Functional Analysis Exercises 4 : Convergence, Cauchy Sequences, and Completeness

Avishek Sen Gupta on 12 October 2021

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This post lists solutions to the exercises in the Convergence, Cauchy Sequences, and Completeness section 1.4 of Erwin Kreyszig’s Introductory Functional Analysis with Applications. This is a work in progress, and proofs may be refined over time.

1.4.1. (Subsequence) If a sequence $(x)$ in a metric space $X$ is convergent and has limit $x$, show that every subsequence $(x_{n_k})$ of $(x_n)$ is convergent and has the same limit $x$.

Proof:

Suppose $(x_n)$ is convergent and $x_n \rightarrow x$. Let $(x_{n_k})$ be a subsequence. Let $x_{n_m}$ correspond to $x_i$.

Since $(x_n)$ is convergent, $\forall \epsilon>0, \exists N$, such that $d(x_n,x)<\epsilon$ for $n>N$. Pick $p \geq N$, such that $x_p$ exists in $(x_{n_k})$ and is identified as $x_{n_m}$.

This implies that $\forall \epsilon>0, \exists M$, such that $d(x_{n_m},x)<\epsilon$.

The above is the definition of the convergence of a sequence to a limit. Thus, $(x_{n_k})$ converges to $x$.

Alternatively, you can prove that the limit of $(x_{n_k})$ is $x$ in the following manner.
Suppose $x_{n_k} \rightarrow y$. Then, by the Triangle Inequality, we have:

$$d(x,y) \leq d(x,x_p) + d(x_p,y) \\ \Rightarrow d(x,y) \leq d(x,x_p) + d(x_{n_m},y)$$

Both $d(x,x_p)$ and $d(x_{n_m},y)$ can be made as small as possible, since $(x_n)$ and $(x_{n_k})$ converge, implying that $d(x,y)$ is smaller than any positive value. Thus:

$$d(x,y) \leq d(x,x_p) + d(x_{n_m},y) \\ \Rightarrow d(x,y) < \epsilon_1 + \epsilon_2, \epsilon_1, \epsilon_2 > 0 \\ \Rightarrow d(x,y)=0$$

Thus, $x=y$, i.e., $x_{n_k}$ has the same limit as $(x_n)$.

$$\blacksquare$$

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1.4.2. If $(x_n)$ is Cauchy and has a convergent subsequence, say, $x_n \rightarrow x$, show that $(x_n)$ is convergent with the limit $x$.

Proof:

Suppose $(x_n)$ is Cauchy. Let $(x_{n_k})$ be a subsequence. Let $x_{n_m}$ correspond to $x_i$.

• Since $(x_{n_k})$ is convergent, $\forall \epsilon>0, \exists M$, such that $d(x_{n_k},x)<\epsilon$ for $n>M$.
• Since $(x_n)$ is convergent, $\forall \epsilon>0, \exists N$, such that $d(x_i,x_j)<\epsilon$ for $i,j>N$.

Pick $N_0=\max(M,N)$. Then the above two statements become:

• Since $(x_{n_k})$ is convergent, $\forall \epsilon>0, \exists N_0$, such that $d(x_{n_i},x)<\epsilon$ for $i>N_0$.
• Since $(x_n)$ is convergent, $\forall \epsilon>0, \exists N_0$, such that $d(x_i,x_j)<\epsilon$ for $i,j>N_0$.

(Note that we picked the same $i$ for both $(x_n)$ and $(x_{n_k})$ because any value greater than $N_0$ will fulfil the above conditions, so we might as well pick the same index. For any index $i$, we can use $x_i$ and $x_{n_i}$ interchangeably, since they index the same element in both the sequence and the subsequence.)

By the Triangle Inequality, we have:

$$d(x_j,x) \leq d(x_j, x_i) + d(x_i,x) \\ \Rightarrow d(x_j,x) \leq d(x_j, x_i) + d(x_{n_i},x) \\ \Rightarrow d(x_j,x) \leq \epsilon + \epsilon \\ \Rightarrow d(x_j,x) < 2\epsilon$$

$\epsilon$ can be made as small as possible, implying that $d(x_j,x)$ is smaller than any positive value. Thus:

$$d(x_j,x)=0$$

Hence, $(x_n)$ is convergent with the limit $x$.

$$\blacksquare$$

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1.4.3. Show that $x_n \rightarrow x$ if and only if for every neighborhood $V$ of $x$ there is an integer $n_0$ such that $x_n \in V$ for all $n > n_0$.

Proof:

Suppose that $x_n \rightarrow x$. Then $\forall \epsilon>0, \exists N_0$, such that $d(x_n,x)<\epsilon$ for all $n>N_0$. This implies that a neighbourhood $V_\epsilon$ of $x$ exists, which contains all $x_{n>N_0}$. Since there are an infinite number of values for $\epsilon$, it follows that this applies to every neighbourhood of $x$.

Conversely, suppose that for every neighborhood $V$ of $x$ there is an integer $n_0$ such that $x_n \in V$ for all $n > n_0$. Assume each neighbourhood has a size of $\epsilon$. Thus, $x_n \in V$ implies that $d(x_n,x)<\epsilon$. Then, we can restate this as the following: $\forall \epsilon>0, \exists N_0$, such that $d(x_n,x)<\epsilon$ for all $n>N_0$.

$$\blacksquare$$

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1.4.4. (Boundedness) Show that a Cauchy sequence is bounded.

Proof:

By definition, for a Cauchy sequence, we have: $\forall \epsilon>0, \exists N_0$, such that $d(x_m,x_n)<\epsilon$ for all $m,n>N_0$$.

Choose $\epsilon=1$. Then, assume the value of $N_0$ to be $N_1$. For any $d(x_a,x_b)$, we have:

• $a<b \leq N_0$: Then $d(x_a,x_b) \leq a = max[d(x_a,x_0), d(x_a,x_1), \cdots, d(x_a,x_{N_1})]$
• $N_0<a<b$: Then $d(x_a,x_b) < \epsilon = 1$
• $a \leq N_0<b$: By the Triangle Inequality, we have: $d(x_a,x_b) \leq d(x_a,x_{N_1}) + d(x_{N_1}, x_b) < a + 1$

Combining these upper bounds, we get: $\sup d(x_a,x_b) < a+1$

$$\blacksquare$$

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1.4.5. Is boundedness of a sequence in a metric space sufficient for the sequence to be Cauchy? Convergent?

Answer:

Consider the discrete metric on $\mathbb{R}$. If we have a sequence $(x_n)=0,1,0,1,\cdots$, then the series is bounded because $\sup d(x_m, x_n)=1$, but for $\epsilon=\frac{1}{2}$, there is no $N$ for which $d(x_m,x_n)<\epsilon$ for $m,n>N$. Thus, the sequence is not Cauchy, though it is bounded.

Convergence is sufficient for a sequence to be Cauchy. For convergence, we have the condition: if $x_n \rightarrow x$, $\forall \epsilon>0, \exists N_0$, such that $d(x_n,x)<\epsilon$ for all $n>N_0$.

Consider $m,n>N_0$. Then, by the Triangle Inequality, we have:

$$d(x_m,x_n) \leq d(x_m,x) + d(x,x_n) < \epsilon + \epsilon = 2 \epsilon$$

thus proving the Cauchy criterion.

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1.4.6. If $(x_n)$ and $(y_n)$ are Cauchy sequences in a metric space $(X, d)$, show that $(a_n)$, where $a_n = d(x_n, y_n)$, converges. Give illustrative examples.

Proof:

$$d(x_m,x_n)<\epsilon \\ d(y_m,y_n)<\epsilon$$

Then, we have:

$$d(x_m,y_m) \leq d(x_m,x_n) + d(x_n,y_n) + d(y_n,y_m) \\ \Rightarrow d(x_m,y_m) - d(x_n,y_n) \leq d(x_m,x_n) + d(y_n,y_m) \\ \Rightarrow d(x_m,y_m) - d(x_n,y_n) < 2 \epsilon$$

Similarly, we have:

$$d(x_n,y_n) \leq d(x_n,x_m) + d(x_m,y_m) + d(y_m,y_n) \\ d(x_n,y_n) - d(x_m,y_m) \leq d(x_n,x_m) + d(y_m,y_n) \\ \Rightarrow d(x_n,y_n) - d(x_m,y_m) < 2 \epsilon$$

The above inequalities imply that: $\vert d(x_m,y_m) - d(x_n,y_n) \vert < 2 \epsilon \\ d[d(x_m,y_m) - d(x_n,y_n)] < 2 \epsilon$

This implies that $a_n=d(x_n,y_n)$ is Cauchy, and thus converges.

$$\blacksquare$$

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1.4.7. Give an indirect proof of Lemma 1.4-2(b).

Lemma 1.4-2(b) is: Let $X=(X,d)$ be a metric space. Then, if $x_n \rightarrow x$ and $y_n \rightarrow y$, then $d(x_n,y_n) \rightarrow d(x,y)$.

Proof:

We have $x_n \rightarrow x$ and $y_n \rightarrow y$. Then $(x_n)$ and $(y_n)$ are Cauchy. Thus the following two statements hold true:

• $\forall \epsilon/2>0, \exists M$ such that $d(x_m,x_n)<\epsilon/2$ for $m,n>M$
• $\forall \epsilon/2>0, \exists N$ such that $d(y_m,y_n)<\epsilon/2$ for $m,n>N$

Taking $N_0=\max(M,N)$, the above statements become:

$\forall \epsilon/2>0, \exists N_0$ such that $d(x_m,x_n)<\epsilon/2$ and $d(y_m,y_n)<\epsilon/2$ for $m,n>N_0$, i.e., $d(x_m,x_n)+d(y_m,y_n)<\epsilon/2+\epsilon/2=\epsilon$

We will prove the result using proof by contradiction.

Suppose $(a_n)=(d(x_n,y_n))$ Suppose the claim is not true. Then, $\require{cancel} a_n \cancel\rightarrow a$, thus $(a_n)$ is not Cauchy. This implies that: $\exists \epsilon$ such that $\forall N$, we have $d(a_m, a_n)>\epsilon$ for all $m,n>N$.

By the Triangle Inequality, we have:

$$d(x_m,y_m) \leq d(x_m,x_n) + d(x_n,y_n) + d(y_n,y_m) \\ \Rightarrow d(x_m,x_n) + d(y_n,y_m) \geq d(x_m,y_m) - d(x_n,y_n) \\ \Rightarrow d(x_m,x_n) + d(y_n,y_m) > \epsilon \\$$

This is then true for arbitrary $\epsilon$. But, this implies that for all $N$, we cannot make $d(x_m,x_n)+d(y_m,y_n)<\epsilon$. This is a contradiction, since by assumption, we have

$$d(x_m,x_n)+d(y_m,y_n)<\epsilon$$

Thus $(a_n)$ is Cauchy, and is thus a convergent sequence.

$$\blacksquare$$

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1.4.8. If $d_1$ and $d_2$ are metrics on the same set $X$ and there are positive numbers $a$ and $b$ such that for all $x, y \in X$, $a.d_1(x,y) \leq d_2(x,y) \leq b.d_1(x,y)$, show that the Cauchy sequences in $(X, d_1)$ and $(X, d_2)$ are the same.

Proof:

We wish to show that if $L_1$ is the limit of a Cauchy sequence in $(X,d_1)$ (call it $x_n(d_1)$)and $L_2$ is the limit of a Cauchy sequence in $(X,d_2)$ (call it $x_n(d_2)$), then $L_1=L_2$.

We have $\forall x, y \in X$, $a.d_1(x,y) \leq d_2(x,y) \leq b.d_1(x,y)$.

Then, the Triangle Inequality gives us:

$$d_1(L_1,L_2) \leq d_1(L_1,x) + d_1(x,L_2)$$

Applying the given metric constraints:

$$d_1(L_1,L_2) \leq d_1(L_1,x) + \frac{1}{a} d_2(x,L_2)$$

We know that $x_n(d_1) \rightarrow L_1$ and $x_n(d_2) \rightarrow L_2$, therefore. If we have the following:

• $\forall \epsilon>0, \exists M$ such that $d_1(x_m(d_1),L_1)<\epsilon$ for $m>M$
• $\forall \epsilon>0, \exists N$ such that $d_1(x_m(d_2),L_2)<\epsilon$ for $m>N$

Pick $N_0=\max(M,N)$, so that the above holds true for $N_0$.

Then $d_1(L_1,x_{N_0})<\epsilon$ and $d_2(L_2,x_{N_0})<\epsilon$, so that we get:

$$d_1(L_1,L_2) < \epsilon \left(1+\frac{1}{a} \right)$$

Since the above is true for all $\epsilon>0$, we can conclude that $d_1(L_1,L_2)=0$. Hence $L_1=L_2$.

The same procedure can also be showing using $d_2$.

$$\blacksquare$$

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1.4.9. Using Prob. 8, show that the metric spaces in Probs. 13 to 15, Sec. 1.2, have the same Cauchy sequences.

The three distance metrics mentioned are:

• $$d_1(x,y)=d(x_1,x_2)+d(y_1,y_2)$$
• $$d_2(x,y)=\sqrt{ {d(x_1,x_2)}^2+{d(y_1,y_2)}^2}$$
• $$d_{max}(x,y)=\max [d(x_1,x_2),d(y_1,y_2)]$$

Proof:

For $d_1$ and $d_2$, let’s determine the conditions.

$$x+y \leq \sqrt{x^2+y^2} \\ x^2+y^2+2xy \leq x^2+y^2$$

This gives us $2xy \leq 0$, so that’s invalid; if we however we introduce a $\sqrt{2}$ on the right hand side, we get:

$$x+y \leq \sqrt{2(x^2+y^2)} \\ x^2+y^2+2xy \leq 2x^2+2y^2 \\ x^2+y^2-2xy \geq 0$$

which works. For the reverse inequality $x+y \geq \sqrt{x^2+y^2}$, note that we immediately get $2xy>0$, which works, so we can write the combined inequalities as:

$$\sqrt{x^2+y^2} \leq x+y \leq \sqrt{2} \sqrt{x^2+y^2} \\ \Rightarrow d_2(x,y) \leq d_1(x,y) \leq \sqrt{2} d_2(x,y)$$

For $d_1$ and $d_max$, note that $x+y> \geq \max(x,y)$ and $2 \max(x,y) \geq x+y$

Then, we get:

$$\max(x,y) \leq x+y \leq 2 \text{ max }(x,y) \\ \Rightarrow d_{max}(x,y) \leq d_1(x,y) \leq 2 d_{ max}(x,y)$$

For $d_2$ and $d_max$, note that $x^2+y^2> \geq {\max(x,y)}^2$ and $2 {\text{ max }(x,y)}^2 \geq x^2+y^2$

Then, we get:

$$\max(x,y) \leq \sqrt{x^2+y^2} \leq 2 \text{ max }(x,y) \\ \Rightarrow d_{max}(x,y) \leq d_2(x,y) \leq 2 d_{max}(x,y)$$ $$\blacksquare$$

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1.4.10. Using the completeness of $\mathbb{R}$, prove completeness of $\mathbb{C}$.

Proof:

Assume $\mathbb{R}$ is complete.

Assume two Cauchy Sequences in $\mathbb{R}$:

• $(x_n)=x_1,x_2,\cdots$ converges to $x$.
• $(y_n)=y_1,y_2,\cdots$ converges to $y$.

Construct a sequence in $\mathbb{C}$, like so:

$$(z_n)=x_1+iy_1,x_2+iy_2,\cdots$$

Assume the distance metric for $\mathbb{Z}$ is $d(z_1,z_2)=\sqrt{ {(x_1-x_2)}^2 + {(y_1-y_2)}^2}$.

• $\forall \epsilon>0, \exists M$ such that $x_m-x<\frac{\epsilon}{\sqrt{2}}$ for $m>M$
• $\forall \epsilon>0, \exists N$ such that $x_m-x<\frac{\epsilon}{\sqrt{2}}$ for $m>N$

Pick $N_0=\max(M,N)$, so that the above holds true for $N_0$.

Pick $z_i$ so that $i>N_0$. Assume $z=x+iy$. Then, we have:

$$d(z_i,z)=\sqrt{ {(x_i-x)}^2 + {(y_i-y)}^2}<\epsilon$$

Then, for an arbitrary $\epsilon$, there exists $N_0$, such that $d(z_i,z)<\epsilon$. Furthermore $z \in \mathbb{C}$. Thus, $\mathbb{C}$ contains the limits of all its Cauchy sequences. Thus, it is a closed set; hence it is a complete metric space.

$$\blacksquare$$

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