Project 1

Mark Sukaiti 100064482

Problem 1

a) Let $X$ be a chess board for any pair of squares $x, y \in X$ , define $d (x, y)$ to be the number of steps that a knight needs to go from $x$ to $y$ . Check if $(X, d)$ is a metric space.
b) Let $X$ be a metric space and the function $d_{1}$ and $d_{2}$ are metrics on $X$ . Check if the following are metrics:

(i)) d_{1} + d_{2} (i i)) min (d_{1}, d_{2}) (i i i) \frac{1}{3} d_{1} + \frac{2}{3} d_{2}

c) If $d (x, y)$ is a metric on $X$ , show that $d_{b} (x, y)$ given by

d_{b} (x, y) = \frac{d (x, y)}{1 + d (x, y)}

is also a metric on $X$ .

Solution

a) We note that a chessboard can be represented as a graph whose squares are the vertices and an edge connected the vertices $v_{1}$ and $v_{2}$ iff a knight can jump from the square associated $v_{1}$ to the square associated to $v_{2}$ .
We assume that $d (x, y)$ is the minimum number of moves a knight needs to take from $x$ to $y$ , then its clear that $d (x, y) > 0$ with $d (x, x) = 0$ . Moreover $d (x, y) = d (y, x)$ is trivially satisfied since if $x = x_{1} x_{2} . . . x_{n} = y$ is a minimum sequence of moves (that is a sequence of moves whose length is the minimum) then by reversing it we get $d (y, x)$ . We only need to show the triangle inequality holds. Assume that $x = x_{1} . . . x_{n} = z$ is a minimum number of moves needed to go from $x$ to $z$ . Now if $y = x_{i}$ for some $1 < i < n$ then $x = x_{1} . . . x_{i} = y$ must be a minimum number of moves from $x$ to $y$ since otherwise this contradicts our assumption that the prior sequence is a minimum sequence from $x$ to $z$ (since we can otherwise take a smwhich immediatelyaller sequence), with a similar reasoning the sequence $y = x_{i} . . . x_{n} = z$ is also a minimum. Thus $d (x, z) = d (x, y) + d (y, z)$ . Now if $y \neq x_{i}, \forall i$ then let $x = a_{1} . . . a_{n_{1}} = y$ be a minimum sequence of moves from $x$ to $y$ and $y = b_{1} . . . b_{n_{2}} = z$ be a minimum sequence of moves from $y$ to $z .$ Now by concatenating these two sequence of moves $x = a_{1} . . . a_{n_{1}} b_{2} . . . b_{n_{2}} = z$ we have a sequence of moves from $x$ to $z$ . Either this new sequence is of the same length as our original or must be greater since otherwise it contradicts our assumption that $x_{1} . . . x_{n}$ is a minimum sequence. Thus $d (x, z) \leq d (x, y) + d (y, z)$ . Thus indeed $d (x, y)$ is a metric.

b) We shall prove $(i)$ and $(i i i)$ by proving the following instead:

\begin{matrix} (*) & D_{1} = α d_{1} + β d_{2} is a metric for α, β > 0 \end{matrix}

To prove $(*)$ since $d_{1}$ and $d_{2}$ are metrics and $α, β > 0$ we have that $D_{1} (x, y) \geq 0$ with equality only attained when $x = y$ , and $D_{1} (x, y) = D_{2} (y, x)$ immediately follows from the fact that $d_{1}, d_{2}$ are symmetric. We just have to show the triangle inequality holds. Let $x, y, z \in X$ :

\begin{aligned} D_{1} (x, z) & = α d_{1} (x, z) + β d_{2} (x, z) \\ \leq α (d_{1} (x, y) + d_{1} (y, z)) + β (d_{2} (x, y) + d_{2} (y, z)) \\ = (α d_{1} (x, y) + β d_{2} (x, y)) + (α d_{1} (y, z) + β d_{2} (y, z)) \\ = D_{1} (x, y) + D_{2} (y, z) \end{aligned}

For $(i i)$ , we shall show it is not a metric using a counter example. Let $d_{1}$ and $d_{2}$ be the following (with $d (x, y) = min (d_{1}, d_{2})$ ):

d_{1} = \sqrt{\sum_{i = 1}^{2} (\frac{1}{10000} x_{i} - y_{i})^{2}}

d_{2} = \sqrt{\sum_{i = 1}^{2} (x_{i} - \frac{1}{10000} y_{i})^{2}}

Then let $x = (1, 0), z = (0, 1), y = (0, 0)$ . Then $d (x, z) > 1$ since $d_{1} (x, z) = d_{2} (x, z) = \sqrt{\frac{1}{10000} + 1} > 1$ . However notice that $d (x, y) = d_{1} (x, y) = 1 / 100$ and likewise for $d (x, z)$ that is:

1 = d (x, z) \geq d (x, y) + d (y, z) = \frac{1}{50}

i.e, the triangle inequality doesn't hold. So in general the minimum of two metrics is not a metric.

To prove $c$ we shall prove instead the following:

Proposition

Let $d (x, y)$ be a metric on $X$ and $f : R \times R \to R$ be a function that satisfies the following properties:

$f (0, 0) = 0$
$f (x, y) \geq 0, \forall x, y \geq 0$
$f (x_{1} + x_{2}, y) = f (x_{1}, y) + f (x_{2}, y)$
$\frac{d}{d x} f (x, x) \geq 0, \forall x \geq 0$
$\frac{\partial}{\partial y} f (x, y) \leq 0, \forall x, y \geq 0$
Then $f (d (x, y), d (x, y))$ is a metric on $X$ .

Proof

Define $d_{f} (x, y) = f (d (x, y), d (x, y))$ , then by the symmetry of $d (x, y)$ we immediately get that $d_{f} (x, y) = d_{f} (y, x)$ , and moreover by property $(1)$ and $(2)$ of $f$ we guarantee that $d_{f} (x, y) \geq 0$ and equality is only attained when $x = y$ . Thus we only have to prove the triangle inequality but this immediately follows since $d$ is a metric we have $d (x, z) \leq d (x, y) + d (y, z)$ thus:

\begin{aligned} d_{f} (x, z) & = f (d (x, z), d (x, z)) \\ \leq f (d (x, y) + d (y, z), d (x, y) + d (y, z)) (Since f (x) = f (x, x) is an non-decreasing function by property (4)) \\ = f (d (x, y), d (x, y) + d (y, z)) + f (d (y, z), d (x, y) + d (y, z)) (Since f is linear in its first slow by property (3)) \\ \leq f (d (x, y), d (x, y)) + f (d (y, z), d (y, z)) (Since f is non-increasing in its second slot by property (5)) \\ ◻ & = d_{f} (x, y) + d_{f} (y, z) \end{aligned}

Corollary

$(c)$ is a metric.

Proof

Take $f (x, y) = \frac{x}{1 + y}$ and use the proposition above.

Problem 2

Show that for all $x, y, z$ in a metric space $(X, d)$

d (x, y) \geq | d (x, z) - d (z, y) |

Proof

Let $x, y, z \in X$ then by the triangle inequality we have:

\begin{matrix} (1) & d (x, z) \leq d (x, y) - d (y, z) ⟹ d (x, y) \geq d (x, z) - d (z, y) \end{matrix}

and

\begin{matrix} (2) & d (z, y) \leq d (z, x) + d (x, y) ⟹ d (x, y) \geq d (z, y) - d (x, z) \end{matrix}

Thus by $(1) + (2)$ we have proven

\begin{matrix} ◻ & d (x, y) \geq | d (x, z) - d (z, y) | \end{matrix}

Problem 3

A set $C$ in a vector space $V$ is called convex if for every pair of points $x, y \in C$ and for every $λ \in [0, 1]$ , it holds:

λ x + (1 - λ) y \in C

Prove that any any ball in a normed space $X$ is convex.

Proof

Let $B_{z} (r)$ be a closed ball in $X$ of radius $r$ centered at a point $z$ . Let $x, y \in B_{z} (r)$ then we have that:

| | z - x | | \leq r & | | z - y | | \leq r

We are interested in the point $λ x + (1 - λ) y$ :

\begin{aligned} | | λ x + (1 - λ) y - z | | & = | | λ (x - z) + (1 - λ) (y - z) | | \\ \leq | | λ (x - z) | | + | | (1 - λ) (y - z) | | \\ = λ | | x - z | | + (1 - λ) | | y - z | | \\ \leq λ r + (1 - λ) r \\ ◻ & = r \end{aligned}

Problem 4

Two norms $∥ \cdot ∥_{1}$ and $∥ \cdot ∥_{2}$ on a vector space $V$ are said to be equivalent if there exists positive constants $c_{1}$ and $c_{2}$ such that for all $x \in V$ :

c_{1} ∥ x ∥_{1} \leq∥ x ∥_{2} \leq c_{2} ∥ x ∥_{1} .

Prove that two any norms in a finite dimensional space $X$ are equivalent.

Proof

Let $X$ be an $n$ -dimensional normed space and let $e_{i}$ be a basis. We fix one of the metrics $| | a | |_{1} = | | \sum_{i = 1}^{n} a_{i} e_{i} | | = \sum_{i = 1}^{n} | a_{i} |$ . Then notice that any other norm $| | \cdot | |_{2}$ is continuous with respect to $| | \cdot | |_{1}$ :
Let $ϵ > 0$ then we have that:

\begin{aligned} | {| | x | |}_{2} - {| | a | |}_{2} | & \leq {| | x - a | |}_{2} \\ \leq \sum_{i = 1}^{n} | x_{i} - a_{i} | {| | e_{i} | |}_{2} \\ \leq {| | x - a | |}_{1} max_{1 \leq i \leq n} (| | e_{i} | |_{2}) \end{aligned}

Thus if we choose $δ = \frac{ϵ}{max_{1 \leq i \leq n} (| | e_{i} | |_{2})}$ , that is if ${| | x - a | |}_{1} < δ$ we have $| | | x | |_{2} - | | a | |_{2} | < ϵ$ . Thus any metric $| | \cdot | |_{2}$ is continuous with respect to our fixed $| | \cdot | |_{1}$ .
Now let $B$ be the unit ball with respect to the first norm, i.e, if $x \in B$ then $| | x | |_{1} = 1$ , then $B$ is compact in a finite dimensional normed space. Thus by Heine Borel, since $| | x | |_{2}$ is continuous it attains a maximum and a minimum on $B$ , i.e if $x \in B$ , $\exists c_{1}, c_{2}$ such that:

c_{1} \leq | | x | |_{2} \leq c_{2}

Now suppose $x \neq 0$ , we can take $x^{'} = \frac{1}{| | x | |_{1}} x$ , then $x^{'} \in B$ thus we obtain:

\begin{array}{r} c_{1} \leq | | x^{'} | |_{2} \leq c_{2} \\ ⟹ c_{1} \leq \frac{1}{| | x | |_{1}} | | x | |_{2} \leq c_{2} \\ ⟹ c_{1} | | x | |_{1} \leq | | x | |_{2} \leq c_{2} | | x | |_{1} \end{array}

Thus every norm is equivalent to $| | \cdot | |_{1}$ . Now all that is left to show is that equivalence of norms forms an equivalence relation.
Let us define $| | \cdot | |_{a} \sim | | \cdot | |_{b}$ iff $| | . | |_{a}$ is equivalent to $| | \cdot | |_{b}$ (as defined above). Then its clear that $| | \cdot | |_{a} \sim | | \cdot | |_{a}$ (reflexivity), and if we have:

c_{1} | | x | |_{a} \leq | | x | |_{b} \leq c_{2} | | x | |_{a}

then:

\frac{1}{c_{2}} | | x | |_{b} \leq | | x | |_{a} \leq \frac{1}{c_{1}} | | x | |_{b}

Thus we shown if $| | \cdot | |_{a} \sim | | \cdot | |_{b}$ then $| | \cdot | |_{b} \sim | | \cdot | |_{a}$ , i.e. the relation is symmetric. Lastly now assume that $| | \cdot | |_{a} \sim | | \cdot | |_{b}$ and $| | \cdot | |_{b} \sim | | \cdot | |_{c}$ . Then we have that:

\begin{array}{r} c_{1} | | x | |_{a} \leq | | x | |_{b} \leq c_{2} | | x | |_{a} \\ d_{1} | | x | |_{b} \leq | | x | |_{c} \leq d_{2} | | x | |_{b} \end{array}

then we have:

c_{1} d_{1} | | x | |_{a} \leq | | x | |_{c} \leq c_{2} d_{2} | | x | |_{a}

Thus $| | \cdot | |_{a} \sim | | \cdot | |_{c}$ , i.e. , the relation is transitive. Thus it indeed is an equivalence relation thus all norms in a finite dimensional normed space is equivalent since they are all equivalent to $| | \cdot | |_{1}$

Problem 5

(a) Prove that the inner product in a normed space $X$ can be recovered from the polarization identity:

⟨ x, y ⟩ = \frac{1}{4} [(| | x + y | |^{2} - | | x - y | |^{2}) - i (| | x + i y | |^{2} - | | x - i y | |^{2})]

(b) Prove that a normed space is an inner product space if and only if the norm satisfies the parallelogram law:

| | x + y | |^{2} + | | x - y | |^{2} = 2 (| | x | |^{2} + | | y | |^{2})

| | f | |_{1} = \int_{0}^{1} | f (x) | d x

Prove that there is no inner product on $C [0, 1]$ that agreed with this norm..

Proof

(a) Let $⟨ x, y ⟩$ be an inner product then:

\begin{aligned} \frac{1}{4} [(| | x + y | |^{2} - | | x - y | |^{2}) - i (| | x + i y | |^{2} - | | x - i y | |^{2})] & = \frac{1}{4} [(⟨ x + y, x + y ⟩ - ⟨ x - y, x - y ⟩) - i (⟨ x + i y, x + i y ⟩ - ⟨ x - i y, x - i y ⟩)] \\ = \frac{1}{4} [2 ⟨ x, y ⟩ - 2 i ⟨ x, i y ⟩] \\ = \frac{1}{4} (4 ⟨ x, y ⟩) \\ ◻ & = ⟨ x, y ⟩ \end{aligned}

(b) ( $⟹$ ) If $X$ is an inner product space then:

\begin{aligned} | | x + y | |^{2} + | | x - y | |^{2} & = ⟨ x + y, x + y ⟩ + ⟨ x - y, x - y ⟩ \\ = 2 (⟨ x, x ⟩ + ⟨ y, y ⟩) \\ = 2 (| | x | |^{2} + | | y | |^{2}) \end{aligned}

( $⟸$ ) Assume that $X$ is a normed space that satisfies the parallelgram law. Then let $⟨ x, y ⟩ = \frac{1}{4} [(| | x + y | |^{2} - | | x - y | |^{2}) - i (| | x + i y | |^{2} - | | x - i y | |^{2})]$ . We shall show this does indeed define an inner product on the space.

Positivity: Let $x \neq 0$ , then

\begin{aligned} ⟨ x, x ⟩ & = \frac{1}{4} [(| | x + x | |^{2} - | | x - x | |^{2}) - i (| | x + i x | |^{2} - | | x - i x | |^{2})] \\ = \frac{1}{4} [4 | | x | |^{2} - i (| 1 + i |^{2} | | x | |^{2} - | 1 - i |^{2} | | x | |^{2})] \\ = | | x | |^{2} > 0 \end{aligned}

Conjugate Symmetry:

\begin{aligned} \overset{―}{⟨ y, x ⟩} & = \overset{―}{\frac{1}{4} [(| | y + x | |^{2} - | | y - x | |^{2}) - i (| | y + i x | |^{2} - | | y - i x | |^{2})]} \\ = \frac{1}{4} [(| | y + x | |^{2} - | | y - x | |^{2}) + i (| | y + i x | |^{2} - | | y - i x | |^{2})] \\ = \frac{1}{4} [(| | y + x | |^{2} - | | y - x | |^{2}) - i (| | y - i x | |^{2} - | | y + i x | |^{2})] \\ = \frac{1}{4} [(| | y + x | |^{2} - | | y - x | |^{2}) - i (| i |^{2} | | i y + x | |^{2} - | i |^{2} | | i y - x | |^{2})] \\ = \frac{1}{4} [(| | x + y | |^{2} - | | x - y | |^{2}) - i (| | x + i y | |^{2} - | | x - i y | |^{2})] \\ = ⟨ x, y ⟩ \end{aligned}

Linearity in the first argument: We first show

\begin{matrix} (*) & ⟨ x + z, y ⟩ = ⟨ x, y ⟩ + ⟨ z, y ⟩ \end{matrix}

Let $x, y, z \in X$ , then its sufficient to show that:

| | x + z + y | |^{2} - | | x + z - y | |^{2} \overset{?}{=} | | x + y | |^{2} - | | x - y | |^{2} + | | z + y | |^{2} - | | z - y | |^{2}

This is equivalent to:

2 (| | x + z + y | |^{2} + | | x - y | |^{2}) - 2 (| | x + z - y | |^{2} + | | x + y | |^{2}) \overset{?}{=} 2 | | z + y | |^{2} - 2 | | z - y | |^{2}

Applying the parallelogram law on the left gives us:

| | 2 x + z | |^{2} + | | 2 y + z | |^{2} - | | 2 x + z | |^{2} - | | z - 2 y | |^{2} \overset{?}{=} 2 | | z + y | |^{2} - 2 | | z - y | |^{2}

Which is indeed true since:

\begin{aligned} | | 2 y + z | |^{2} - | | z - 2 y | |^{2} + z^{2} - z^{2} & = (| | 2 y + z | |^{2} + | | z | |^{2}) - (| | z - 2 y | |^{2} + | | z | |^{2}) \\ = 2 (| | z + y | |^{2} + | | y | |^{2}) - 2 (| | z - y | |^{2} + | | y | |^{2}) \\ = 2 | | z + y | |^{2} - 2 | | z - y | |^{2} \end{aligned}

To prove $⟨ α x, y ⟩ = α ⟨ x, y ⟩$ , it can be derived from $(*)$ . From $(*)$ , it holds if $α \in Z$ (using induction), but since it holds for $Z$ then it holds for $α \in Q$ since if $α = \frac{n}{m}$ then:

m ⟨ \frac{n}{m} x, y ⟩ = n ⟨ x, y ⟩

However since $| | \cdot | |$ is continuous then $⟨ \cdot, \cdot ⟩$ is as well, thus $⟨ α x, y ⟩ = α ⟨ x, y ⟩$ must hold for all $α \in R$ since it holds for all $α \in Q$ and $⟨ \cdot, \cdot ⟩$ is continuous. $◻$
(c) Its sufficient to find a counter example, take $f (x) = 1 - 2 x$ , and $g (x) = 1$ then:

\begin{aligned} 2 | | f | |^{2} & = 2 {(\int_{0}^{1} | f (x) | d x)}^{2} = 0.5 \\ 2 | | g | |^{2} & = 2 {(\int_{0}^{1} | g (x) | d x)}^{2} = 2 \\ | | f + g | |^{2} & = {(\int_{0}^{1} | f (x) + g (x) | d x)}^{2} = 1 \\ | | f - g | |^{2} & = {(\int_{0}^{1} | f (x) - g (x) | d x)}^{2} = 1 \end{aligned}

Thus the parallelogram law doesn't hold.

Problem 6

Least square approximation. (Reed-Simon II.5) Let $X$ be an inner product space and let $x_{1}, . . ., x_{N}$ be an orthonormal set. Prove that

| | x - \sum_{n = 1}^{N} c_{n} x_{n} | |

is minimized by choosing $c_{n} = ⟨ x, x_{n} ⟩$

Proof

For ease of notation, we shall use Einstein summation. So I am looking to show:

| | x - c^{i} x_{i} | |

is minimized when $c^{i} = ⟨ x, x_{i} ⟩$ . Thus:

\begin{aligned} | | x - c^{i} x_{i} | |^{2} & = ⟨ x - c^{i} x_{i}, x - c^{j} x_{j} ⟩ \\ = ⟨ x, x ⟩ - ⟨ x, c^{i} x_{i} ⟩ - ⟨ c^{j} x_{j}, x ⟩ + ⟨ c^{j} x_{j}, c^{i} x_{i} ⟩ \\ = ⟨ x, x ⟩ - \overset{―}{c^{i} ⟨ x_{i}, x ⟩} - c^{j} ⟨ x_{j}, x ⟩ + δ_{i j} c^{j} \overset{―}{c^{i}} \\ = | | x | |^{2} - \overset{―}{c^{i}} ⟨ x, x_{i} ⟩ - c^{i} \overset{―}{⟨ x, x_{i} ⟩} + \sum_{i = 1}^{N} | c^{i} |^{2} \end{aligned}

To minimize this we will differentiate with respect to $c^{j}$ then to $\overset{―}{c^{j}}$ and set it to 0.
(Note: here we are treating $c_{j}$ independent from $\overset{―}{c_{j}}$ , that is we are studying the real valued function of the form $f (x^{1}, . . . x^{N}, y^{1}, . . ., y^{N}) = | | x | |^{2} - x^{i} ⟨ x, x_{i} ⟩ - y^{i} \overset{―}{⟨ x, x_{i} ⟩} + \sum_{i = 1}^{N} x^{i} y^{i}$ and are trying to find the global minimum, we note that our function is unbounded above that is if $x^{i}, y^{i} \to \infty$ then $f (x^{i}, y^{i}) \to \infty$ so its sufficient just to find a point that sets our partials to zero and note that it should be the minimum.)
That is we get a system of $2 N$ equations of the form:

\begin{aligned} \frac{\partial}{\partial c^{j}} (| | x | |^{2} - \overset{―}{c^{i}} ⟨ x, x_{i} ⟩ - c^{i} \overset{―}{⟨ x, x_{i} ⟩} + \sum_{i = 1}^{N} c^{i} \overset{―}{c^{i}}) & = 0 & \frac{\partial}{\partial \overset{―}{c^{j}}} (| | x | |^{2} - \overset{―}{c^{i}} ⟨ x, x_{i} ⟩ - c^{i} \overset{―}{⟨ x, x_{i} ⟩} + \sum_{i = 1}^{N} c^{i} \overset{―}{c^{i}}) = 0 \\ ⟹ \overset{―}{⟨ x, x_{j} ⟩} & = \overset{―}{c^{j}} & ⟨ x, x_{j} ⟩ = c^{j} \end{aligned}

That is indeed $c^{i} = ⟨ x, x_{i} ⟩$ minimizes our norm. Note if our field is real then $⟨ x, x_{i} ⟩ = ⟨ x_{i}, x ⟩$ .

Problem 7

Show that $l^{\infty}$ equipped with a the norm $| | \cdot | |_{\infty}$ is a Banach space.

Proof

Note that $l^{\infty}$ equipped with the $| | \cdot | |_{\infty}$ is indeed a normed space. So we just have to check whether $l^{\infty}$ is complete.
Let $x_{n} = (x_{1, n}, x_{2, n}, . . .)$ be a Cauchy Sequence, that is $\forall ϵ > 0, \exists N \in N$ such that

| | x_{n} - x_{m} | |_{\infty} < ϵ, \forall n, m > N ⟹ \underset{k \geq 1}{s u p} | x_{k, n} - x_{k, m} | < ϵ, \forall n, m > N

This implies that for a fixed $k_{0}$ we have: $| x_{k_{0}, n} - x_{k_{0}, m} | < ϵ, \forall n, m > N$ which is just a sequence in $R$ which is a complete space thus $\exists y_{k_{0}} \in R$ such that $x_{k_{0}, n} \to y_{k_{0}}$ . Then all thats left to show is that the element $y = (y_{1}, y_{2}, . . .) \in l^{\infty}$ , that is $sup_{k} (y_{n}) < \infty$ :

| | y | |_{\infty} \leq | | x_{n} - y | |_{\infty} + | | x_{n} | |_{\infty} \leq ϵ + | | x_{n} | |_{\infty} < \infty

Problem 8

Prove that:
(a) If $1 \leq p < q < \infty$ , then $l^{p} \subset l^{q}$ and $| | x | |_{q} \leq | | x | |_{p}$
(b) If $x \in ⋃_{1 \leq p < \infty} l^{p}$ then $| | x | |_{p} \to | | x | |_{\infty}$ as $p \to \infty$ .

Proof

(a) We shall first show $| | x | |_{q} \leq | | x | |_{p}, \forall x = (x_{1}, x_{2}, . . .) \in l^{p}$ .

\begin{aligned} 1 & = \sum_{i = 1}^{\infty} {(\frac{x_{i}}{| | x | |_{p}})}^{p} \geq \sum {(\frac{x_{i}}{| | x | |_{p}})}^{q} = {(\frac{| | x | |_{q}}{| | x | |_{p}})}^{q} \end{aligned}

Thus if $x \in l^{p}$ then $| | x | |_{q} \leq | | x | |_{p} < \infty ⟹ x \in l^{q}$

(b) Let $x \in ⋃_{1 \leq p < \infty} l^{p}$ then $x \in l^{q}$ for some $q > 0$ , thus $| | x | |_{n} < \infty$ for all $n > q$ and:

| | x | |_{k} \leq | | x | |_{q}, \forall k \geq q

Thus $| | x | |_{\infty} \leq \underset{p \to \infty}{lim inf} | | x | |_{p}$ . However notice that:

| | x | |_{p} = lim_{n \to \infty} {(\sum_{i = 1}^{n} | x_{i} |^{p})}^{\frac{1}{p}} \leq lim_{n \to \infty} n^{\frac{1}{p}} sup_{i} (| x_{i} |) = sup_{i} (| x_{i} |) = | | x | |_{\infty}

That is, $\underset{p \to \infty}{lim sup} | | x | |_{p} \leq | | x | |_{\infty}$ . Thus $\underset{p \to \infty}{lim sup} | | x | |_{p} \leq | | x | |_{\infty} \leq \underset{p \to \infty}{lim inf} | | x | |_{p}$ i.e $lim_{p \to \infty} | | x | |_{p} = | | x | |_{\infty}$