intro to mathematical analysis unit 12 study guides

What's Uniform Convergence?

• Uniform convergence is a stronger form of convergence for sequences and series of functions
• Ensures the limit function is well-behaved and the convergence is uniform across the entire domain
• Guarantees that the limit of the sequence of functions is continuous if each function in the sequence is continuous
• Allows for the interchange of limits and certain operations, such as integration and differentiation
• Plays a crucial role in various branches of mathematical analysis, including real analysis, complex analysis, and functional analysis
• Helps to establish the validity of certain approximations and numerical methods
• Provides a rigorous foundation for the study of infinite series of functions and their properties

Key Concepts and Definitions

• Pointwise convergence: a sequence of functions ${f_n}$ converges pointwise to a limit function $f$ if, for each fixed $x$ in the domain, the sequence of real numbers ${f_n(x)}$ converges to $f(x)$
• Uniform convergence: a sequence of functions ${f_n}$ converges uniformly to a limit function $f$ on a set $E$ if, for every $\varepsilon > 0$, there exists an $N(\varepsilon)$ such that $|f_n(x) - f(x)| < \varepsilon$ for all $n \geq N(\varepsilon)$ and all $x \in E$
  • The key difference is that $N(\varepsilon)$ depends only on $\varepsilon$ and not on $x$
• Cauchy criterion for uniform convergence: a sequence of functions ${f_n}$ converges uniformly on a set $E$ if and only if, for every $\varepsilon > 0$, there exists an $N(\varepsilon)$ such that $|f_n(x) - f_m(x)| < \varepsilon$ for all $n, m \geq N(\varepsilon)$ and all $x \in E$
• Uniform convergence of series: a series of functions $\sum_{n=1}^{\infty} f_n(x)$ converges uniformly on a set $E$ if the sequence of partial sums ${S_n(x)}$, where $S_n(x) = \sum_{k=1}^{n} f_k(x)$, converges uniformly on $E$
• Weierstrass M-test: if ${f_n}$ is a sequence of functions on a set $E$ and there exists a sequence of positive real numbers ${M_n}$ such that $|f_n(x)| \leq M_n$ for all $x \in E$ and $\sum_{n=1}^{\infty} M_n$ converges, then $\sum_{n=1}^{\infty} f_n(x)$ converges uniformly on $E$
• Dini's theorem: if ${f_n}$ is a sequence of continuous functions on a compact set $K$ that converges pointwise to a continuous function $f$ and $f_n(x) \geq f_{n+1}(x)$ for all $n$ and $x \in K$, then the convergence is uniform on $K$

Comparing Pointwise and Uniform Convergence

• Uniform convergence implies pointwise convergence, but the converse is not always true
  • Example: the sequence of functions $f_n(x) = x^n$ on $[0, 1)$ converges pointwise to the function $f(x) = 0$ for $x \in [0, 1)$ and $f(1) = 1$, but the convergence is not uniform on $[0, 1]$
• Pointwise convergence does not guarantee the continuity of the limit function, even if all functions in the sequence are continuous
  • Example: the sequence of continuous functions $f_n(x) = x^n$ on $[0, 1]$ converges pointwise to a discontinuous function
• Uniform convergence preserves continuity, meaning if each $f_n$ is continuous and ${f_n}$ converges uniformly to $f$, then $f$ is also continuous
• Uniform convergence allows for the interchange of limits and operations, such as integration and differentiation, under certain conditions
  • Example: if ${f_n}$ converges uniformly to $f$ on $[a, b]$ and each $f_n$ is Riemann integrable, then $\lim_{n \to \infty} \int_a^b f_n(x) dx = \int_a^b f(x) dx$
• Pointwise convergence does not generally allow for the interchange of limits and operations without additional conditions

Criteria for Uniform Convergence

• Cauchy criterion: ${f_n}$ converges uniformly on $E$ if and only if for every $\varepsilon > 0$, there exists an $N(\varepsilon)$ such that $|f_n(x) - f_m(x)| < \varepsilon$ for all $n, m \geq N(\varepsilon)$ and all $x \in E$
  • Useful for proving uniform convergence without explicitly finding the limit function
• Weierstrass M-test: if $|f_n(x)| \leq M_n$ for all $x \in E$ and $\sum_{n=1}^{\infty} M_n$ converges, then $\sum_{n=1}^{\infty} f_n(x)$ converges uniformly on $E$
  • Provides a sufficient condition for uniform convergence of series
• Dini's theorem: if ${f_n}$ is a sequence of continuous functions on a compact set $K$ that converges pointwise to a continuous function $f$ and $f_n(x) \geq f_{n+1}(x)$ for all $n$ and $x \in K$, then the convergence is uniform on $K$
  • Establishes uniform convergence for monotonically decreasing sequences of continuous functions on compact sets
• Uniform boundedness: if ${f_n}$ converges uniformly on $E$, then there exists an $M > 0$ such that $|f_n(x)| \leq M$ for all $n$ and all $x \in E$
  • A necessary condition for uniform convergence
• Continuity of the limit function: if ${f_n}$ is a sequence of continuous functions that converges uniformly to $f$ on $E$, then $f$ is continuous on $E$
  • Uniform convergence preserves continuity

Examples and Counterexamples

• Example of uniform convergence: the sequence of functions $f_n(x) = \frac{1}{n} \sin(nx)$ on $\mathbb{R}$ converges uniformly to the zero function
  • Proof: $|f_n(x)| = |\frac{1}{n} \sin(nx)| \leq \frac{1}{n}$ for all $x \in \mathbb{R}$, and $\lim_{n \to \infty} \frac{1}{n} = 0$
• Counterexample to uniform convergence: the sequence of functions $f_n(x) = x^n$ on $[0, 1)$ converges pointwise to the function $f(x) = 0$ for $x \in [0, 1)$ and $f(1) = 1$, but the convergence is not uniform on $[0, 1]$
  • Proof: for any $N \in \mathbb{N}$, choose $x_N = (1 - \frac{1}{N})^{1/N}$. Then, $|f_N(x_N) - f(x_N)| = |x_N^N - 0| = (1 - \frac{1}{N}) > \frac{1}{2}$ for all $N > 2$, violating the definition of uniform convergence
• Example of the Weierstrass M-test: the series $\sum_{n=1}^{\infty} \frac{\sin(nx)}{n^2}$ converges uniformly on $\mathbb{R}$
  • Proof: $|\frac{\sin(nx)}{n^2}| \leq \frac{1}{n^2}$ for all $x \in \mathbb{R}$, and $\sum_{n=1}^{\infty} \frac{1}{n^2}$ converges by the p-series test
• Counterexample to the interchange of limits and integration: the sequence of functions $f_n(x) = n x e^{-nx}$ on $[0, 1]$ converges pointwise to the zero function, but $\lim_{n \to \infty} \int_0^1 f_n(x) dx = \lim_{n \to \infty} (1 - e^{-n}) = 1 \neq 0 = \int_0^1 \lim_{n \to \infty} f_n(x) dx$
  • This is because the convergence is not uniform on $[0, 1]$

Applications in Analysis

• Uniform convergence is essential for the rigorous development of power series, Fourier series, and other infinite series expansions
  • Example: the Weierstrass approximation theorem states that any continuous function on a closed interval can be uniformly approximated by polynomials
• Uniform convergence allows for the interchange of limits and differentiation under certain conditions
  • If ${f_n}$ is a sequence of differentiable functions on $[a, b]$ that converges uniformly to $f$ and ${f_n'}$ converges uniformly to $g$, then $f$ is differentiable and $f' = g$
• Uniform convergence is used to establish the continuity and differentiability of functions defined by infinite series or integrals
  • Example: the uniform convergence of the series $\sum_{n=1}^{\infty} \frac{\sin(nx)}{n}$ on $\mathbb{R}$ implies that its sum is a continuous function
• Uniform convergence plays a role in the study of functional spaces, such as the space of continuous functions or the space of integrable functions
  • It helps to characterize the completeness and compactness properties of these spaces
• Uniform convergence is used in the construction of solutions to differential equations and integral equations
  • Example: the Picard-Lindelöf theorem uses uniform convergence to prove the existence and uniqueness of solutions to initial value problems

Common Pitfalls and Misconceptions

• Mistakenly assuming that pointwise convergence implies uniform convergence
  • Counterexample: $f_n(x) = x^n$ on $[0, 1)$ converges pointwise but not uniformly
• Forgetting to check the uniform convergence of the series when applying the Weierstrass M-test
  • The M-test provides a sufficient condition, but not a necessary one, for uniform convergence
• Incorrectly interchanging limits and operations without verifying uniform convergence
  • Example: interchanging the limit and integral for $f_n(x) = n x e^{-nx}$ on $[0, 1]$ leads to an incorrect result
• Confusing uniform convergence with other types of convergence, such as pointwise convergence or convergence in measure
  • Each type of convergence has its own definition and properties
• Misapplying uniform convergence criteria, such as Dini's theorem, without verifying all the necessary conditions
  • Example: applying Dini's theorem to a sequence of functions that is not monotonically decreasing

Practice Problems and Solutions

1. Determine whether the sequence of functions $f_n(x) = \frac{x}{1 + nx}$ converges uniformly on $[0, 1]$.

   • Solution: The sequence converges pointwise to the zero function on $[0, 1]$. To prove uniform convergence, we use the definition: $|f_n(x) - 0| = |\frac{x}{1 + nx}| \leq \frac{1}{n}$ for all $x \in [0, 1]$. Given $\varepsilon > 0$, choose $N = \lceil \frac{1}{\varepsilon} \rceil$. Then, for all $n \geq N$ and all $x \in [0, 1]$, $|f_n(x) - 0| \leq \frac{1}{n} \leq \frac{1}{N} < \varepsilon$. Thus, the convergence is uniform on $[0, 1]$.

2. Prove that the series $\sum_{n=1}^{\infty} \frac{x^2}{n(1 + nx^2)}$ converges uniformly on $\mathbb{R}$.

   • Solution: We use the Weierstrass M-test. Observe that $|\frac{x^2}{n(1 + nx^2)}| \leq \frac{1}{n^2}$ for all $x \in \mathbb{R}$, as $\frac{x^2}{1 + nx^2} \leq \frac{1}{n}$. The series $\sum_{n=1}^{\infty} \frac{1}{n^2}$ converges by the p-series test. Therefore, by the Weierstrass M-test, the given series converges uniformly on $\mathbb{R}$.

3. Let $f_n(x) = \frac{\sin(nx)}{n}$ on $[0, \pi]$. Show that ${f_n}$ converges pointwise to the zero function, but the convergence is not uniform.

   • Solution: For pointwise convergence, fix $x \in [0, \pi]$. Then, $\lim_{n \to \infty} f_n(x) = \lim_{n \to \infty} \frac{\sin(nx)}{n} = 0$, as $|\sin(nx)| \leq 1$ for all $n$ and $x$. To show that the convergence is not uniform, consider $x_n = \frac{\pi}{2n}$. Then, $|f_n(x_n) - 0| = |\frac{\sin(nx_n)}{n}| = |\frac{\sin(\pi/2)}{n}| = \frac{1}{n}$. For any $N \in \mathbb{N}$, choose $\varepsilon = \frac{1}{2N}$. Then, for $n = N$, we have $|f_N(x_N) - 0| = \frac{1}{N} > \frac{1}{2N} = \varepsilon$, violating the definition of uniform convergence.