Question

Give an example of a sequence of functions (f) which converges uniformly, but does not converge in the mean or in the mean square. Hint. According to Problem \(7 \mathrm{a}\), we must have \(\mu(X)=\infty .\) Let $$ f_{n}(x)= \begin{cases}\frac{1}{\sqrt{n}} & \text { if } \quad|x| \leq n \\ 0 & \text { if } \quad|x|>n\end{cases} $$

Short answer

The sequence converges uniformly to 0, but not in the mean or mean square.

Step-by-step solution

Introduce the Sequence of Functions

The sequence of functions given is \( f_n(x) = \begin{cases}\frac{1}{\sqrt{n}} & \text{if } |x| \leq n \ 0 & \text{if } |x| > n \end{cases}\). This means for each \( n \), \( f_n(x) \) is constant \( \frac{1}{\sqrt{n}} \) within the interval \([-n, n]\) and zero elsewhere.

Analyze Uniform Convergence

To check for uniform convergence, we evaluate the limit as \( n \to \infty \). For any \( x \), if \( |x| \leq n \), then \( |f_n(x)| = \frac{1}{\sqrt{n}} \) and as \( n \to \infty \), \( \frac{1}{\sqrt{n}} \to 0 \). Hence, \( f_n(x) \to 0 \) uniformly for all \( x \) with respect to \( n \) since \( \frac{1}{\sqrt{n}} \to 0 \) does not depend on \( x \). Therefore, the sequence converges uniformly to the zero function \( f(x) = 0 \).

Investigate Convergence in Mean

To determine if \( f_n(x) \) converges in the mean, we compute the integral \( \int_{-\infty}^{\infty} |f_n(x) - 0| \, dx \). This evaluates to \( 2n \cdot \frac{1}{\sqrt{n}} = 2\sqrt{n} \). As \( \sqrt{n} \to \infty \), the mean convergence condition is not met; hence \( f_n(x) \) does not converge in the mean.

Investigate Convergence in Mean Square

For mean square convergence, we compute the integral of the square, \( \int_{-\infty}^{\infty} |f_n(x) - 0|^2 \, dx \), which gives \( 2n \cdot \frac{1}{n} = 2 \). This does not approach zero as \( n \to \infty \); hence \( f_n(x) \) does not converge in mean square.

Key concepts

Sequence of Functions

A sequence of functions is essentially a list of functions ordered in a specific sequence. In mathematical terms, you can think of it as a series like \(f_1(x), f_2(x), f_3(x), \ldots, f_n(x)\). In our problem, the sequence is defined as \( f_n(x) = \begin{cases}\frac{1}{\sqrt{n}} & \text{if } |x| \leq n \ 0 & \text{if } |x| > n \end{cases}\).

• The function \(f_n(x)\) is constant within the interval \([-n, n]\), having a value of \(\frac{1}{\sqrt{n}}\).
• Outside this interval, the function is zero. This characteristic makes it easy to understand its behavior across different domains for each \(n\).

As \(n\) increases, the interval \([-n, n]\) broadens, yet the value \(\frac{1}{\sqrt{n}}\) diminishes. This constant value within the expanding boundaries plays a crucial role in evaluating how the function behaves in terms of convergence as \(n\) approaches infinity.

Convergence in the Mean

Convergence in the mean involves looking at how the average values of a sequence of functions behave as \(n\) becomes very large. In essence, do these average values approach zero as \(n\) goes to infinity?

• To assess this, we calculate the integral of the absolute difference between the function and its limit over the entire real line. The formula is \( \int_{-\infty}^{\infty} |f_n(x) - 0| \, dx \).
• This integral evaluates to \(2n \cdot \frac{1}{\sqrt{n}} = 2\sqrt{n}\). Since \(\sqrt{n}\) increases indefinitely as \(n\) grows, the result does not tend towards zero.

Therefore, the sequence \(f_n(x)\) does not converge in the mean because the average deviation from zero actually becomes larger over time rather than shrinking. This indicates that despite the pointwise approach to zero, the spread of \(f_n(x)\) in terms of average value does not head towards zero as \(n\) increases.

Mean Square Convergence

Mean square convergence is another layer of evaluating how a sequence of functions behaves as \(n\) increases, but this time, by squaring the differences.

• To check for mean square convergence, we compute the square of \(f_n(x)\) and then find its integral over the whole real number line. This is expressed formally as \( \int_{-\infty}^{\infty} |f_n(x) - 0|^2 \, dx \).
• The result of this integral is \(2n \cdot \frac{1}{n} = 2\). As you can see, the value 2 does not lean towards zero as \(n\) progresses.

This means that although each function \(f_n(x)\) gets closer to zero in a pointwise sense, the total energy or the so-called 'squared difference' does not decrease, hindering mean square convergence. This non-convergence in the mean square sense underscores the complexity of confirming that a function sequence becomes zero through different mathematical lenses.