Problem 43 - A Course in Calculus and Real Analysis

Question

Let \(f_{1}, \ldots, f_{m}:[a, b] \rightarrow \mathbb{R}\) be integrable functions and let \(r_{j}:=\int_{a}^{b} f_{j}(x) d x\) for \(j=1, \ldots, m .\) Show that the function \(\sqrt{f_{1}^{2}+\cdots+f_{m}^{2}}\) is integrable and $$ \sqrt{r_{1}^{2}+\cdots+r_{m}^{2}} \leq \int_{a}^{b} \sqrt{f_{1}^{2}(x)+\cdots+f_{m}^{2}(x)} d x $$ (Hint: Note that \(\sum_{j=1}^{m} r_{j}^{2}=\sum_{j=1}^{m} r_{j} \int_{a}^{b} f_{j}(x) d x=\int_{a}^{b}\left(\sum_{j=1}^{m} r_{j} f_{j}(x)\right) d x\) and use Proposition 1.12.)

Step-by-Step Solution

Answer

In this problem, we need to prove the integrability of the function \(\sqrt{f_1^2 + \cdots + f_m^2}\) on the interval \([a, b]\) and show that \(\sqrt{r_{1}^{2}+\cdots+r_{m}^{2}} \leq \int_{a}^{b} \sqrt{f_{1}^{2}(x)+\cdots+f_{m}^{2}(x)} d x\). To do this, we first show that the function \(\sqrt{f_1^2 + \cdots + f_m^2}\) is integrable, since all \(f_j\) are integrable functions on \([a, b]\). Next, we follow the provided hint and use Proposition 1.12 to derive the desired inequality. This method allows us to obtain a clear expression involving \(\sum_{j=1}^{m} r_j^2\), which we then manipulate to arrive at the inequality: \(\sqrt{r_{1}^{2}+\cdots+r_{m}^{2}} \leq \int_{a}^{b} \sqrt{f_{1}^{2}(x)+\cdots+f_{m}^{2}(x)} d x\).

Step 1: Prove the integrability of \(\sqrt{f_1^2 + \cdots + f_m^2}\)

Since all \(f_j\) are integrable functions on \([a, b]\), their squares \(f_j^2\) are also integrable. The sum of integrable functions is integrable, so \(f_1^2 + \cdots + f_m^2\) is an integrable function. Now, since \(f_1^2 + \cdots + f_m^2 \geq 0\), the square root function is continuous, and hence, the composition of a continuous function with an integrable function (\(\sqrt{f_1^2 + \cdots + f_m^2}\)) is also integrable on \([a, b]\).

Step 2: Express \(\sum_{j=1}^{m} r_j^2\) in terms of integral

Given the hint, we can write \(\sum_{j=1}^{m} r_j^2\) as: $$\sum_{j=1}^{m} r_j^2 =\sum_{j=1}^{m} r_j \int_{a}^{b} f_j(x) d x =\int_{a}^{b}\left(\sum_{j=1}^{m} r_j f_j(x)\right) d x$$

Step 3: Apply Proposition 1.12

Proposition 1.12 states that if \(g\) is integrable and nonnegative on an interval \([c, d]\) and \(g(x) \leq h(x)\) for all \(x \in [c, d]\), then \(\int_{c}^{d} g(x) dx \leq \int_{c}^{d} h(x) dx\). Now, let \(g(x) = \sum_{j=1}^{m} r_j f_j(x)\) and \(h(x) = \sqrt{f_{1}^{2}(x)+\cdots+f_{m}^{2}(x)}\). Notice that by the Cauchy-Schwarz inequality for finite sequences, we have $$g(x)^2 = \left(\sum_{j=1}^{m} r_j f_j(x)\right)^2 \leq \left(\sum_{j=1}^{m} r_j^2\right)\left(\sum_{j=1}^{m} f_j^2(x)\right) = h(x)^2 \sum_{j=1}^{m} r_j^2$$ Taking the square root of both sides, we get \(g(x) \leq h(x) \sqrt{\sum_{j=1}^{m} r_j^2}\) for all \(x \in [a, b]\). As the inequality holds for all \(x\) on the interval, we can apply Proposition 1.12, which gives $$\int_{a}^{b} g(x) dx \leq \sqrt{\sum_{j=1}^{m} r_j^2} \int_{a}^{b} h(x) dx$$ Since we already showed in Step 2 that $$\int_{a}^{b} g(x) dx = \sum_{j=1}^{m} r_j^2,$$ we can substitute this into the inequality above, which gives $$\sum_{j=1}^{m} r_j^2 \leq \sqrt{\sum_{j=1}^{m} r_j^2} \int_{a}^{b} h(x) dx$$

Step 4: Obtain the final inequality

Since \(\sum_{j=1}^{m} r_j^2 \geq 0\), we can safely divide both sides of the inequality obtained in Step 3 by \(\sqrt{\sum_{j=1}^{m} r_j^2}\) and get the desired inequality: $$\sqrt{r_{1}^{2}+\cdots+r_{m}^{2}} \leq \int_{a}^{b} \sqrt{f_{1}^{2}(x)+\cdots+f_{m}^{2}(x)} d x$$