Problem 64 - A Course in Calculus and Real Analysis

Question

Let \(\Gamma_{\ell}:(0, \infty) \rightarrow \mathbb{R}\) be defined by \(\Gamma_{\ell}(s)=\ln \Gamma(s)\). Show that \(\Gamma_{\ell}\) is a convex function. (Hint: Use Exercise 55 (iv) of Chapter 7 to show that if \(p, q \in(1, \infty)\) with \(\frac{1}{p}+\frac{1}{q}=1\), then \(\Gamma((s / p)+(u / q)) \leq \Gamma(s)^{1 / p} \Gamma(u)^{1 / q}\) for all \(\left.s, u \in(0, \infty) .\right)\)

Step-by-Step Solution

Answer

Using Hölder's Inequality and properties of the gamma function, we show that \[\Gamma((s/p)+(u/q)) \leq \Gamma(s)^{\frac{1}{p}}\Gamma(u)^{\frac{1}{q}}.\] for all \(s, u \in (0, \infty)\) and for all \(p, q \in (1, \infty)\) such that \(\frac{1}{p}+\frac{1}{q}=1\). Then, we take the natural logarithm of both sides to obtain \[\Gamma_{\ell}((s/p) + (u/q)) \leq \frac{1}{p}\Gamma_{\ell}(s) + \frac{1}{q}\Gamma_{\ell}(u),\] which proves that \(\Gamma_{\ell}\) is a convex function.

Step 1: Hölder's Inequality

Recall Hölder's Inequality. It states that for any non-negative sequences \((a_k)\) and \((b_k)\), and for real numbers \(p, q \in (1, \infty)\) such that \(\frac{1}{p} + \frac{1}{q} = 1\), we have: \(\sum_k {a_k b_k} \leq \left(\sum_k a_k^p\right)^{1/p} \left(\sum_k b_k^q\right)^{1/q}\).

Step 2: Use Hölder's Inequality with the gamma function and its properties

Suppose \(s, u > 0\) are any positive numbers, and define two sequences \((a_k)_k\) and \((b_k)_k\) as: \(a_k = \frac{k^{s-1}e^{-k}}{k^s}\) and \(b_k = \frac{k^{u-1}e^{-k}}{k^u}\). Now, since the gamma function is defined as \(Γ(t) = \int_{0}^{\infty} x^{t-1} e^{-x} dx\), we can use this definition along with Hölder's Inequality to get: \(\sum_k a_kb_k \leq \left(\sum_k a_k^p\right)^{1/p} \left(\sum_k b_k^q\right)^{1/q}\) which is equivalent to, \(\sum_k \frac{k^{s-1}e^{-k}k^{u-1}e^{-k}}{k^{s+u}} \leq \left(\sum_k M(k, s)^p\right)^{1/p} \left(\sum_k M(k, u)^q\right)^{1/q}\), where \(M(k, t) = \frac{k^{t-1}e^{-k}}{k^t}\).

Step 3: Simplify the inequality

By simplifying the inequality, we get: \[\sum_k k^{s+u-2}e^{-2k} \leq \Gamma(s)^{\frac{1}{p}}\Gamma(u)^{\frac{1}{q}}.\] Now integrate both sides with respect to \(s\) and \(u\), we obtain \[ \int_{0}^{\infty} k^{(s/p)+(u/q)-1} e^{-k} dk \leq \Gamma(s)^{\frac{1}{p}}\Gamma(u)^{\frac{1}{q}}.\] Notice that the left side of the inequality is precisely the definition of the gamma function for \(((s/p)+(u/q))\). Hence, we have \[\Gamma((s/p)+(u/q)) \leq \Gamma(s)^{\frac{1}{p}}\Gamma(u)^{\frac{1}{q}}.\]

Step 4: Take the logarithm on both sides

Now, take the natural logarithm (ln) on both sides of the inequality: \[\ln \Gamma((s/p)+(u/q)) \leq \frac{1}{p} \ln \Gamma(s) + \frac{1}{q} \ln \Gamma(u),\] which can be rewritten as: \[\Gamma_{\ell}((s/p) + (u/q)) \leq \frac{1}{p}\Gamma_{\ell}(s) + \frac{1}{q}\Gamma_{\ell}(u),\] where \(\Gamma_{\ell}(t) = \ln \Gamma(t)\). This completes the proof that \(\Gamma_{\ell}\) is convex.

Key Concepts

Gamma Function

The Gamma Function is a vital part of mathematics that extends the concept of factorials beyond the natural numbers. Usually, the factorial of a positive integer \(n\) is \(n!\), which means \(n \times (n-1) \times (n-2) \ldots \times 1\). But what if you wanted to find the factorial for a number that's not an integer, like \(3.5!\)? Here’s where the Gamma Function steps in. The Gamma Function \(\Gamma(t)\) is defined as: \[\Gamma(t) = \int_{0}^{\infty} x^{t-1} e^{-x} \, dx\] This integral converges for any positive real \(t\). So, how does the Gamma function capture the factorial? For positive integers \(n\), it turns out that:

• \(\Gamma(n) = (n-1)!\)

This property elegantly extends the notion of factorials to the real and complex numbers, except negative integers.To sum up, the Gamma Function is essential for generalizing factorial definitions and appears in many areas such as probability, statistics, and various fields of physics.

Hölder's Inequality

Hölder's Inequality is a powerful tool in analysis that is used to bound the absolute value of an integral or sum. In simple terms, it's about controlling the size of combined sequences or functions using separate sizes.The inequality states that for any non-negative sequences \((a_k)\) and \((b_k)\), and real numbers \(p, q \in (1, \infty)\) with \(\frac{1}{p} + \frac{1}{q} = 1\), the following holds: \[\sum_k {a_k b_k} \leq \left(\sum_k a_k^p\right)^{1/p} \left(\sum_k b_k^q\right)^{1/q}\]Here’s the takeaway:

• It connects the product of sums to the individual sums raised to certain powers.
• It ensures that you can't "oversize" a product sum compared to the individual sums under weighted power conditions.

When we use it with the Gamma Function, as seen in the original exercise, Hölder's Inequality aids in proving other sophisticated properties like convexity by allowing one to break down complexities into manageable parts.

Logarithmic Functions

Logarithmic functions are the inverses of exponential functions and are crucial for dealing with exponential growth or decay situations. The natural logarithm, denoted as \(\ln(x)\), is the logarithm to the base \(e\), where \(e\) is an irrational and transcendental number roughly equal to 2.71828.Some key aspects of logarithmic functions to remember are:

• They transform multiplication into addition. For instance, \(\ln(ab) = \ln(a) + \ln(b)\).
• The function \(\ln(x)\) grows very slow as \(x\) increases.
• They simplify the process of dealing with products, powers, and roots. For example, \(\ln(x^y) = y\ln(x)\).

In the context of the exercise, the logarithm of the Gamma Function, expressed as \(\Gamma_\ell(s) = \ln \Gamma(s)\), was used to demonstrate convexity. This is because taking logarithms can simplify inequalities and bring out their true nature, as with the inequality \(\Gamma((s/p)+(u/q)) \leq \Gamma(s)^{1/p} \Gamma(u)^{1/q}\). Logarithms are inherently linked to concepts of growth and scaling, making them invaluable for mathematical proofs and analysis.