Kreyszig 2.3, Further Properties of Normed Spaces

Problem 1. Show that $c \subset l^{\infty}$ is a vector subspace of $l^{\infty}$ and so is $C_0$, the space of all sequences of scalars converging to zero.

Solution:

The space $l^{\infty}$ is defined as the set of all bounded sequences of real (or complex) numbers. A sequence $(a_n)$ is in $l^{\infty}$ if there exists a real number $M$ such that for every term $a_n$ in the sequence, $|a_n| \leq M$.

The space $c$ denotes the set of all convergent sequences. A sequence $(a_n)$ is in $c$ if it converges to some limit $L$ in the real (or complex) numbers.

The space $C_0$, or $c_0$ as it is often denoted, is the set of all sequences that converge to zero.

To show that $c$ and $C_0$ are subspaces of $l^{\infty}$, we must verify the following properties for each:

1. Non-emptiness: The subspace must contain the zero vector.

2. Closed under vector addition: If two vectors $x$ and $y$ are in the subspace, then their sum $x + y$ must also be in the subspace.

3. Closed under scalar multiplication: If a vector $x$ is in the subspace and $\alpha$ is any scalar, then the product $\alpha x$ must also be in the subspace.

For $c$ (all convergent sequences):

1. Non-emptiness: The zero sequence is in $c$ since it converges to zero, and it is clearly bounded.

2. Closed under vector addition: If $x, y \in c$, both converge to some limits $L_x$ and $L_y$, and their sum $x + y$ converges to $L_x + L_y$. Also, the sum of two bounded sequences is bounded.

3. Closed under scalar multiplication: For any $x \in c$ and scalar $\alpha$, the sequence $\alpha x$ converges to $\alpha L_x$ and is bounded if $x$ is bounded.

For $C_0$ (sequences converging to zero):

1. Non-emptiness: $C_0$ contains the zero sequence.

2. Closed under vector addition: The sum of two sequences in $C_0$ also converges to zero.

3. Closed under scalar multiplication: A scalar multiple of a sequence in $C_0$ also converges to zero and is bounded.

Since both $c$ and $C_0$ satisfy these properties, they are both subspaces of $l^{\infty}$.

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Problem 2. Show that $c_0$ in Problem 1 is a closed subspace of $l^{\infty}$, so that $c_0$ is complete by: (a) Theorem (Complete subspace): A subspace $M$ of a complete metric space $X$ is itself complete if and only if the set $M$ is closed in $X$. (b) Completeness of $l^{\infty}$: The space $l^{\infty}$, is complete.

Solution:

To show that $c_0$ from Problem 1 is a closed subspace of $l^{\infty}$, we will use the theorem provided and the fact that $l^{\infty}$ is complete.

Step 1: Use the Theorem (Complete Subspace)

The theorem states that a subspace $M$ of a complete metric space $X$ is complete if and only if $M$ is closed in $X$. Therefore, we must demonstrate that $c_0$ is closed in $l^{\infty}$.

Step 2: Show that $c_0$ is closed in $l^{\infty}$

A subset of a metric space is closed if it contains all of its limit points. To prove that $c_0$ is closed, we need to show that if a sequence of elements in $c_0$ converges to some limit within $l^{\infty}$, then this limit is also in $c_0$.

Suppose $(x_n)$ is a sequence of sequences in $c_0$ that converges to some sequence $x$ in $l^{\infty}$. We need to show that $x$ is also in $c_0$. This means that $x$ must converge to zero.

Since $(x_n)$ converges to $x$ in $l^{\infty}$, for every $\epsilon > 0$, there exists an $N$ such that for all $n \geq N$, the sequences $x_n$ are within $\epsilon$ of $x$ in the supremum norm, i.e.,

Each $x_n$ is in $c_0$, meaning that for each $x_n$ and for every $\epsilon > 0$, there exists an $M$ (which can depend on $n$) such that for all $k \geq M$, $|(x_n)_k| < \epsilon$.

As $n \rightarrow \infty$, $x_n$ converges to $x$ and since each $x_n$ gets arbitrarily close to zero for large enough indices, $x$ must also get arbitrarily close to zero for large enough indices. This means that $x$ converges to zero and thus $x \in c_0$.

Step 3: Apply the Completeness of $l^{\infty}$

Since $l^{\infty}$ is a complete metric space and $c_0$ is closed in $l^{\infty}$, by the theorem, $c_0$ is also complete.

By showing that $c_0$ is a closed subset of the complete space $l^{\infty}$, we have shown that $c_0$ is a complete subspace of $l^{\infty}$.

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Problem 3. Problem Statement: In $l^{\infty}$, let $Y$ be the subset of all sequences with only finitely many nonzero terms. Show that $Y$ is a subspace of $l^{\infty}$ but not a closed subspace.

Solution:

To demonstrate that $Y$ is a subspace of $l^{\infty}$, we must verify that $Y$ satisfies the three properties of a vector subspace:

1. Non-emptiness: $Y$ contains the zero vector.

2. Closed under vector addition: If two vectors $x$ and $y$ are in $Y$, then their sum $x + y$ must also be in $Y$.

3. Closed under scalar multiplication: If a vector $x$ is in $Y$ and $\alpha$ is any scalar, then the product $\alpha x$ must also be in $Y$.

Let's examine each property:

1. Non-emptiness: The zero sequence, where every term is zero, is a sequence with finitely many nonzero terms (specifically, none), so $Y$ contains the zero vector.

2. Closed under vector addition: If $x$ and $y$ are in $Y$, they each have only finitely many nonzero terms. The sum $x + y$ will also have only finitely many nonzero terms because the nonzero terms can only occur at the indices where $x$ or $y$ (or both) have nonzero terms. Therefore, $x + y$ is also in $Y$.

3. Closed under scalar multiplication: If $x$ is in $Y$ and $\alpha$ is any scalar, multiplying $x$ by $\alpha$ will not introduce any new nonzero terms beyond those already present in $x$. Therefore, $\alpha x$ will also have only finitely many nonzero terms and is in $Y$.

Since $Y$ satisfies all three properties, it is a subspace of $l^{\infty}$.

To show that $Y$ is not a closed subspace, we need to find a sequence of elements in $Y$ that converges to a limit not in $Y$. This limit will be a sequence with infinitely many nonzero terms, demonstrating that $Y$ does not contain all its limit points, and hence it is not closed.

Consider the sequence of sequences $(y^{(n)})$ defined by:

Each $y^{(n)}$ is in $Y$ because it has only $n$ nonzero terms. Now, consider the sequence $y$ defined by:

The sequence $y$ is not in $Y$ because it has infinitely many nonzero terms. However, $(y^{(n)})$ converges to $y$ in the $l^{\infty}$ norm because for every $\epsilon > 0$, there exists an $N$ such that for all $n \geq N$, the tail of the sequence $y$ (from $n$ onward) is bounded above by $\epsilon$.

Therefore, the limit of the convergent sequence $(y^{(n)})$ is not in $Y$, showing that $Y$ is not closed in $l^{\infty}$.

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Problem 4. In a normed space $X$, show that vector addition and multiplication by scalars are continuous operations with respect to the norm; that is, the mappings defined by $(x, y) \mapsto x+y$ and $(\alpha, x) \mapsto \alpha x$ are continuous.

Solution:

Continuity of Vector Addition

Let $(x_n)$ and $(y_n)$ be sequences in $X$ such that $x_n \to x$ and $y_n \to y$ as $n \to \infty$. We need to show that $x_n + y_n \to x + y$. By the definition of convergence in a normed space, for every $\epsilon > 0$, there exist $N_1, N_2 \in \mathbb{N}$ such that for all $n \geq N_1$, $\|x_n - x\| < \frac{\epsilon}{2}$ and for all $n \geq N_2$, $\|y_n - y\| < \frac{\epsilon}{2}$.

Let $N = \max\{N_1, N_2\}$. Then for all $n \geq N$, we have:

The inequality follows from the triangle inequality of the norm. Since $\epsilon$ was arbitrary, this shows that $x_n + y_n \to x + y$, and thus vector addition is continuous.

Continuity of Scalar Multiplication

Let $(\alpha_n)$ be a sequence of scalars converging to $\alpha$, and let $(x_n)$ be a sequence in $X$ such that $x_n \to x$. We need to show that $\alpha_n x_n \to \alpha x$. For every $\epsilon > 0$, there exist $N_1, N_2 \in \mathbb{N}$ such that for all $n \geq N_1$, $|\alpha_n - \alpha| < \frac{\epsilon}{2(\|x\|+1)}$ and for all $n \geq N_2$, $\|x_n - x\| < \frac{\epsilon}{2(\|\alpha\|+1)}$.

Let $N = \max\{N_1, N_2\}$. Then for all $n \geq N$, we have:

Using the properties of the norm and the convergence of $\alpha_n$ and $x_n$, we further obtain:

Since $\epsilon$ was arbitrary, this shows that $\alpha_n x_n \to \alpha x$, and thus scalar multiplication is continuous.

Hence, in a normed space $X$, both vector addition and scalar multiplication are continuous with respect to the norm.

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Problem 5. Show that $x_n \to x$ and $y_n \to y$ implies $x_n + y_n \to x + y$. Show that $\alpha_n \to \alpha$ and $x_n \to x$ implies $\alpha_n x_n \to \alpha x$.

Solution:

Continuity of Vector Addition

Given $x_n \to x$ and $y_n \to y$, we need to demonstrate that $x_n + y_n \to x + y$.

By the definition of convergence, for every $\epsilon > 0$, there exists an $N_1$ such that for all $n \geq N_1$, $\|x_n - x\| < \frac{\epsilon}{2}$. Similarly, there exists an $N_2$ such that for all $n \geq N_2$, $\|y_n - y\| < \frac{\epsilon}{2}$.

Let $N = \max(N_1, N_2)$. Then for all $n \geq N$:

This proves that $x_n + y_n \to x + y$, confirming the continuity of vector addition.

Continuity of Scalar Multiplication

Given $\alpha_n \to \alpha$ and $x_n \to x$, we need to show that $\alpha_n x_n \to \alpha x$.

For every $\epsilon > 0$, there exists an $N_1$ such that for all $n \geq N_1$, $|\alpha_n - \alpha| < \frac{\epsilon}{2(\|x\| + 1)}$ (assuming $x \neq 0$, otherwise the result is trivial). Also, there exists an $N_2$ such that for all $n \geq N_2$, $\|x_n - x\| < \frac{\epsilon}{2(|\alpha| + 1)}$.

Let $N = \max(N_1, N_2)$. Then for all $n \geq N$:

Using the convergence criteria and the norm properties, we get:

and

Summing these inequalities gives:

This confirms that $\alpha_n x_n \to \alpha x$, establishing the continuity of scalar multiplication.

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Problem 6. Show that the closure $\bar{Y}$ of a subspace $Y$ of a normed space $X$ is again a vector subspace.

Solution:

To show that the closure $\overline{Y}$ of a subspace $Y$ is a vector subspace, we need to verify that it satisfies the properties of a vector subspace:

Non-emptiness: The closure $\overline{Y}$ must contain the zero vector. Since $Y$ is a subspace, it contains the zero vector $0$. The closure of a set contains all the limit points of that set, and since $0$ is in $Y$ and is its own limit, $0$ is also in $\overline{Y}$.

Closed under vector addition: If $x$ and $y$ are in $\overline{Y}$, then $x + y$ must also be in $\overline{Y}$. Let $x$ and $y$ be in $\overline{Y}$. By the definition of closure, for every $\epsilon > 0$, there exist points $x' \in Y$ and $y' \in Y$ such that $\|x - x'\| < \frac{\epsilon}{2}$ and $\|y - y'\| < \frac{\epsilon}{2}$. Since $Y$ is a subspace and therefore closed under addition, $x' + y'$ is in $Y$.

Consider $x + y$ and $x' + y'$. We have:

This inequality shows that for every point $x + y$ in $\overline{Y}$, we can find a point $x' + y'$ in $Y$ such that $x + y$ is as close as we wish to $x' + y'$, which means $x + y$ is a limit point of $Y$ and hence in $\overline{Y}$.

Closed under scalar multiplication: If $x$ is in $\overline{Y}$ and $\alpha$ is a scalar, then $\alpha x$ must also be in $\overline{Y}$. Let $x$ be in $\overline{Y}$ and let $\alpha$ be any scalar. By the definition of closure, for every $\epsilon > 0$, there exists a point $x' \in Y$ such that $\|x - x'\| < \frac{\epsilon}{|\alpha|}$ if $\alpha \neq 0$ (if $\alpha = 0$, the result is trivial since $0 \cdot x = 0$ is in $Y$ and hence in $\overline{Y}$).

Since $Y$ is a subspace, it is closed under scalar multiplication, so $\alpha x'$ is in $Y$. Consider $\alpha x$ and $\alpha x'$. We have:

This inequality shows that for every point $\alpha x$ in $\overline{Y}$, we can find a point $\alpha x'$ in $Y$ such that $\alpha x$ is as close as we wish to $\alpha x'$, which means $\alpha x$ is a limit point of $Y$ and hence in $\overline{Y}$.

Therefore, the closure $\overline{Y}$ of a subspace $Y$ of a normed space $X$ satisfies all the properties of a vector subspace and is thus itself a vector subspace of $X$.

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Problem 7. Show that convergence of $\|\mathbf{y}_1\| + \|\mathbf{y}_2\| + \|\mathbf{y}_3\| + \ldots$ may not imply convergence of $\mathbf{y}_1 + \mathbf{y}_2 + \mathbf{y}_3 + \ldots$. Hint: Consider $\mathbf{y}$ in Prob. 3 and $(\mathbf{y}_n)$, where $\mathbf{y}_n = (\eta_j^{(n)})$, $\eta_n^{(n)} = 1/n^2$, $\eta_j^{(n)} = 0$ for all $j \neq n$.

Solution:

To demonstrate the statement, we'll consider a sequence in the space $l^\infty$ of all bounded sequences of scalars, which is the space mentioned in Problem 3.

We'll construct a specific example using the hint provided, which involves sequences with only one non-zero term whose magnitude is $\frac{1}{n^2}$. This example will show that the series of norms converges (absolute convergence), but the series of vectors does not converge in the $l^\infty$ space.

Construction:

Let $y_n$ be a sequence in $l^\infty$ defined by $y_n = (\eta_j^{(n)})$ where:

This sequence $y_n$ has only the $n$-th term non-zero and equal to $\frac{1}{n^2}$, and all other terms are zero.

Absolute convergence of norms:

Consider the series of norms $\sum_{n=1}^\infty \|y_n\|$. Since $\|y_n\| = \frac{1}{n^2}$ for each $n$, the series is:

The series $\sum_{n=1}^\infty \frac{1}{n^2}$ is known to converge (it's a p-series with $p = 2$, which converges for $p > 1$).

Lack of convergence of the vector series:

Now consider the series of vectors $\sum_{n=1}^\infty y_n$. The $n$-th partial sum of this series is:

Each partial sum $S_n$ is a sequence in $l^\infty$ where the first $n$ terms are the reciprocals of the squares of the natural numbers, and the rest are zeros.

The limit of the partial sums $S_n$ as $n \to \infty$, if it exists, would be the sequence:

The sequence $S$ represents the harmonic series of squares, which does not converge in the $l^\infty$ space, because it's not a bounded sequence. Each term in the sequence $S$ is a positive number, and there are infinitely many terms, so the sequence does not converge to a point in $l^\infty$ (which requires boundedness).

Conclusion:

We have shown that while the series of norms $\sum_{n=1}^\infty \|y_n\|$ converges, the series of vectors $\sum_{n=1}^\infty y_n$ does not converge in the $l^\infty$ space. This example illustrates that absolute convergence of the norms does not imply convergence of the series of vectors in the $l^\infty$ space.

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Problem 8. Problem Statement: In a normed space $X$, if absolute convergence of any series always implies convergence of that series, show that $X$ is complete.

Proof:

1. Absolute Convergence Implies Convergence: By hypothesis, if a series $\sum_{n=1}^\infty x_n$ in $X$ is absolutely convergent, meaning that $\sum_{n=1}^\infty \|x_n\|$ converges, then the series $\sum_{n=1}^\infty x_n$ itself converges in $X$.

2. Cauchy Criterion for Series: A series $\sum_{n=1}^\infty x_n$ converges if and only if the sequence of partial sums $S_m = \sum_{n=1}^m x_n$ is a Cauchy sequence.

3. Absolute Convergence and Cauchy Sequences: Suppose $\sum_{n=1}^\infty x_n$ is absolutely convergent. Then for every $\varepsilon > 0$, there exists $N \in \mathbb{N}$ such that for all $m > n \geq N$, we have $\sum_{k=n}^m \|x_k\| < \varepsilon$ because the series of norms is convergent and hence satisfies the Cauchy criterion.

4. Implication for Partial Sums: The property above implies that the sequence of partial sums $(S_m)$ is Cauchy. To see this, note that for $m > n \geq N$,

   $\|S_m - S_n\| = \left\|\sum_{k=n+1}^m x_k\right\| \leq \sum_{k=n+1}^m \|x_k\| < \varepsilon.$

   This inequality holds because the norm is subadditive (it satisfies the triangle inequality).

5. Completeness of X: If $(S_m)$ is a Cauchy sequence in $X$ and $X$ is a space where absolute convergence implies convergence, then $(S_m)$ must converge in $X$ because it is absolutely convergent.

6. Conclusion: Since every Cauchy sequence in $X$ converges in $X$, $X$ is complete. Hence, $X$ is a Banach space.

The key point here is the equivalence of the Cauchy criterion for series convergence and the completeness of the space. The hypothesis that absolute convergence implies convergence ensures that Cauchy sequences of partial sums always converge, which is precisely the definition of a complete space. Solution:

The key point here is the equivalence of the Cauchy criterion for series convergence and the completeness of the space. The hypothesis that absolute convergence implies convergence ensures that Cauchy sequences of partial sums always converge, which is precisely the definition of a complete space.

Completeness of $X$:

To say that $X$ is complete means that every Cauchy sequence in $X$ converges to a limit within $X$. Now, let's consider any Cauchy sequence $(x_n)$ in $X$. By the property of normed spaces, we can form a series $\sum_{n=1}^\infty (x_{n+1} - x_n)$. This series is absolutely convergent if the series of norms $\sum_{n=1}^\infty \|x_{n+1} - x_n\|$ converges.

Since $(x_n)$ is a Cauchy sequence, for every $\varepsilon > 0$, there exists an $N$ such that for all $m > n \geq N$, the distance between $x_n$ and $x_m$ is less than $\varepsilon$. Formally, $\|x_m - x_n\| < \varepsilon$.

The property of being a Cauchy sequence suggests that the series of differences $(x_{n+1} - x_n)$ has terms that become arbitrarily small as $n$ increases. In other words, the series $\sum_{n=N}^\infty \|x_{n+1} - x_n\|$ has terms that decrease and approach zero, implying that the series of norms is convergent.

Given that absolute convergence of a series in $X$ implies its convergence, we can conclude that the series $\sum_{n=1}^\infty (x_{n+1} - x_n)$ converges in $X$. The convergence of this series means that the sequence of partial sums, which corresponds to the sequence $(x_n)$ up to an initial segment, converges to a limit in $X$.

Therefore, the original Cauchy sequence $(x_n)$ must also converge in $X$, because its behavior at infinity is captured by the series formed by its successive differences. Since every Cauchy sequence in $X$ has a limit in $X$, we conclude that $X$ is complete.

In summary, the condition that absolute convergence implies convergence in $X$ allows us to transform the Cauchy criterion for sequences into a condition on series. Since this condition guarantees convergence for all absolutely convergent series—and hence for all Cauchy sequences—it follows that $X$ is a complete normed space, or a Banach space.

Solution:

This proof leverages the fundamental property of normed spaces: a space is complete if every Cauchy sequence converges within the space. The given condition, that absolute convergence implies convergence, is used to show that Cauchy sequences, constructed from series of vectors in the space, converge. This implies that the space must be complete, as all such Cauchy sequences have a limit in the space, satisfying the definition of a Banach space.

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Problem 9. Show that in a Banach space, an absolutely convergent series is convergent.

Detailed Proof:

Let $(X, \|\cdot\|)$ be a Banach space. Suppose we have a series $\sum_{n=1}^\infty x_n$ in $X$ that is absolutely convergent. By definition, this means that the series $\sum_{n=1}^\infty \|x_n\|$ converges in the real numbers.

1. Definition of Absolute Convergence: The series $\sum_{n=1}^\infty \|x_n\|$ is said to be absolutely convergent if the sum of the norms, which are real numbers, is a convergent series in $\mathbb{R}$, i.e., there exists a real number $L$ such that for every $\epsilon > 0$, there is an integer $N$ such that for all $n \geq N$, it holds that

   $\left|\sum_{k=N+1}^n \|x_k\| - L\right| < \epsilon.$

2. Partial Sums as a Sequence: Define the $n$-th partial sum $S_n$ of the series $\sum_{n=1}^\infty x_n$ by $S_n = \sum_{k=1}^n x_k$. The sequence $(S_n)$ is a sequence of elements in $X$.

3. Partial Sums Are Cauchy: To show that $(S_n)$ is a Cauchy sequence, consider any $\epsilon > 0$. Since the series of norms converges, there exists an integer $N$ such that for all $m, n \geq N$ with $m < n$, we have

   $\sum_{k=m+1}^n \|x_k\| < \epsilon.$

   Now, consider the difference between the $n$-th and $m$-th partial sums:

   $\|S_n - S_m\| = \left\|\sum_{k=m+1}^n x_k\right\| \leq \sum_{k=m+1}^n \|x_k\|,$

   where we used the triangle inequality for norms. Given our choice of $N$, for $m, n \geq N$, this implies

   $\|S_n - S_m\| < \epsilon.$

   This is the Cauchy criterion for sequences in a normed space: for any $\epsilon > 0$, there exists an $N$ such that for all $m, n \geq N$, the norm of the difference between the $n$-th and $m$-th terms of the sequence is less than $\epsilon$.

4. Convergence of Cauchy Sequences in Banach Spaces: A Banach space is, by definition, a complete normed vector space. Completeness means that every Cauchy sequence in the space converges to a limit within the space. Since we have established that $(S_n)$ is a Cauchy sequence, it must converge to some limit $S$ in $X$.

5. Conclusion: The limit $S$ to which the sequence $(S_n)$ converges is the sum of the series $\sum_{n=1}^\infty x_n$. Therefore, the series converges in $X$, and we have demonstrated that an absolutely convergent series in a Banach space is indeed convergent.

Solution:

This detailed proof walks through the concepts of absolute convergence, the properties of Cauchy sequences, and the completeness of Banach spaces to conclusively show that an absolutely convergent series in a Banach space must converge. This result is a cornerstone of functional analysis and underscores the robustness of Banach spaces for analytical purposes.