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Normed Vector Space

Last updated Nov 1, 2022

# Definition

X

equipped with a Norm

||\cdot||

Backlinks

A Normed Vector Space is Complete iff all Absolutely Convergent Series Converge

A Normed Vector Space is Complete iff all Absolutely Convergent Series Converge

Statement Let $(X, ||\cdot||)$ be a . Then $X$ is a for all $({x}{n}){n=1}^{\infty} \subset X$ such that $\sum\limits{n=1 }^{\infty} ||x{n}||$ converge, $\sum\limits{n=1}^{\infty} x{n}$ converge in $X$ . Proof ...

11/7/2022
A Normed Vector Space is a Topological Vector Space

A Normed Vector Space is a Topological Vector Space

Statement Let $(X, ||\cdot||)$ be a . Then $X$ is a . Proof put results about algebraic operations in the page below, and reference them from pages like this one. [ ] ...

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Absolute Convergence

Absolute Convergence

Definition Let $(X, ||\cdot||)$ be a and let $({x}{n}){n=1}^{\infty} \subset X$ . We say the $\sum\limits{n=1}^{\infty} x{n}$ converges absolutel if $\sum\limits{n=1}^{\infty} ||x{n}||$ converge in $\mathbb{R}$ . Other Outlinks ...

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Banach Space

Banach Space

Definition A is a that is a ....

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Bounded Continuous Function Space

Bounded Continuous Function Space

...called the from $X$ to $Y$ . Proof Note that $C{b}(X, Y)$ is simply the restrcition of the $C(X, Y)$ to those with finit . Since , we have that $C{b}(X, Y)$ is a . $\blacksquare$ Other Outlinks ......

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Bounded Function Space

Bounded Function Space

...||\cdot||)$ be a . Then $B(X, Y) := \{f: X \to Y : \sup\limits_{x \in X} ||f(x)|| < \infty\}$ is a with the $||\cdot||_{\infty}$ . Proof This follows from noting that the is an , and .......

11/7/2022
Bounded Set

Bounded Set

...[s,t] $. That is,$ s \geq a \geq t $for all$ a \in A$. Remarks Definitions (1) and (2) are carrying definition (3) over to sets that are not s by a into $\mathbb{R}$ (namely $d$ and $||\cdot||$ ). Other Outlinks......

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Closed Unit Ball

Closed Unit Ball

...of radius $1$ about $0 \in X$ . It is denoted $\overline{B(X)}$ . Remarks is because . Let $r > 0$ and $x \in X$ . Then $\overline{B_{r}(x)} = r \overline{B(X)} + x$ . Proof is almost identical to Remark 2 in......

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Closure of Open Unit Ball is Closed Unit Ball

Closure of Open Unit Ball is Closed Unit Ball

... $x{n} = (1-\frac{1}{n})x$ . Since $||x{n}|| = (1-\frac{1}{n})||x|| = \left(1-\frac{1}{n}\right)< 1$ , $x{n} \in B(X)$ as we desired. Then, $||x{n} - x|| = \frac{1}{n}||x|| = \frac{1}{n} \to 0$ , so $x{n} \to x$ . Because , $x \in \text{cl}B(X)$ . $\blacksquare$ Other Outlinks ......

11/7/2022
Compact Function Space

Compact Function Space

...\subset \overline{\mathbb{R}{\geq 0}}$ is a for all $f \in C(K,Y)$ . Therefore $\sup\limits ||f(K)|| =: ||f||{\infty}$ exists and is finit. Thus, $C(K,Y) = C_{b}(K, Y)$ . Since the is a , the is also one. $\blacksquare$ Other Outlinks ......

11/7/2022
Direct Product Normed Space

Direct Product Normed Space

...X \times Y \to \mathbb{R}{\geq 0} $defined as$ $||(x, y)||_{p} := (||x||^{p} + ||y||^{p})^{\frac{1}{p}}$ $ forms a . It is known as the . Remarks Sometimes, this space is called the " $l{p}$ sum of $X$ and $Y$ " and written $X \oplus{p}......

11/7/2022
Equivalent Conditions for Convexity

Equivalent Conditions for Convexity

...+ (1 - \lambda) (v, f(v)) = \Big(\lambda u + (1- \lambda)v, \lambda f(u) + (1- \lambda)f(v)\Big)$$ Thus, by definition of , we have that $f(\lambda u + (1-\lambda)v) \leq \lambda f(u) + (1-\lambda) f(v)$ and $f$ is a .......

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...because and . Definition (as a ) Taking the as the described above, we can endow it with the . Although it might seem that this depends on the used above, , so any ......

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Finite Normed Vectors form a Normed Vector Space

Finite Normed Vectors form a Normed Vector Space

...and let $||\cdot||$ be an on $V$ . Then $X \subset V$ defined as $X := \{x \in V : ||x|| < \infty\}$ is a . Proof , this basically requires applying the and properties of an ......

11/7/2022
Frechet Derivative

Frechet Derivative

...0} Th = \mathbf{0}_{W}\\ &\Rightarrow \lim\limits{h \to 0} \big( f(v + h) - f(v) \big) = \mathbf{0}{W} \end{align*}$$ Where the last line follows because (how to prove this?). The last line is the definition of continuit and $f$ is......

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...2 \text{Re}\langle \mathbf{x}, \mathbf{y} \rangle + ||\mathbf{y}||^{2}}\\ &\leq\sqrt{||\mathbf{x}||^{2} + 2 |\langle \mathbf{x}, \mathbf{y} \rangle| + ||\mathbf{y}||^{2}}\\ &\leq\sqrt{||\mathbf{x}||^{2} + 2 ||\mathbf{x}|| ||\mathbf{y}|| + ||\mathbf{y}||^{2}} &\text{(Cauchy Schwarz)}\\ &= ||\mathbf{x}|| + ||\mathbf{y}|| \end{align*}$$ Thus $||\cdot||$ is indeed a , making $X$ a .......

11/7/2022
Lp Space

Lp Space

...broken into the following pieces [x] Lp is a [x] is well defined [ ] is a Remarks The definition of $||\cdot||_\infty$ hinges on the fact that $\mathbb{K}$ is a . Other Outlinks ......

11/7/2022
Magnitude is a Norm

Magnitude is a Norm

Statement $\mathbb{C}$ equipped with $|\cdot|$ is a . Proof Follows from as an and ....

11/7/2022
Norm Topology

Norm Topology

Definition Let $(X, ||\cdot||)$ be a . The on $X$ is the of the metric induced b $||\cdot||$ Remarks When we talk about a , we implicitly endow it with a ....

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Normed Vector Space

Normed Vector Space

Definition A is a $X$ equipped with a $||\cdot||$ ....

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Vector Subspace is a Normed Vector Subspace

Vector Subspace is a Normed Vector Subspace

...If $W \subset V$ is a , then $(W, ||\cdot|| {\big|}_{W})$ is a called a . Proof $W$ is a by defnition of . The s easily follow because they already hold $V$ and $W \subset V$ .......

11/7/2022
Norms are Continuous

Norms are Continuous

... $X$ . Then $(x{n}, 0) \to (x, 0)$ in $X \times X$ because . Because and , $||x{n}|| = ||x{n} - 0|| \to ||x - 0|| = ||x||$ . By the definition of in a , $||\cdot||$ is a .......

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...\in X $,$ $\begin{align*} d(\mathbf{x}, \mathbf{z}) &= ||\mathbf{x} - \mathbf{z}||\\ &= ||\mathbf{x} - \mathbf{y} + \mathbf{y} - \mathbf{z}||\\ &\leq ||\mathbf{x} - \mathbf{y}|| + ||\mathbf{y} - \mathbf{z}||\\ &=d(\mathbf{x}, \mathbf{y}) + d(\mathbf{y}, \mathbf{z}) \end{align*}$ $Thus$ d$ is a and $X$ is a......

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Open Unit Ball

Open Unit Ball

...\in B(X) $. Thus$ B{r}(X) \subset rB(X) + x $. On the other hand, let$ x' \in B(X) $. Then$ ||(rx' + x) - x|| = ||rx'|| = r||x'|| < r $. Thus,$ rx' + x \in B{r}(X) $and$ B{r}(X) \supset rB(X) + x$.......

11/7/2022
Quotient Norm

Quotient Norm

...y|| = \text{dist}\{x, Y\}$$ for $x \in X$ is a on $X / Y$ . Proof Remarks The requirement that $Y$ is [Closed]] is very important. See [StackExchange on why the subspace needs to be closed. Other Outlinks ......

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Quotient Normed Vector Space

Quotient Normed Vector Space

...Let $(X, ||\cdot||)$ be a and let $Y \subset X$ be a . Then we call the $X / Y$ endowed with the $||[\cdot]||$ . Remarks The requirement that $Y$ is is very important. See......

11/7/2022
Restricting an Extended Norm to elements of Finite Norm form a Normed Vector Space

Restricting an Extended Norm to elements of Finite Norm form a Normed Vector Space

...X $and$ X$ is a because . This also makes $X$ a in its own right. It follows that $X$ is a because finitenes is preserved by the 3 defining s of a . $\blacksquare$ Other Outlinks......

11/7/2022
Supremum Norm

Supremum Norm

...||x + y|| &= \sup\limits{i \in I} ||x{i} + y{i}||{i} \\ &\leq \sup\limits{i \in I} (||x{i}||{i} + ||y{i}||_{i}) \\ &\leq \sup\limits{i \in I} ||x{i}||{i} + \sup\limits{i \in I} ||y{i}||{i} \\ &= ||x + y||. \checkmark \end{align*}$$ $\blacksquare$ Other Outlinks ......

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Tightly Bounded Set

Tightly Bounded Set

...for any $x \in A$ . If $A$ is a , this is equivalent $\sup\limits A \in A$ and $\inf\limits A \in A$ . This is because $A$ contains is and its if it is . Other Outlinks ......

11/7/2022
Total Subset

Total Subset

... from our is in $\text{span}A$ . This means for each $x \in X$ , we have a in $\text{span}A$ that converge to $x$ . Thus, by equivalent conditions on density on metric space, $\text{span}A$ is and $A$ is a......

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Unit Sphere

Unit Sphere

... of radius $1$ about $0 \in X$ . It is denoted $S(X)$ . Remarks is because . Let $r > 0$ and $x \in X$ . Then $S_{r}(x) = rS(X) + x$ . Proof is almost identical to Remark 2 in......

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Unit Vector

Unit Vector

... $x \in X$ is a if $||x|| = 1$ . Remarks The is the of s. For any $x \in X$ so that $x \neq 0$ , we can rescale it to be a by taking $\hat{x} :=......

11/7/2022

Normed Vector Space

# Definition

Normed Vector Space

Vector Space

Norm

Backlinks

A Normed Vector Space is Complete iff all Absolutely Convergent Series Converge

A Normed Vector Space is a Topological Vector Space

Absolute Convergence

Banach Space

Bounded Continuous Function Space

Bounded Function Space

Bounded Set

Closed Unit Ball

Closure of Open Unit Ball is Closed Unit Ball

Compact Function Space

Direct Product Normed Space

Equivalent Conditions for Convexity

Finite Normed Vectors form a Normed Vector Space

Frechet Derivative

Lp Space

Magnitude is a Norm

Norm Topology

Normed Vector Space

Vector Subspace is a Normed Vector Subspace

Norms are Continuous

Open Unit Ball

Quotient Norm

Quotient Normed Vector Space

Restricting an Extended Norm to elements of Finite Norm form a Normed Vector Space

Supremum Norm

Tightly Bounded Set

Total Subset

Unit Sphere

Unit Vector

Interactive Graph