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Solution to Principles of Mathematical Analysis Chapter 4 Part A

Chapter 4 Continuity

Exercise 1

(By ghostofgarborg) No. As an example, take the function
\[ f(x) = \begin{cases} 1 & x = 0 \\ 0 & \text{otherwise} \end{cases} \]which is discontinuous at $0$, although $\lim_{h \to 0} [f(x+h) – f(x-h)] = 0$ everywhere.

Exercise 2

(By ghostofgarborg) We have\[ f^{-1}(\overline{f(E)}) = f^{-1} \Big( \bigcap_{\substack{V \supseteq f(E) \\ V \text{ closed}}} V \Big)
= \bigcap_{\substack{V \supseteq f(E) \\ V \text{ closed}}} f^{-1}(V) \supseteq \bigcap_{\substack{W \supseteq E \\ W \text{ closed}}} W = \bar E \]Consequently, $f(\bar E) \subseteq \overline{f(E)}$. To see that the inclusion can be proper, let $X = (0,1)$ seen as a subspace of $\mathbb R$, and $f$ the inclusion into $\mathbb R$. If we let $E = X$, then
\[ f(\bar E) = f((0,1)) = (0,1) \subsetneq \overline{(0,1)} = [0,1] \]

Exercise 3

(By ghostofgarborg) Note that $\{ 0 \}$ is closed, so that $Z(f) = f^{-1}(\{ 0 \})$ is closed by the continuity of $f$.

Exercise 4

(By ghostofgarborg) Let $U \subset f(X)$ be open. Then $f^{-1}(U)$ is open, and since $E$ is dense, it contains points of $E$. Therefore $U$ contains points of $f(E)$. This makes $f(E)$ dense in $f(X)$.

Let $p \in X$, and $p_n$ a sequence in $E$ that converges to $p$. By theorem 4.6 and the equality of $f$ and $g$ on $E$ \[ f(p) = \lim_{n \to \infty} f(p_n) = \lim_{n \to \infty} g(p_n) = g(p) \] so that $f$ and $g$ agree on all of $X$.

Exercise 5

(By analambanomenos) By Exercise 2.29, the open complement of $E$ is a countable collection of disjoint open intervals $\cup_i(a_i,b_i)$. If $b_i=\infty$ define $g_i$ on $[a_i,\infty)$ to take the constant value $f(a_i)$. Similarly, if $a_i=-\infty$, define $g_i$ on $(-\infty,b_i]$ to take the constant value $f(b_i)$. Otherwise, on $[a_i,b_i]$, let $g_i$ be the linear function $$g_i(x)=\frac{f(b_i)-f(a_i)}{b_i-a_i}x+\frac{f(a_i)b_i-f(b_i)a_i}{b_i-a_i}.$$ Note that $g_i(a_i)=f(a_i)$, $g_i(b_i)=f(b_i)$, and the values of $g_i$ lie between $f(a_i)$ and $f(b_i)$ on $(a_i,b_i)$.

Let $g(x)=f(x)$ for $x\in E$ and $g(x)=g_i(x)$ for $x\in(a_i,b_i)$. It is clear that $g$ is continuous at any $x\in(a_i,b_i)$, so suppose $x\in E$ and let $\varepsilon>0$. Then there is a $\delta>0$ such that $\big|f(x)-f(y)\big|<\varepsilon$ for $y\in(x-\delta,x+\delta)\cap E$. If $x-\delta\notin E$, then $x\in(a_i,b_i)$ for some $i$, and we can replace $x-\delta$ with $x-\delta_1=b_i\in E$. Similarly, if $x+\delta\notin E$, we can replace $x+\delta$ with some $x+\delta_2=a_j\in E$. Hence, if any of the open intervals $(a_i,b_i)$ intersect $(x-\delta_1,x+\delta_2)$, then both $a_i$ and $b_i$ must be in $(x-\delta_1,x+\delta_2)$. By the construction of the $g_i$, we must have $\big|g(x)-g(y)\big|<\varepsilon$ for $y\in(x-\delta_1,x+\delta_2)$.

If $\delta_1=0$, so that $x=b_i$ for some $i$, then by the linearity of $g_i$ we can increase $\delta_1$ by an amount small enough so that $\big|g(x)-g(y)\big|=
\big|g_i(x)-g_i(y)\big|<\varepsilon$ for $y\in(x-\delta_1,x)$. Similarly, if $\delta_2=0$ so that $x=a_j$ for some $j$, we can increase $\delta_2$ by an amount small enough so that $\big|g(x)-g(y)\big|<\varepsilon$ for $y\in(x,x+\delta_2)$. Hence both $\delta_1>0$ and $\delta_2>0$, so we can conclude that $g$ is continuous at $x$.

Let $f(x)=x^{-1}$ for $x\in(0,1)$. There can be no extension of $f$ to all of $\mathbf R$ since $\lim_{x\rightarrow 0+}f(x)=\infty$.

If $\mathbf f(x)=\bigl(f_1(x),\ldots,f_k(x)\bigr)$ is a continuous map from a closed set $E$ in $\mathbf R$ into $\mathbf R^k$, then each of the component functions $f_n$ are continuous functions on $E$ by Theorem 4.10(a). Extend each of the $f_n$ to a continuous function $g_n$ defined on all of $\mathbf R$. Then, also by Theorem 4.10(a), the vector-valued function $\mathbf g(x)=\bigl(g_1(x),\ldots,g_k(x)\bigr)$ is a continuous extension of $\mathbf f$ to all of $\mathbf R$.

Exercise 6

(By analambanomenos) Assume $f : X \to Y$ is continuous. Then the map $\Gamma(x) = (x, f(x))$ is also continuous by thm. 4.10a. Since $E$ is compact, so is $\Gamma(E)$, which is the graph of $F$.

If $\Gamma(E)$ is compact, let $V$ be a closed subset of $Y$. The set
$$V^\prime = (X \times V) \cap \Gamma(E)$$ is closed in $\Gamma(E)$, hence compact. The projection $ \pi : X \times Y \to X$ is continuous, so $ f^{-1}(V) = \pi(V^\prime) $ is compact, hence closed (since $X$ is a metric space and therefore Hausdorff). This makes $f$ continuous.

Exercise 7

(By analambanomenos) Note that both $f$ and $g$ are equal to 0 on the $x$ and $y$ axes.

For $k\in\mathbf R$, $f(ky^2,y)=k/(k^2+1)$ is constant for $y\neq0$. Since the value of $f$ along the parabola $(ky^2,y)$ drops from $k/(k^2+1)$ to 0 at $y=0$, $f$ is not continuous at $(0,0)$. These parabolas sweep out $\mathbf R^2$ except for the $x$ axis, and the values of $f$ along these parabolas reach a maximum value of $1/2$ for $k=1$, so the values of $f$ lie in $[0,1/2]$.

For $y\neq 0$, $g(ky^3,y)=k/\bigl((k^3+1)y\bigr)\rightarrow\infty$ as $y\rightarrow 0$, so $g$ is unbounded in every neighborhood of $(0,0)$.

Since $f$ and $g$ are continuous away from the origin, their restrictions to any line which doesn’t intersect the origin is also continuous. And since $$f(x,kx)=\frac{k^2x}{1+k^4x^2}\quad\quad g(x,kx)=\frac{k^2x}{1+k^6x^4}$$ restrictions of $f$ and $g$ to lines which go through the origin are also continuous.

Exercise 8

(By analambanomenos) The quickest solution uses Exercise 13 below. Note that the closure $\bar E$ is also bounded, since if $E\subset[m,M]$ and $x>M$, then $x$ has a neighborhood which doesn’t intersect $E$ so $x\notin\bar E$, similarly $y\notin\bar E$ if $y<m$, so that $\bar E\subset[m,M]$ also. Hence $\bar E$ is compact. By Exercise 13, $f$ can be extended to a continuous function $\bar f$ on $\bar E$ whose range is also compact. Hence $\bar f$ is bounded on $\bar E$, so $f$ is bounded on $E$.

The identity function $f(x)=x$ on $E=\mathbf R$ (which is not bounded) is uniformly continuous but not bounded on $E$.

Exercise 9

(By analambanomenos) Suppose $f$ is a uniformly continuous function from the metric space $X$ to the metric space $Y$. Let $\varepsilon>0$. Then there is a $\delta>0$ such that $$d_Y\bigl(f(p),
f(q)\bigr)<\varepsilon$$ if $d_X(p,q)<\delta$. Let $E\subset X$ with $\mathop{\rm diam}E<\delta$. If $p,q\in E$, then $d_X(p,q)\le\mathop{\rm diam}E<\delta$, so $d_Y\bigl(f(p),f(q)\bigr)<\varepsilon$. Hence $\mathop{\rm diam}f(E)<\varepsilon$.

Conversely, let $f$ be a function from the metric space $X$ to the metric space $Y$ with the diameter property. Let $\varepsilon>0$ and $\delta>0$ such that $$\mathop{\rm diam}f(E)<\varepsilon$$ for all $E\subset X$ with $\mathop{\rm diam}E<\delta$. Let $p,q\in X$ such that $d_X(p,q)<\delta$. Letting $E=\{p,q\}$ we have $\mathop{\rm diam}E<\delta$, so $\mathop{\rm diam}f(E)=d_Y\bigl(f(p),f(q)\bigr)<\varepsilon$. Hence $f$ is uniformly continuous.

Baby Rudin 数学分析原理完整第四章习题解答


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This Post Has One Comment

  1. Hi,

    For exercise 3, it isn't stated that f is one-to-one, so we cannot assume that its inverse exists. How do we prove this without this assumption?


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