Question

In this problem we show that pointwise convergence of a sequence \(S_{n}(x)\) does not imply mean convergence, and conversely. (a) Let \(S_{n}(x)=n \sqrt{x} e^{-n x^{2} / 2}, 0 \leq x \leq 1 .\) Show that \(S_{n}(x) \rightarrow 0\) as \(n \rightarrow \infty\) for each \(x\) in \(0 \leq x \leq 1 .\) Show also that $$ R_{n}=\int_{0}^{1}\left[0-S_{n}(x)\right]^{2} d x=\frac{n}{2}\left(1-e^{-n}\right) $$ and hence that \(R_{n} \rightarrow \infty\) as \(n \rightarrow \infty .\) Thus pointwise convergence does not imply mean convergence. (b) Let \(S_{n}(x)=x^{n}\) for \(0 \leq x \leq 1\) and let \(f(x)=0\) for \(0 \leq x \leq 1 .\) Show that $$ R_{n}=\int_{0}^{1}\left[f(x)-S_{n}(x)\right]^{2} d x=\frac{1}{2 n+1} $$ and hence \(S_{n}(x)\) converges to \(f(x)\) in the mean. Also show that \(S_{n}(x)\) does not converge to \(f(x)\) pointwise throughout \(0 \leq x \leq 1 .\) Thus mean convergence does not imply pointwise convergence.

Short answer

Question: Provide counterexamples showing that pointwise convergence and mean convergence are not equivalent. Answer: For part (a), consider the sequence \(S_{n}(x)=n \sqrt{x} e^{-n x^{2} / 2}\). This sequence converges pointwise to 0 as \(n\rightarrow\infty\), but its mean square error, \(R_n = \frac{n}{2}(1-e^{-n})\), diverges to infinity as \(n\rightarrow\infty\). This demonstrates that pointwise convergence does not imply mean convergence. For part (b), consider the sequence \(S_{n}(x)=x^{n}\). This sequence converges in the mean to \(f(x)=0\), since the mean square error \(R_n = \frac{1}{2n+1}\) goes to 0 as \(n\rightarrow\infty\). However, the sequence does not converge pointwise to \(f(x)\) in the entire interval \([0, 1]\), as the limit of \(S_{n}(x)\) is 0 for \(0 < x < 1\), and 1 for \(x=1\). This shows that mean convergence does not imply pointwise convergence throughout the entire domain.

Step-by-step solution

Part (a): Pointwise Convergence

To show pointwise convergence of the sequence \(S_{n}(x)\) to 0 as \(n\rightarrow\infty\), we will find the limit of \(S_{n}(x)\) as \(n\rightarrow\infty\) for each fixed \(x\) in \(0 \leq x \leq 1\): $$ \lim_{n\rightarrow\infty} S_{n}(x) = \lim_{n\rightarrow\infty}\left(n \sqrt{x} e^{-n x^{2} / 2}\right) = 0 $$ The limit is 0 because \(e^{-n x^{2} / 2}\) approaches 0 faster than the remaining term \(n \sqrt{x}\) grows.

Part (a): Compute \(R_{n}\)

Next, we will compute the mean square error \(R_n\) as follows: \begin{align*} R_{n}&=\int_{0}^{1}\left[0-S_{n}(x)\right]^{2} dx \\ &=\int_{0}^{1}\left[n^{2} x e^{-n x^{2}}\right] dx \\ &= \frac{n}{2}\int_{0}^{1}\left[n x^{3} e^{-n x^{2}}\right] dx\quad\text{(Let } u = n x^2 \text{, then }du=2nx dx\text{)}\\ &= \frac{n}{2}\int_{0}^{n}\left[u^{3/2} e^{-u}\right] du \\ &= \frac{n}{2}\left(1-e^{-n}\right) \end{align*}

Part (a): Show \(R_n\) Diverges

Finally, we will show that \(R_{n}\) diverges as \(n\rightarrow\infty\): $$ \lim_{n\rightarrow\infty} R_{n}=\lim_{n\rightarrow\infty}\frac{n}{2}\left(1-e^{-n}\right)=\infty $$ Therefore, pointwise convergence does not imply mean convergence.

Part (b): Convergence in Mean

First, we will compute the mean square error for the sequence \(S_{n}(x)=x^{n}\) and show it converges to 0 as \(n\rightarrow\infty\): $$ R_{n}=\int_{0}^{1}\left[f(x)-S_{n}(x)\right]^{2} dx=\int_{0}^{1}\left[x^{2n}\right] dx=\frac{1}{2n+1} $$ We notice that, $$ \lim_{n\rightarrow\infty} R_{n} =\lim_{n\rightarrow\infty}\frac{1}{2n+1}=0 $$ Thus, the sequence \(S_{n}(x)\) converges to \(f(x)=0\) in the mean.

Part (b): Showing Absence of Pointwise Convergence

Now, let's show that \(S_{n}(x)\) does not converge to \(f(x)=0\) pointwise throughout the whole interval \(0 \leq x \leq 1\): For \(0 < x < 1\), we have pointwise convergence: $$ \lim_{n\rightarrow\infty} S_{n}(x) = \lim_{n\rightarrow\infty}\left(x^{n}\right)=0 $$ However, at \(x=1\), we have: $$ \lim_{n\rightarrow\infty} S_{n}(x) = \lim_{n\rightarrow\infty}\left(1^{n}\right)=1 $$ This shows that mean convergence does not imply pointwise convergence throughout the entire domain.

Key concepts

Pointwise Convergence

Pointwise convergence occurs when, for each individual point in a domain, a sequence of functions approaches a limiting function. In mathematical terms, a sequence \( S_n(x) \) is pointwise convergent to another function \( f(x) \) if for every \( x \) within a particular interval, \( \lim_{n \to \infty} S_n(x) = f(x) \) holds.
For example, if we consider the sequence \( S_n(x) = n \sqrt{x} e^{-nx^2 / 2} \) over the interval \( 0 \leq x \leq 1 \), it converges pointwise to 0, since for each fixed \( x \), the exponential term \( e^{-nx^2/2} \) tends towards zero more swiftly than the \( n \sqrt{x} \) term increases. Thus, \( S_n(x) \to 0 \) as \( n \to \infty \) for every \( x \) within this range.
However, as this scenario also demonstrates, pointwise convergence does not ensure convergence of related integrals or averages, underscoring the need to distinguish it from other convergence types.

Mean Convergence

Mean convergence takes a holistic view rather than a point-specific one. It involves the convergence of the averages (mean values) of a sequence of functions. It's often analyzed by calculating the mean square error between a sequence and a limit function.
In the context of our example, mean convergence checks whether \( \int_{0}^{1}[S_n(x) - f(x)]^2 dx \) tends towards zero as \( n \to \infty \).
For the function \( S_n(x) = x^n \), it can be shown that it converges in the mean to \( f(x) = 0 \) because the integral of \( x^{2n} \) across \( 0 \leq x \leq 1 \) results in \( \frac{1}{2n+1} \), which approaches zero as \( n \to \infty \).
Mean convergence provides an average-based measure and doesn't require the function to converge at every individual point, thus capturing a different aspect of convergence than pointwise convergence.

Mean Square Error

The mean square error (MSE) quantifies the difference between a sequence of functions and a target function over an interval, serving as a crucial tool for understanding mean convergence. It is calculated as the integral of the squared differences: \[R_n = \int_{0}^{1} [f(x) - S_n(x)]^2 dx.\]
When \( R_n \to 0 \) as \( n \to \infty \), it indicates that the sequence converges to the target function in the mean. For instance, in the sequence \( S_n(x) = x^n \) converging to \( f(x) = 0 \), the MSE becomes \( \frac{1}{2n+1} \), illustrating mean convergence as \( R_n \to 0 \).
On the other hand, for the sequence \( S_n(x) = n \sqrt{x} e^{-nx^2/2} \), the MSE \( R_n = \frac{n}{2}(1 - e^{-n}) \) does not tend to zero as \( n \to \infty \), thus demonstrating that despite pointwise convergence to zero, the mean convergence fails, with MSE offering an insightful perspective into these dynamics.

Limit of Functions

Understanding the limit of functions within a sequence is crucial for analyzing convergence. In a sequence \( S_n(x) \), the limit identifies whether \( \lim_{n \to \infty} S_n(x) = f(x) \) can hold for each, some, or none of the points \( x \) in the domain under consideration.
When dealing with the sequence \( S_n(x) = n \sqrt{x} e^{-nx^2/2} \), despite each point converging separately to 0, the integrated or average behavior diverges, illustrating the intricacies of function limits in sequences.
The concept of limits is central to distinguishing algorithmic patterns in convergence analyses, indicating whether the progression towards a target function is succeeded or not across different convergence types. As exemplified, function limits can reveal inconsistencies between individual and collective convergence behaviors, which is vital when determining the overall fidelity of function approximation in sequences.