8. Sequences of Functions

Let $A=[0,1]$ and let $f_n(x)=x^n$ for $n\in \mathbb{N}$ and $x\in A$. Then $(f_n)=(f_1,f_2,f_3,\dots )$ is a sequence of functions on $A$. As another example, for $n\in \mathbb{N}$ and $x\in A$ let $g_n(x)=nx(1-x^2{)}^n$. Then $(g_n)=(g_1,g_2,g_3,\dots )$ is a sequence of functions on $A$. Or how about $$f_n(x)=a_n\cos (nx)+b_n\sin (nx)$$ where $a_n,b_n\in \mathbb{R}$ and $x\in [-\pi ,\pi ]$.

Consider the sequence $(f_n)$ defined on $\mathbb{R}$ by $f_n(x)=(2xn+(-1{)}^n{x}^2)/n$. For fixed $x\in \mathbb{R}$ we have $$\underset{n\to \mathrm{\infty }}{lim}{f}_n(x)=\underset{n\to \mathrm{\infty }}{lim}\frac{2xn+(-1{)}^n{x}^2}{n}=2x.$$ Hence, $(f_n)$ converges pointwise to $f(x)=2x$ on $\mathbb{R}$. In Figure 8.1, we graph $f_n(x)$ for the values $n=1,2,3,4$ and the function $f(x)=2x$. Notice that $f_n^{\prime }(x)=(2n+2(-1{)}^{n}x)/n$ and therefore $\underset{n\to \mathrm{\infty }}{lim}{f}_n^{\prime }(x)=2$, and for the limit function $f(x)=2x$ we have $f^{\prime }(x)=2$. Hence, the sequence of derivatives $(f_n^{\prime })$ converges pointwise to $f^{\prime }$. Also, after some basic computations, $$\begin{array}{rl}{\int }_{-1}^1{f}_n(x)dx& ={\int }_{-1}^1\frac{2xn+(-1{)}^n{x}^2}{n}dx\\ & =\frac{2(-1{)}^n}{3n}\end{array}$$ and therefore $$\begin{array}{rl}\underset{n\to \mathrm{\infty }}{lim}{\int }_{-1}^1{f}_n(x)dx& =\underset{n\to \mathrm{\infty }}{lim}\frac{2(-1{)}^n}{3n}\\ & =0.\end{array}$$ On the other hand it is clear that ${\int }_{-1}^{1}f(x)dx=0$.

Consider the sequence $$ defined on $$ by $$. For all $$ we have $$ and therefore $$. On the other hand if $$ then $$ Therefore, $$ converges pointwise on $$ to the function $$ In Figure 8.2, we graph $$ for various values of $$. Consider a fixed $$. Since $$ it follows that for $$ there exists $$ such that $$ for all $$. For $$, in order for $$ we can choose $$. Notice that $$ clearly depends on both $$ and $$, and in particular, as $$ get closer to $$ then a larger $$ is needed. We note that each $$ is continuous while $$ is not.

Consider the sequence $$ defined on $$ by $$. For fixed $$ we have $$ Hence, $$ converges pointwise on $$ to the function $$. Notice that each function $$ is continuous on $$ and the pointwise limit $$ is also continuous. After some basic calculations we find that $$ and $$ exists for each $$, in other words, $$ is differentiable on $$. However, $$ is not differentiable on $$ since $$ does not have a derivative at $$. In Figure 8.3, we graph $$ for various values of $$.

Consider the sequence $$ on $$ defined by $$. For fixed $$ we find (using l'H\^{o}pital's rule) that $$ Hence, $$ converges pointwise to $$ on $$. Consider $$ and therefore $$ On the other hand, $$. Therefore, $$ or another way to write this is $$

Suppose that $$ is a sequence of functions such that $$ is increasing for each $$. Suppose that $$ exists for each $$. Is $$ an increasing function?

Let $$ be a sequence of positive numbers and define $$ as $$

1. Find the pointwise limit $$ of the sequence $$.
2. Find $$.
3. If $$, find $$

Recall that $$ is countable and thus there exists a bijection $$. Define the sequence $$ by letting $$. Now define $$ as $$

1. Find the pointwise limit $$ of the sequence $$.
2. Is $$ Riemann integrable? Explain.
3. Is $$ Riemann integrable? Explain.

Let $$ and let $$ be the sequence of functions on $$ defined by $$. Prove that $$ converges uniformly to $$.

Show that the sequence of functions $$ converges uniformly to $$ on $$.

Let $$ be a continuous function on $$. Prove that there exists a sequence of step functions $$ on $$ that converges uniformly to $$ on $$.

Let $$ be a sequence of functions converging uniformly to $$. Let $$ be a function and let $$ for each $$. Under what condition on $$ does the sequence $$ converge uniformly? Prove it. What is the uniform limit of $$?

Prove that if $$ converges uniformly to $$ on $$ and $$ converges uniformly to $$ on $$ then $$ converges uniformly to $$ on $$.

Let $$ be the sequence defined in Exercise 8.1.2. Show that if $$ then $$ converges uniformly.

Let $$ for $$. Prove that $$ converges uniformly on $$.

Give an example of a set $$ and a sequence of functions $$ on $$ such that $$ is bounded for each $$, $$ converges pointwise to $$ but $$ is not uniformly bounded on $$.

Let $$ for $$ and $$. Each function $$ is clearly continuous. Now $$ and thus $$. If $$ then $$ Therefore, $$ converges pointwise to the function $$ The function $$ is discontinuous and therefore, by Theorem 8.3.3, $$ does not converge uniformly to $$.

Consider the sequence of functions $$ defined on $$ given by $$

1. Draw a typical function $$.
2. Prove that $$ converges pointwise.
3. Use Theorem 8.3.5 to show that the convergence is not uniform.

Consider the sequence $$ defined on $$ by $$. We compute that $$ and clearly $$ is continuous on $$ for each $$. Now $$ for all $$ and thus $$ converges pointwise to $$. To prove that the convergence is uniform we note that $$ Therefore, $$ converges uniformly to $$ on $$. Now $$ and thus $$ converges to $$. By Theorem 8.3.8, $$ converges uniformly to say $$ with $$ and $$. Now by the FTC, $$ and since $$ then $$.

Give an example of a set $$ and a sequence of functions $$ on $$ such that $$ is bounded for each $$, $$ converges pointwise to $$ but $$ is not uniformly bounded on $$.

Suppose that $$ uniformly on $$ and $$ uniformly on $$. Prove that if $$ and $$ are uniformly bounded on $$ then $$ converges uniformly to $$ on $$. Then give an example to show that if one of $$ or $$ is not uniformly bounded then the result is false.

Let $$ for $$ and $$.

1. Show that $$ converges pointwise on $$.
2. Show that $$ does not converge uniformly on any closed interval containing $$.
3. Show that $$ converges uniformly on any closed interval not containing $$. For instance, take $$ with $$.

Suppose that $$ has that property that $$ for all $$ and some $$. Prove that if $$ converges uniformly on $$ to $$ then the sequence $$ converges uniformly to $$ on $$. Note: $$ and $$ are compositions of functions and not function multiplication.

Let $$ for $$ and $$ where $$.

1. Prove directly that the sequence $$ is uniformly Cauchy.
2. If $$ is the uniform limit of $$, find $$ without computing $$.

Consider the sequence of functions $$ on $$ defined as follows: $$

1. Prove that $$ converges uniformly to $$ on $$.
2. For each fixed $$, find the improper integral $$ and show that $$
3. The results above seem to contradict Theorem 8.3.5. Explain why there is no contradiction.

Prove that $$

Consider the function $$ whose graph is given in Figure 8.5; one can write down an explicit expression for $$ but the details are unimportant. Consider the series $$ Since $$ and $$ converges, then by the $$-test the above series converges uniformly on any interval $$. Let $$ be the function defined by the series on $$. Now, on $$, the function $$ has only a finite number of discontinuities and thus $$ is Riemann integrable. Therefore, by Theorem 8.3.5, the function $$ is Riemann integrable. The graph of $$ is shown in Figure 8.6. One can show that $$ has discontinuities at the rational points $$ where $$.

Recall that a power series is a series of the form $$ where $$ and $$. Hence, in this case if we write the series as $$ then $$ for each $$ and $$. In calculus courses, the main problem you were asked to solve is to find the interval of convergence of the given power series. The main tool is to apply the Ratio test (Theorem 3.7.23): $$ Suppose that $$ exists and is non-zero and $$ (a similar argument can be done when the limit is zero). Then by the Ratio test, the power series converges if $$, that is, if $$. The number $$ is called the radius of convergence and the interval $$ is the interval of convergence (if $$ is zero then $$ can be chosen arbitrarily and the argument that follows is applicable). Let $$ and consider the closed interval $$. Then if $$ then $$ Now if $$ then by assumption the series $$ converges, and in particular the sequence $$ is bounded, say by $$. Therefore, $$ Since $$, the geometric series $$ converges. Therefore, by the $$-test, the series $$ converges uniformly on the interval $$. Let $$ for $$. Now consider the series of the derivatives $$ Applying the Ratio test again we conclude that the series of the derivatives converges for each $$ and a similar argument as before shows that the series of derivatives converges uniformly on any interval $$ where $$. It follows from the Term-by-Term Differentiation theorem that $$ is differentiable and $$. By the Term-by-Term Integration theorem, we can also integrate the series and $$ where $$ is any closed and bounded interval.

Consider the power series $$

1. Prove that the series converges at every $$.
2. Let $$ denote the function defined by the series on the left and let $$ denote the function defined by the series on the right. Justifying each step, show that $$ exists and that $$.
3. Similarly, show that $$ exists and $$.

A Fourier series is a series of the form $$ where $$.

1. Suppose that for a given $$ and $$, the associated Fourier series converges pointwise on $$ and let $$ be the pointwise limit. Prove that in fact the Fourier series converges on $$. Hint: For any $$ there exists $$ such that $$.
2. Prove that if $$ and $$ are convergent series then the associated Fourier series converges uniformly on $$.
3. Suppose that for a given $$ and $$, the associated Fourier series converges uniformly on $$ and let $$ be the uniform limit. Prove the following: $$ You will need the following identities: $$

1. By an open cover of an interval $$, we mean a collection of open intervals $$ such that $$ for each $$ and $$ Here $$ is some set, possibly uncountable. Prove that if $$ is any open cover of $$ then there there exists finitely many $$ such that $$
2. Let $$ be continuous and suppose that $$ for all $$ and all $$. Suppose that $$ converges pointwise to a continuous function $$. Prove that the convergence is actually uniform. Give an example to show that if $$ is not continuous then we only have pointwise convergence.

Let $$ be continuous and suppose that $$ for all $$. Prove that if $$ converges pointwise to a continuous function $$ on $$ then in fact the convergence is uniform.

Show that $$ converges uniformly on $$ for every $$ such that $$. Then show that the given series does not converge uniformly on $$. Hint: This is an important series and you should know what function the series converges uniformly to.

If $$ prove that $$ converges uniformly on $$.

Prove, justifying each step, that $$

For any number $$ let $$ be the function defined as $$ if $$ and $$ if $$. Let $$ be an enumeration of the rational numbers. Define $$ as $$ Find the pointwise limit $$ of the sequence $$. Is the convergence uniform? Explain.