, power functions, and exponential functions. While a simple linear function does not satisfy the condition, functions that decrease more rapidly, such as $\frac{1}{n^p}$ with $p > 1$ or $e^{-n}$, show promise. The key challenge lies in ensuring that the ratio $\frac{\frac{1}{n+1} + f(\frac{1}{n+1})}{\frac{1}{n} + f(\frac{1}{n})}$ is strictly less than 1 for all $n$ and that the supremum of this ratio is also strictly less than 1. Further investigation, possibly involving calculus or a recursive definition of $f$, is needed to definitively determine whether such a function exists and, if so, to construct it explicitly. The problem highlights the interplay between sequence behavior, functional inequalities, and the concept of supremum. While we have not yet arrived at a definitive answer, the exploration has provided valuable insights into the properties that such a function must possess. We need a function that decreases at a rate that balances the decrease in $\frac{1}{n}$ as $n$ increases, ensuring the ratio stays bounded below 1. The quest for this function continues to be an intriguing journey in the realm of real analysis.