Further Topics

Describe the Bernstein polynomials for the two functions $f(x)=x$ and $g(x)=x^{2}$ on $[0,1]$.

Prove that $(1+x)^{n} \geq 1+n x$ for all $x \geq -1$ and each positive integer $n$.

In the Weierstrass approximation theorem, it is crucial that the interval is compact, as addressed in Stone's generalization. One of our favorite noncompact subsets of $\mathbb {R}$ is $\mathbb {R}$ itself. Show that uniform polynomial approximation is not always guaranteed on $\mathbb {R}$. Here are two approaches:

Deduce the Weierstrass approximation theorem from the Stone-Weierstrass theorem (thus showing that the former is indeed a special case of the latter).

Consider the collection $\mathcal{P}_{e}$ of all even polynomials.

Let $A$ be as in the hypotheses of Bishop's theorem.

Find (in another book) another proof of the Weierstrass approximation theorem. Write up your own account of this proof. For example, if you have read the final section of the chapter, you may want to find the proof that makes use of the Fourier transform.

Show that $X$ is nowhere dense in $M$ if and only if its closure has empty interior, $(\bar{X})^{\circ }=\emptyset$.

Show that any finite subset of $\mathbb {R}$ is nowhere dense in $\mathbb {R}$. Give examples to show that a countable set can be nowhere dense in $\mathbb {R}$, but that this is not always the case.

Let $f(x)=\sin \left(\frac{1}{x}\right)$ for $x \neq 0$, and $f(0)=0$. Compute $\omega_{f}(0)$. How does this tie in with what you know about the continuity of this function?

First category sets are in some sense "small," while second category sets are "large." Another notion of set size that we have discussed is "measure zero": "Small" sets are of measure zero, while "large" sets are not. Is there a connection between these two ways of describing the size of a set? The answer is no. Please show this by doing two things:

Prove that the function defined on $(0,1)$ by

is continuous at each irrational point of $(0,1)$ and discontinuous at each rational point of $(0,1)$.

The goal of this exercise is to show that "most" continuous functions are nowhere differentiable. Let $E_{n}$ denote the set of all $f \in C([0,1])$ for which there exists an $x_{f} \in [0,1]$ satisfying

for every $x \in [0,1]$.

Let $X$ and $Y$ be Banach spaces. By considering the right shift on $\ell^{2}$ into itself defined by $S\left(x_{1}, x_{2}, \ldots \right)=\left(0, x_{1}, x_{2}, \ldots \right)$, we see that the property of being one-to-one is not enough to guarantee that a bounded linear operator be invertible (compare what happens in the finite-dimensional case). However, if a one-to-one bounded linear operator is also onto, then it must be invertible. Use the open mapping theorem to prove that the inverse is also bounded.

Let $X$ be a Banach space in the two different norms $\left\| \cdot \right\|_{1}$ and $\left\| \cdot \right\|_{2}$.

Show that $x \sim y$ if $x-y \in \mathbb {Q}$ defines an equivalence relation on the open interval $(0,1)$.

Let $X = (X, d)$ be a complete metric space and $T: X \rightarrow X$ a contraction, i.e., there exists $M \in [0,1)$ such that $d(Tx, Ty) \leq M d(x, y)$ for all $x, y \in X$. Theorem 6.11 (the Banach contraction mapping principle) states that there exists a unique point $x \in X$ with $Tx = x$; its proof chooses any $x_{0} \in X$ and puts $x_{1} = T x_{0}$, $x_{2} = T x_{1} = T^{2} x_{0}$, and so on, $x_{n} = T^{n} x_{0}$.

Give an inductive proof of the inequality $d\left(x_{n+1}, x_{n}\right) \leq M^{n} d\left(x_{1}, x_{0}\right)$, $n=1,2, \ldots$, and thereby complete the proof of the contraction mapping theorem.

Show that the method of successive approximations applied to the differential equation $f^{\prime }=f$ with $f(0)=1$ yields the usual formula for $e^{x}$.

For each of the following sets give an example of a continuous mapping of the set into itself that has no fixed point:

Give an example of a mapping of $[0,1]$ into itself that is not continuous and has no fixed points.

Consider the function $f: \mathbb {R} \rightarrow \mathbb {R}$ given by

Show that $\left|f(x)-f(y)\right|<\left|x-y\right|$ for all $x$ and $y$, yet $f$ has no fixed point. Does this contradict the contraction mapping principle? Explain.

For $f$ and $g$ in $C([a, b])$ define $f+g$ and $f \cdot g$ by

Show that $C([a, b])$ is a ring with identity.

Prove that a proper ideal of a ring with identity contains no invertible elements.

If $\mathcal{J}$ is an ideal in a Banach algebra, then $\overline{\mathcal{J}}$ is also an ideal. (In this statement, "Banach algebra" can be replaced by "normed ring.")

Prove that $\mathcal{J}_{x}$ is a proper ideal of $C([a, b])$.

Prove that for $x, y \in [a, b], \mathcal{J}_{x}=\mathcal{J}_{y}$ if and only if $x=y$. (In proving this, you may find yourself assuming that $x \neq y$ and constructing an actual function $f$ that is in one of the ideals but is not in the either. It is this step that requires Urysohn's lemma when $[a, b]$ is replaced by a compact Hausdorff space; see the paragraph preceding Theorem 6.14. In fact, in that case $f$ is not actually constructed but only its existence implied.)

Theorem 6.14: Each ideal $\mathcal{J}_{x}$ is a maximal ideal of $C([a, b])$, and moreover, every maximal ideal of $C([a, b])$ is of this form. Finally, $\mathcal{J}_{x} = \mathcal{J}_{y}$ if and only if $x = y$.

Fix a positive real number $a$ and let $\chi_{[-a, a]}$ denote the characteristic function

Show that the Fourier transform of $\chi_{[-a, a]}$ is given by

As in Example 1, fix real numbers $\theta$ and $a$, and let

Show that $\int_{-\infty }^{\infty }|f(t)|^{2} d t=1$ and

Use that $e^{i \theta }=\cos \theta +i \sin \theta$ for every real number $\theta$.

In the section we claimed that

The point of this exercise is to verify the middle equality, and hence all three equalities. Note that

Use integration by parts to show that this equals $i w \hat{f}(w)$, and hence that the equation holds.

Verify the formula

for $T=T_{b \ell } T_{t \ell }$, as used in the section.

Show that (6.8) (in the text of the section) is true whenever $\alpha^{2}+\beta^{2} \leq 1$.