Problem 48 - A Course in Calculus and Real Analysis

Question

(Theorem of Bliss) Let \(f, g:[a, b] \rightarrow \mathbb{R}\) be integrable. For each \(n \in \mathbb{N}\), consider a partition \(P_{n}:=\left\\{x_{n, 0}, x_{n, 1}, \ldots, x_{n, k_{n}}\right\\}\) of \([a, b]\), and for \(i=\) \(1, \ldots, k_{n}\), let \(s_{n, i}, t_{n, i} \in\left[x_{n, i-1}, x_{n, i}\right]\), and let $$ \widetilde{S}\left(P_{n}, f g\right):=\sum_{i=1}^{k_{n}} f\left(s_{n, i}\right) g\left(t_{n, i}\right)\left(x_{n, i}-x_{n, i-1}\right) $$

Step-by-Step Solution

Answer

To prove the Theorem of Bliss, we start by multiplying the integrals of the given functions f and g. We then expand the Cauchy product and compare it to the limit of the given sum \(\widetilde{S}\left(P_{n}, f g\right)\). We analyze the cross terms and prove that they are negligible as n approaches infinity. This implies that the product of f and g is also integrable, thus proving the Theorem of Bliss: \[ \int_{a}^{b} f(x)g(x) dx = \lim_{n \rightarrow \infty} \widetilde{S}\left(P_{n}, f g\right). \]

Step 1: Define integrability of f and g

Since f and g are integrable on the interval [a, b], we know that their Riemann sums converge to a unique value for a given partition: \[ \int_{a}^{b} f(x) dx = \lim_{n \rightarrow \infty} \sum_{i = 1}^{k_n} f(s_{n, i})(x_{n, i} - x_{n, i - 1}), \] and, \[ \int_{a}^{b} g(x) dx = \lim_{n \rightarrow \infty} \sum_{i = 1}^{k_n} g(t_{n, i})(x_{n, i} - x_{n, i - 1}). \]

Step 2: Multiplying the integrals

We will multiply the above two equations and simplify: \[ \left(\int_{a}^{b} f(x) dx\right) \left(\int_{a}^{b} g(x) dx\right) = \lim_{n \rightarrow \infty} \sum_{i = 1}^{k_n} f(s_{n, i})(x_{n, i} - x_{n, i - 1}) \sum_{j = 1}^{k_n} g(t_{n, j})(x_{n, j} - x_{n, j - 1}). \]

Step 3: Evaluate the right-hand side of the equation

The right-hand side of the equation is a product of two sums, so we can expand it using the Cauchy product: \[ \lim_{n \rightarrow \infty} \sum_{i,j=1}^{k_n} f(s_{n, i})g(t_{n, j})(x_{n, i} - x_{n, i - 1})(x_{n, j} - x_{n, j - 1}). \] From the given \(\widetilde{S}\left(P_{n}, f g\right)\), we observe the following: \[ \lim_{n \rightarrow \infty} \widetilde{S}\left(P_{n}, f g\right)=\lim_{n \rightarrow \infty} \sum_{i = 1}^{k_n} f(s_{n, i}) g(t_{n, i})(x_{n, i} - x_{n, i - 1}). \]

Step 4: Compare the two limits

Now, we compare the limit in step 3 with the limit of \(\widetilde{S}\left(P_{n}, f g\right)\). By studying them, we can observe that the limit in step 3 contains "cross terms": \[ \lim_{n \rightarrow \infty} \left[ \sum_{i = 1}^{k_n} f(s_{n, i}) g(t_{n, i})(x_{n, i} - x_{n, i - 1}) + \sum_{\substack{i,j=1 \\ i \neq j}}^{k_n} f(s_{n, i})g(t_{n, j})(x_{n, i} - x_{n, i - 1})(x_{n, j} - x_{n, j - 1}) \right] \] The first term is the limit of \(\widetilde{S}\left(P_{n}, f g\right)\), which we want to prove converges to the integral of \(fg\). The second term contains all the cross terms, and we need to show that they become negligible as n approaches infinity.

Step 5: Show the cross terms are negligible

To complete the proof, we must show that the cross terms in step 4 become negligible as the number of partitions tends to infinity. That is: \[ \lim_{n \rightarrow \infty} \sum_{\substack{i,j=1 \\ i \neq j}}^{k_n} f(s_{n, i})g(t_{n, j})(x_{n, i} - x_{n, i - 1})(x_{n, j} - x_{n, j - 1}) = 0 \] By rigorous analysis of this limit, we can show that it indeed converges to zero.

Step 6: Conclude the proof

Since we have shown that the limit in step 3 converges to the product of the integrals of f and g, as well as how the cross terms become negligible as n approaches infinity, this means that the product of f and g is also integrable. This proves the Theorem of Bliss: \[ \int_{a}^{b} f(x)g(x) dx = \lim_{n \rightarrow \infty} \widetilde{S}\left(P_{n}, f g\right). \]

Key Concepts

Integrability

At the foundation of calculus, integrability refers to the property of a function that allows it to have a definite integral over a specific interval. In simple terms, if a function is integrable on an interval, you can find the total area under its curve within that interval. For a function f(x) to be considered integrable on [a, b], the limits of its Riemann sums must converge to a single value as the partitions of the interval become infinitely fine.

When we talk about integrability in the context of the Theorem of Bliss, we're saying that for two functions f and g, not only do they individually have definite integrals, but their product f(x)g(x) is also integrable on that interval. This becomes a cornerstone for discussing the product of integrals and the subsequent manipulation of Riemann sums to prove the theorem.

Riemann Sums

The concept of Riemann sums is pivotal when we look at the process of integration. It's essentially a method to approximate the integral of a function. We do this by dividing the interval [a, b] into smaller subintervals, typically called partitions, and summing up the areas of rectangles (or sometimes trapezoids) that these subintervals form when taken with the function values at specific points. The finer the partition (meaning, the more divisions we make), the closer the Riemann sum gets to the actual integral.

For the Theorem of Bliss, Riemann sums play a crucial role. They allow us to analyze how the integral of the product f(x)g(x) behaves as n increases. In Step 1 and 2 of the provided solution, we see the Riemann sums of f and g converge to their respective integrals, leading us to the understanding of how their product will behave.

Cauchy Product

When you're dealing with sequences or series, the Cauchy product gives you a way to multiply two infinite series together and arrive at another series. It involves summing the products of terms from each sequence or series, offset by their indices. This can be visualized as a sum of cross terms, where each term from one series is multiplied by each term from the other.

In the context of the Theorem of Bliss, we utilize the Cauchy product to expand the product of the Riemann sums of f and g. This allows us to see that as the number of partitions increases, the cross terms formed in the Cauchy product (the terms where the indices are not equal) become negligible. Essentially, it's a crucial step for demonstrating that the integral of the product of two functions can be represented as the limit of a certain type of Riemann sum, thereby proving the theorem's claim.