HW 2 - Finishing Duality, Review of Inner Product Spaces

3.F: 12, 14, 15, 20, 22, 27, 35, 36
6.B: 5, 7, 15

3.F: Duality (cont.)

12

Theorem

The dual map of the identity map on $V$ is the identity map on $V^{'}$ . Namely, where $i : V \to V$ defined as:

i (v) = v

then the dual map of $i$ is the identity $i^{'} : V^{'} \to V^{'}$ given by:

i^{'} (φ) = φ

Proof
Notice that $i \in L (V)$ so then the dual map of $i$ is the linear map $i^{'} \in L (V^{'})$ defined by:

i^{'} (φ) = φ \circ i = φ

as we expected.
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14

Define $T : P (R) \to P (R)$ by $(T p) (x) = x^{2} p (x) + p^{″} (x)$ for $x \in R$ .

a

Suppose $φ \in P (R)^{'}$ is defined by $φ (p) = p^{'} (4)$ . Then notice that the linear function $T^{'} (φ)$ on $P (R)$ is:

T^{'} (φ) = φ \circ T

where if we plug in some $p \in P (R)$ then:

T^{'} (φ) (p) = φ (T p) = φ (x^{2} p (x) + p^{″} (x))

notice that since $φ$ is linear, then:

\begin{aligned} T^{'} (φ) (p) & = φ (x^{2} p (x)) + φ (p^{″} (x)) \\ = \frac{d}{d x} (x^{2} p (x)) |_{x = 4} + p^{‴} (4) \\ = x^{2} p^{'} (x) + 2 x p (x) |_{x = 4} + p^{‴} (4) \\ = 16 p^{'} (4) + 8 p (4) + p^{‴} (4) \end{aligned}

So then $T^{'} (φ)$ essentially takes $φ$ and does extra "stuff", namely $T$ on it, and gives a new transformation! Namely $T^{'} (φ) : P (R) \to R$ (ie: it's a functional) where:

T^{'} (φ) (p) = 8 p (4) + 16 p^{'} (4) + p^{‴} (4)

b

Suppose $φ \in P (R)^{'}$ is defined by $φ (p) = \int_{0}^{1} p (x) d x$ . Then we'll find what $(T^{'} (φ)) (x^{3})$ is. Notice for any $p$ like we did before that:

\begin{aligned} (T^{'} (φ)) (p) & = (φ \circ T) p \\ = φ (T p) \\ = φ (x^{2} p (x) + p^{″} (x)) \\ = φ (x^{2} p (x)) + φ (p^{″} (x)) \\ = \int_{0}^{1} x^{2} p (x) d x + \int_{0}^{1} p^{″} (x) d x \\ = \int_{0}^{1} x^{2} p (x) d x + p^{'} (1) - p^{'} (0) \end{aligned}

Now plug in $p = x^{3}$ as we needed:

\begin{aligned} (T^{'} (φ)) (x^{3}) & = \int_{0}^{1} x^{2} \cdot x^{3} d x + 3 x^{2} |_{0}^{1} \\ = \frac{x^{6}}{6} |_{0}^{1} + 3 \\ = \frac{19}{6} \end{aligned}

15

Theorem

Suppose $W$ is finite-dimensional and $T \in L (V, W)$ . Then $T^{'} = 0$ iff $T = 0$ .

Proof
Notice that it's possible that $V$ is not finite dimensional, so we cannot use properties of the dimensions of these spaces here.

As such, first consider $(\leftarrow)$ . Suppose that $T = \vec{0}$ that maps all vectors from $V$ to the zero vector ${\vec{0}}_{W}$ , namely:

T v = {\vec{0}}_{W}

for all $v \in V$ . Then the map $T^{'} \in L (W^{'}, V^{'})$ is given by the following, where $φ \in W^{'}$ :

T^{'} (φ) = φ \circ T = φ \circ \vec{0} = 0

where $0$ is justified because, if we have any arbitrary vector $v \in V$ then:

T^{'} (φ) (v) = (φ \circ T) v = φ (T v) = φ ({\vec{0}}_{W}) = 0

due to $φ$ 's linearity. As a result, then $T^{'} = 0$ as we expected, as $v$ was arbitrary.

Now for $(\to)$ . Suppose $T^{'} = 0$ , namely that for all $v \in V$ and any $φ \in W^{'}$ that:

T^{'} (φ) (w) = (φ \circ T) v = φ (T v) = 0

Now notice that since $W$ is finite-dimensional, then there is some basis $w_{1}, . . ., w_{n}$ for an $n$ -dimensional $W$ . As such, then $T v = \sum_{i = 1}^{n} a_{i} w_{i}$ for all $a_{i} \in F$ . Then:

T^{'} (φ) (v) = 0 = φ (\sum_{i = 1}^{n} a_{i} w_{i}) = \sum_{i = 1}^{n} a_{i} φ (w_{i})

but notice that we get free reign to choose our $φ$ as we so please. As such, choose $φ$ to be the functional from the dual basis $φ_{1}, . . ., φ_{n}$ (since $dim (W^{'}) = n$ ) that does:

φ_{j} (w_{i}) = χ (i = j)

for each basis vector $w_{i}$ . As such, then:

T^{'} (φ) (v) = \sum_{i = 1}^{n} a_{i} φ_{j} (w_{i}) = a_{j} = 0

thus for all $j$ then $a_{j} = 0$ and therefore, $T$ is the zero-map.
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20

Theorem

Suppose $U, W$ are subsets of $V$ with $U \subset W$ . Then $W^{0} \subset U^{0}$ .

Proof
Recall that:

U^{0} = {φ \in V^{'} : φ (u) = 0 \forall u \in U}

and similar for $W^{0}$ . As such, let $φ \in W^{0}$ be arbitrary, so then $φ \in V^{'}$ and $φ (w) = 0$ for all $w \in W$ . We want to show that:

φ \in U^{0} = {φ \in V^{'} : φ (u) = 0 \forall u \in U}

as such, let $u \in U$ be arbitrary. Clearly we already have $φ \in V^{'}$ , but since $U \subset W$ then $u \in W$ . as such, then by $φ \in W^{0}$ then $φ (u) = 0$ as we needed, hence $φ \in U^{0}$ as desired.
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22

Theorem

Suppose $U, W$ are subspaces of $V$ . Then $(U + W)^{0} = U^{0} \cap W^{0}$ .

Proof
Consider $(\subseteq)$ . Let $φ \in (U + W)^{0} \subseteq V^{'}$ , so then for all vectors $\vec{x} \in U + W$ then $φ (\vec{x}) = 0$ . Notice then that if we have vector $u + w \in U + W$ then:

\begin{aligned} φ (u + w) & = 0 \\ φ (u) + φ (w) & = 0 \end{aligned}

But notice that $u = u + {\vec{0}}_{W} \in U + W$ and $w = w + {\vec{0}}_{U} \in U + W$ . Thus, then $φ (u) = 0$ and $φ (w) = 0$ by our definition of $φ$ . As such, then $φ \in U^{0}$ and $φ \in W^{0}$ as $u, w$ are still arbitrary. Hence $φ \in U^{0} \cap w^{0}$ and thus we have a subset.

Consider $(\supseteq)$ . Let $φ \in U^{0} \cap W^{0}$ . Then $φ \in U^{0}$ and $φ \in W^{0}$ . Thus, if we let $u + w \in U + W$ be arbitrary, then notice that:

φ (u + w) = φ (u) + φ (w) = 0 + 0 = 0

Thus $φ \in (U + W)^{0}$ , as desired.
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27

Theorem

Suppose $T \in L (P_{5} (R), P_{5} (R))$ and $null (T^{'}) = span (φ)$ , where $φ$ the linear functional on $P_{5} (R)$ defined by $φ (p) = p (8)$ . Then:

range (T) = {p \in P_{5} (R) : p (8) = 0}

Proof
Notice that our vector space $P_{5} (R)$ is finite dimensional with a dimension 6. As such, then we know that:

null (T^{'}) = (range (T))^{0} = {ψ \in P_{5} (R)^{'} : \forall p \in range (T) (φ (p) = 0)} = span (φ)

expanding these definitions:

\begin{aligned} {ψ \in P_{5} (R)^{'} : \forall p \in range (T) (p (8) = 0)} & = span (φ) \end{aligned}

Now, just let $p \in range (T)$ be arbitrary. Clearly we know that $p \in P_{5} (R)$ , satisfying the first condition of our theorem. For the other condition, we know that and $ϕ \in span (φ)$ then, using our equality above, then has it that $p (8) = 0$ . Thus, this shows at least that $range (T) \subseteq$ to the right set.

Now if we just show that their dimensions are the same, then we are done and get equality. Notice that we are dealing with finite-dimensional vector spaces here:

\begin{aligned} dim (range (T)) & = dim (P_{5} (R)) - dim ((range (T))^{0}) \\ = 6 - dim (span (φ)) \\ = 6 - 1 \\ = 5 \end{aligned}

since $range (T)$ is finite dimensional and is a subspace of $P_{5} (R)$ :

\begin{aligned} dim ({p \in P_{5} (R) : p (8) = 0}) & = 5 \end{aligned}

because the set ${(x - 8), (x - 8)^{2}, (x - 8)^{3}, (x - 8)^{4}, (x - 8)^{5}}$ is a valid spanning set that is LI. This is because, like we did in a previous HW, for any:

0 = α_{1} (x - 8) + \dots + α_{5} (x - 8)^{5} \in P_{5} (R)

clearly if it equals 0, then the $(x - 8)^{5}$ term is the only one with an $x^{5}$ terms, so then it $α_{5} = 0$ . You can repeat this process all the way down to $α_{1} = 0$ , thus showing LI. For span, notice that each basis vector is validly within the set, and any any vector in the set is of the form:

(x - 8) p

where $p \in P_{4} (R)$ , where as a result then any of our basis vectors contains this new polynomial, showing span. Thus, the dimension is of the right value as we have 5 basis vectors for our space.
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35

Theorem

$(P (R))^{'}$ is isomorphic to $R^{\infty}$ .

Proof
Define the map $Γ : R^{\infty} \to (P (R))^{'}$ by the following:

Γ (x_{0}, x_{1}, . . .) = φ (p)

where all $x_{i} \in R$ . We define $φ : P (R) \to R$ as follows from our $x_{i}$ 's:

φ (b_{0} + b_{1} x + \dots) = \sum_{i = 0}^{\infty} x_{i} b_{i}

Let's go through the motions. First, let's show that $Γ$ is linear:

$Γ \in L (R^{\infty}, (P (R)))$

Let $\vec{x} = (x_{0}, x_{1}, . . .), \vec{y} = (y_{0}, y_{1}, . . .) \in R^{\infty}$ be arbitrary, and $λ \in R$ is arbitrary.

For additivity, let $p = b_{0} + b_{1} x + \dots \in P (R)$ be arbitrary:

\begin{aligned} Γ (\vec{x} + \vec{y}) p & = Γ (x_{0} + y_{0}, . . .) p \\ = φ_{x + y} (b_{0} + b_{1} x + \dots) \\ = \sum_{i = 0}^{\infty} (x_{i} + y_{i}) b_{i} \\ = \sum_{i = 0}^{\infty} x_{i} b_{i} + \sum_{i = 0}^{\infty} y_{i} b_{i} \\ = φ_{x} (p) + φ_{y} (p) \\ = Γ (\vec{x}) p + Γ (\vec{y}) p \\ = (Γ (\vec{x}) + Γ (\vec{y})) p \end{aligned}

and for homogeneity:

\begin{aligned} Γ (λ \vec{x}) p & = Γ (λ x_{0}, λ x_{1}, . . .) p \\ = φ_{x} (b_{0} + b_{1} x + \dots) \\ = \sum_{i = 0}^{\infty} (λ x_{i}) b_{i} \\ = λ \sum_{i = 0}^{\infty} x_{i} b_{i} \\ = λ Γ (\vec{x}) p \end{aligned}

Thus $Γ \in L (R^{\infty}), (P (R)^{'})$ .

Showing Injectivity and Surjectivity

For injectivity, we'll show that $null (Γ) = {\vec{0}}$ where here the zero vector is the vector $(0, 0, . . .)$ .

As such, suppose that $\vec{x} \in null (Γ)$ . Then we know that:

Γ (\vec{x}) p = φ (p) = 0 = \sum_{i = 0}^{\infty} x_{i} b_{i}

for any $p \in P (R)$ . But notice if we let $p_{j}$ incrementally be the polynomial where:

p_{j} = x^{j}

then put that in, then we get:

Γ (\vec{x}) p_{j} = 0 = \sum_{i = 0}^{\infty} x_{i} b_{i} = x_{j} \underset{= 1}{\underset{⏟}{b_{j}}} = x_{j}

Thus if we let $j \in Z^{+}$ sweep, then we get that all $x_{j} = 0$ , so then $\vec{x} = \vec{0}$ this whole time. Thus $null (Γ) = {\vec{0}}$ as desired, so $Γ$ is injective.

For surjective, let $φ \in (P (R))^{'}$ be arbitrary. We'll show that we can construct a vector $\vec{x}$ such that:

Γ (\vec{x}) = φ

to do this, let $p_{j}$ be arbitrary. Then notice that any generic polynomial $p$ can be written as:

p = α_{0} p_{0} + α_{1} p_{1} + \dots

Thus, I claim that choosing $\vec{x} = (α_{0}, α_{1}, . . .)$ works:

\begin{aligned} Γ (\vec{x}) p_{j} & = Γ (α_{0}, α_{1}, . . .) p_{j} \\ = \sum_{i = 0}^{\infty} α_{i} β_{i} \\ = α_{j} \\ = φ (p_{j}) \end{aligned}

Notice that since $α_{i} \in R$ then our $φ = Γ (\vec{x})$ as desired.
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36

Theorem

Suppose $U$ is a subspace of $V$ . Let $i : U \to V$ be the inclusion map defined by $i (u) = u$ . Thus, $i^{'} \in L (V^{'}, U^{'})$ . Then:

$null (i^{'}) = U^{0}$
If $V$ is finite-dimensional, then $range (i^{'}) = U^{'}$ .
If $V$ is finite-dimensional, $\tilde{i^{'}}$ is an isomorphism from $V^{'} / U^{0}$ onto $U^{'}$ .

Proof

a

We'll show $null (i^{'}) = U^{0}$ . First, for $(\subseteq)$ . Let $φ \in null (i^{'}) \subseteq V^{'}$ be arbitrary. Then, if we have some $u \in U$ then:

(φ \circ i) u = 0

by $φ \in null (i^{'})$ . Then then $φ \circ i = i^{'} \in U^{0}$ . Notice if $u = \vec{0}$ then clearly $φ (u) = 0$ and if $u \neq \vec{0}$ then:

φ (i u) = φ (v) = 0

Where $i u = v$ so then since $u \neq \vec{0} \leftrightarrow v \neq \vec{0}$ . As such, no matter what then $φ$ sends vectors in $U$ to 0, so then $φ \in U^{0}$ , showing $(\subseteq)$ .

For $(\supseteq)$ , let $φ \in U^{0} = {ψ \in V^{'} : ψ (u) = 0 \forall u \in U}$ , so then for all $u \in U$ we have it that $φ (u) = 0$ . Notice:

i u = u

for all $u$ so then $φ (u) = φ (i u) = (φ \circ i) u = i^{'} (φ) u = 0$ . Thus $φ \in null (i^{'})$ , showing $(\subseteq)$ .

b

Suppose $n = dim (V)$ . By the FTOLM:

\begin{aligned} dim (range (i^{'})) & = n - dim (null (i^{'})) \\ = n - dim (U^{0}) \\ = n - (dim (V) - dim (U)) \\ = n - n + dim (U) \\ = dim (U) \\ = dim (U^{'}) \end{aligned}

Thus $range (i^{'}) ≃ U^{'}$ so then these two spaces are isomorphic. Hence, since $range (i^{'}) \subseteq U^{'}$ , then they must be equal, because we can create the map the basis $φ_{1}, . . ., φ_{n}$ for $range (i^{'})$ to the basis $ψ_{1}, . . ., ψ_{n}$ for $U^{'}$ , and thus their spans are the same and span their respective subspaces. Thus, they are equal.

c

Using (b), recall that $\tilde{i^{'}} : V^{'} / (null (i^{'})) \to U^{'}$ where:

\tilde{i^{'}} (φ_{v} + null (i^{'})) = i^{'} (φ_{v})

for any $φ_{v} \in V^{'}$ . So then:

\begin{aligned} dim (V^{'} / U^{0}) & = dim (V^{'}) - dim (U^{0}) \\ = dim (V^{'}) - dim (null (i^{'})) & (See (a)) \\ = n - (n + dim (U)) & (See (b)) \\ = dim (U^{'}) \end{aligned}

Thus $V^{'} / (null (i^{'})) ≃ U^{'}$ .

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6.B: Review of Orthonormal Bases

5

Theorem

On $P_{2} (R)$ , consider the inner product given by:

⟨ p, q ⟩ = \int_{0}^{1} p (x) q (x) d x

Apply the Gram-Schmidt Procedure to the basis $1, x, x^{2}$ to product an orthonormal basis of $P_{2} (R)$ .

Proof
Start with $e_{1} = 1$ first. We need to normalize this:

‖ 1 ‖^{2} = \int_{0}^{1} 1^{2} d x = 1

Thus $e_{1} = 1$ is our first basis vector.

\begin{aligned} e_{2} & = x - ⟨ x, e_{1} ⟩ e_{1} \\ = x - 1 \int_{0}^{1} x d x \\ = x - \frac{x^{2}}{2} |_{0}^{1} \\ = x - \frac{1}{2} \end{aligned}

Now to normalize it:

{‖ x - \frac{1}{2} ‖}^{2} = \int_{0}^{1} {(x - \frac{1}{2})}^{2} d x = \frac{u^{3}}{3} |_{- \frac{1}{2}}^{\frac{1}{2}} = \frac{1}{4 \cdot 3} = \frac{1}{12}

Thus have $e_{2} = 2 \sqrt{3} (x - \frac{1}{2})$ . For $e_{3}$ :

\begin{aligned} e_{3} & = x^{2} - ⟨ x^{2}, e_{1} ⟩ e_{1} - ⟨ x^{2}, e_{2} ⟩ e_{2} \\ = x^{2} - 1 \int_{0}^{1} x^{2} d x - 2 \sqrt{3} (x - \frac{1}{2}) \int_{0}^{1} 2 \sqrt{3} (x - \frac{1}{2}) x^{2} d x \\ = x^{2} - \frac{x^{3}}{3} |_{0}^{1} - 12 (x - \frac{1}{2}) \int_{0}^{1} x^{3} - \frac{1}{2} x^{2} d x \\ = x^{2} - \frac{1}{3} - 12 (x - \frac{1}{2}) {[\frac{x^{4}}{4} - \frac{x^{3}}{6}]}_{0}^{1} \\ = x^{2} - \frac{1}{3} - (x - \frac{1}{2}) \\ = x^{2} - x + \frac{1}{6} \end{aligned}

Then normalize:

\begin{aligned} {‖ x^{2} - x + \frac{1}{6} ‖}^{2} & = \int_{0}^{1} {(x^{2} - x + \frac{1}{6})}^{2} d x \\ = \frac{1}{180} = \frac{1}{6^{2} \cdot 5} \end{aligned}

Thus $e_{3} = 6 \sqrt{5} (x^{2} - x + \frac{1}{6})$

Thus our basis is:

β = {1, 2 \sqrt{3} (x - \frac{1}{2}), 6 \sqrt{5} (x^{2} - x + \frac{1}{6})}

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7

Theorem

Find a polynomial $q \in P_{2} (R)$ such that:

p (1 / 2) = \int_{0}^{1} p (x) q (x) d x

for every $p \in P_{2} (R)$ .

Proof
Define $φ : P_{2} (R) \to R$ by:

φ (p) = p (1 / 2)

let's first show that $φ$ is linear. For additivity, let $p, q \in P_{2} (R)$ be arbitrary. Then:

φ (p + q) = (p + q) (1 / 2) = p (1 / 2) + q (1 / 2) = φ (p) + φ (q)

and for homogeneity, let $λ \in R$ be arbitrary. Then:

φ (λ p) = (λ p) (1 / 2) = λ p (1 / 2) = λ φ (p)

thus $φ$ is linear. As such, apply Chapter 6 (cont.) - Finishing Inner Product Spaces#^a857c6, using our basis from HW 2 - Finishing Duality, Review of Inner Product Spaces#5. So have:

\begin{aligned} u (x) & = \sum_{i = 1}^{3} \overset{―}{φ (e_{i})} e_{i} \\ = φ (1) 1 + φ (2 \sqrt{3} (x - \frac{1}{2})) 2 \sqrt{3} (x - \frac{1}{2}) + φ (6 \sqrt{5} (x^{2} - x + \frac{1}{6})) 6 \sqrt{5} (x^{2} - x + \frac{1}{6}) \\ = 1 + 0 + 180 \cdot ({(\frac{1}{2})}^{2} - \frac{1}{2} + \frac{1}{6}) (x^{2} - x + \frac{1}{6}) \\ = 1 + 180 (- \frac{1}{4} + \frac{1}{6}) (x^{2} - x + \frac{1}{6}) \\ = 1 + 180 \cdot \frac{- 1}{12} (x^{2} - x + \frac{1}{6}) \\ = 1 - 15 (x^{2} - x + \frac{1}{6}) \\ = - 15 x^{2} + 15 x - \frac{3}{2} \end{aligned}

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15

Theorem

Suppose $C_{R} ([- 1, 1])$ is the vector space of continuous real-valued functions on the interval $[- 1, 1]$ with the inner product given by:

⟨ f, g ⟩ = \int_{- 1}^{1} f (x) g (x) d x

for $f, g \in C_{R} ([- 1, 1])$ . Let $φ$ be the linear functional on $C_{R} ([- 1, 1])$ defined by $φ (f) = f (0)$ . Then there does not exist $g \in C_{R} ([- 1, 1])$ such that:

φ (f) = ⟨ f, g ⟩

for every $f \in C_{R} ([- 1, 1])$ .

Proof
We cannot just use Riesz here since we are actually proving the opposite of what that theorem says, as well as since our vector space here is infinite-dimensional. We are really proving here that, for all $g \in C_{R} ([- 1, 1])$ that there exists some $f \in C_{R} ([- 1, 1])$ where:

φ (f) = f (0) \neq ⟨ f, g ⟩

As such, let $g$ here be arbitrary. Choose the function $f_{n}$ such that:

f_{n} (x) = (1 - | x |)^{n}

for some natural number $n \in N$ . Notice that:

φ (f_{n}) = f_{n} (0) = (1 - | 0 |)^{n} = 1^{n} = 1

for all $n$ . However, consider $⟨ f_{n}, g ⟩$ :

\begin{aligned} ⟨ f_{n}, g ⟩ & = \int_{- 1}^{1} (1 - | x |)^{n} g (x) d x \\ \leq ‖ (1 - | x |)^{n} ‖ ‖ g (x) ‖ & (Cauchy Swhartz) \end{aligned}

But notice that:

\begin{aligned} ‖ (1 - | x |)^{n} ‖ & = \sqrt{\int_{- 1}^{1} (1 - | x |)^{2 n} d x} \\ = \sqrt{2 \int_{0}^{1} (1 - x)^{2 n} d x} \\ = \sqrt{- 2 \frac{(1 - x)^{2 n + 1}}{2 n + 1} |_{0}^{1}} \\ = \sqrt{\frac{2}{2 n + 1}} \end{aligned}

As such, then if $‖ g (x) ‖ = c \geq 0$ where $c \in R$ then:

⟨ f_{n}, g ⟩ \leq c \sqrt{\frac{2}{2 n + 1}} = \sqrt{\frac{2 c^{2}}{2 n + 1}}

We can show that, if given some $c \in R$ based on $g$ , I can find some $n$ such that $⟨ f_{n}, g ⟩ \neq φ (f_{n}) = 1$ . As such, if we show that $⟨ f_{n}, g ⟩ < 1$ then we've shown this as then we get inequality. As such, then choose:

n = ⌈ c^{2} ⌉

Because then:

\begin{aligned} ⟨ f_{n}, g ⟩ & \leq \sqrt{\frac{2 c^{2}}{2 ⌈ c^{2} ⌉ + 1}} \\ < \sqrt{\frac{2 c^{2}}{2 c^{2} + 1}} \\ < \sqrt{\frac{2 c^{2}}{2 c^{2}}} \\ = 1 \end{aligned}

Thus $⟨ f_{n}, g ⟩ < 1$ and thus $⟨ f_{n}, g ⟩ \neq φ (f_{n})$ as desired, for our chosen $n$ .
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3.F: Duality (cont.)

12

14

a

b

15

20

22

27

35

Γ∈L(R∞,(P(R)))

Showing Injectivity and Surjectivity

36

a

b

c

6.B: Review of Orthonormal Bases

5

7

15

$Γ \in L (R^{\infty}, (P (R)))$