# The completeness of the quotient space (topological vector space)

## The Goal

We are going to show the completeness of $X/N$ where $X$ is a TVS and $N$ a closed subspace. Alongside, a bunch of useful analysis tricks will be demonstrated (and that's why you may find this blog post a little tedious.). But what's more important, the theorem proved here will be used in the future.

## The main process

To make it clear, we should give a formal definition of $F$-space.

A topological space $X$ is an $F$-space if its topology $\tau$ is induced by a complete invariant metric $d$.

A metric $d$ on a vector space $X$ will be called invariant if for all $x,y,z \in X$, we have $d(x+z,y+z)=d(x,y).$ By complete we mean every Cauchy sequence of $(X,d)$ converges.

### Defining the quotient metric $\rho$

The metric can be inherited to the quotient space naturally (we will use this fact latter), that is

If $X$ is a $F$-space, $N$ is a closed subspace of a topological vector space $X$, then $X/N$ is still a $F$-space.

Suppose $d$ is a complete invariant metric compatible with $\tau_X$. The metric on $X/N$ is defined by $\boxed{\rho(\pi(x),\pi(y))=\inf_{z \in N}d(x-y,z)}$ ### $\rho$ is a metric

Proof. First, if $\pi(x)=\pi(y)$, that is, $x-y \in N$, we see $\rho(\pi(x),\pi(y))=\inf_{z \in N}d(x-y,z)=d(x-y,x-y)=0.$ If $\pi(x) \neq \pi(y)$ however, we shall show that $\rho(\pi(x),\pi(y))>0$. In this case, we have $x-y \notin N$. Since $N$ is closed, $N^c$ is open, and $x-y$ is an interior point of $X-N$. Therefore there exists an open ball $B_r(x-y)$ centered at $x-y$ with radius $r>0$ such that $B_r(x-y) \cap N = \varnothing$. Notice we have $d(x-y,z)>r$ since otherwise $z \in B_r(x-y)$. By putting $r_0=\sup\{r:B_r(x-y) \cap N = \varnothing\},$ we see $d(x-y,z) \geq r_0$ for all $z \in N$ and indeed $r_0=\inf_{z \in N}d(x-y,z)>0$ (the verification can be done by contradiction). In general, $\inf_z d(x-y,z)=0$ if and only if $x-y \in \overline{N}$.

Next, we shall show that $\rho(\pi(x),\pi(y))=\rho(\pi(y),\pi(x))$, and it suffices to assume that $\pi(x) \neq \pi(y)$. Sgince $d$ is translate invariant, we get \begin{aligned} d(x-y,z)&=d(x-y-z,0) \\ &=d(0,y-x+z) \\ &=d(-z,y-x) \\ &=d(y-x,-z). \end{aligned} Therefore the $\inf$ of the left hand is equal to the one of the right hand. The identity is proved.

Finally, we need to verify the triangle inequality. Let $r,s,t \in X$. For any $\varepsilon>0$, there exist some $z_\varepsilon$ and $z_\varepsilon'$ such that $d(r-s,z_\varepsilon)<\rho(\pi(r),\pi(s))+\frac{\varepsilon}{2},\quad d(s-t,z'_\varepsilon)<\rho(\pi(s),\pi(t))+\frac{\varepsilon}{2}.$ Since $d$ is invariant, we see \begin{aligned} d(r-t,z_\varepsilon+z'_\varepsilon)&=d((r-s)+(s-t)-(z_\varepsilon+z'_\varepsilon),0) \\ &=d([(r-s)-z_\varepsilon]+[(s-t)-z'_\varepsilon],0) \\ &=d(r-s-z_\varepsilon,t-s+z'_\varepsilon) \\ &\leq d(r-s-z_\varepsilon,0)+d(t-s+z'_\varepsilon,0) \\ &=d(r-s,z_\varepsilon)+d(s-t,z'_\varepsilon) \end{aligned} (I owe [@LeechLattice](https://onp4.com/@leechlattice) for the inequality above.)

Therefore \begin{aligned} d(r-t,z_\varepsilon+z'_\varepsilon)&\leq d(r-s,z_\varepsilon)+d(s-t,z'_\varepsilon) \\ &<\rho(\pi(r),\pi(s))+\rho(\pi(s),\pi(t))+\varepsilon. \end{aligned} (Warning: This does not imply that $\rho(\pi(r),\pi(s))+\rho(\pi(s),\pi(t))=\inf_z d(r-t,z)$ since we don't know whether it is the lower bound or not.)

If $\rho(\pi(r),\pi(s))+\rho(\pi(s),\pi(t))<\rho(\pi(r),\pi(t))$ however, let $0<\varepsilon<\rho(\pi(r),\pi(t))-(\rho(\pi(r),\pi(s))+\rho(\pi(s),\pi(t)))$ then there exists some $z''_\varepsilon=z_\varepsilon+z'_\varepsilon$ such that $d(r-t,z''_\varepsilon)<\rho(\pi(r),\pi(t))$ which is a contradiction since $\rho(\pi(r),\pi(t)) \leq d(r-t,z)$ for all $z \in N$.

(We are using the $\varepsilon$ definition of $\inf$. See here.)

### $\rho$ is translate invariant

Since $\pi$ is surjective, we see if $u \in X/N$, there exists some $a \in X$ such that $\pi(a)=u$. Therefore \begin{aligned} \rho(\pi(x)+u,\pi(y)+u) &=\rho(\pi(x)+\pi(a),\pi(y)+\pi(a)) \\ &=\rho(\pi(x+a),\pi(y+a)) \\ &=\inf_{z \in N}d(x+a-y-a,z) \\ &=\rho(\pi(x),\pi(y)). \end{aligned}

### $\rho$ is well-defined

If $\pi(x)=\pi(x')$ and $\pi(y)=\pi(y')$, we have to show that $\rho(\pi(x),\pi(y))=\rho(\pi(x'),\pi(y'))$. In fact, \begin{aligned} \rho(\pi(x),\pi(y)) &\leq \rho(\pi(x),\pi(x'))+\rho(\pi(x'),\pi(y'))+\rho(\pi(y'),\pi(y)) \\ &=\rho(\pi(x'),\pi(y')) \end{aligned} since $\rho(\pi(x),\pi(x'))=0$ as $\pi(x)=\pi(x')$. Meanwhile \begin{aligned} \rho(\pi(x'),\pi(y')) &\leq \rho(\pi(x'),\pi(x)) + \rho(\pi(x),\pi(y)) + \rho(\pi(y),\pi(y')) \\ &= \rho(\pi(x),\pi(y)). \end{aligned} therefore $\rho(\pi(x),\pi(y))=\rho(\pi(x'),\pi(y'))$.

### $\rho$ is compatible with $\tau_N$

By proving this, we need to show that a set $E \subset X/N$ is open with respect to $\tau_N$ if and only if $E$ is a union of open balls. But we need to show a generalized version:

If $\mathscr{B}$ is a local base for $\tau$, then the collection $\mathscr{B}_N$, which contains all sets $\pi(V)$ where $V \in \mathscr{B}$, forms a local base for $\tau_N$.

Proof. We already know that $\pi$ is continuous, linear and open. Therefore $\pi(V)$ is open for all $V \in \mathscr{B}$. For any open set around $E \subset X/N$ containing $\pi(0)$, we see $\pi^{-1}(E)$ is open, and we have $\pi^{-1}(E)=\bigcup_{V\in\mathscr{B}}V$ and therefore $E=\bigcup_{V \in \mathscr{B}}\pi(V).$

Now consider the local base $\mathscr{B}$ containing all open balls around $0 \in X$. Since $\pi(\{x:d(x,0)<r\})=\{u:\rho(u,\pi(0))<r\}$ we see $\rho$ determines $\mathscr{B}_N$. But we have already proved that $\rho$ is invariant; hence $\mathscr{B}_N$ determines $\tau_N$.

### If $d$ is complete, then $\rho$ is complete.

Once this is proved, we are able to claim that, if $X$ is a $F$-space, then $X/N$ is still a $F$-space, since its topology is induced by a complete invariant metric $\rho$.

Proof. Suppose $(x_n)$ is a Cauchy sequence in $X/N$, relative to $\rho$. There is a subsequence $(x_{n_k})$ with $\rho(x_{n_k},x_{n_{k+1}})<2^{-k}$. Since $\pi$ is surjective, we are able to pick some $z_k \in X$ such that $\pi(z_k) = x_{n_k}$ and such that $d(z_{k},z_{k+1})<2^{-k}.$ (The existence can be verified by contradiction still.) By the inequality above, we see $(z_k)$ is Cauchy (can you see why?). Since $X$ is complete, $z_k \to z$ for some $z \in X$. By the continuity of $\pi$, we also see $x_{n_k} \to \pi(z)$ as $k \to \infty$. Therefore $(x_{n_k})$ converges. Hence $(x_n)$ converges since it has a convergent subsequence. $\rho$ is complete.

## Remarks

This fact will be used to prove some corollaries in the open mapping theorem. For instance, for any continuous linear map $\Lambda:X \to Y$, we see $\ker(\Lambda)$ is closed, therefore if $X$ is a $F$-space, then $X/\ker(\Lambda)$ is a $F$-space as well. We will show in the future that $X/\ker(\Lambda)$ and $\Lambda(X)$ are homeomorphic if $\Lambda(X)$ is of the second category.

There are more properties that can be inherited by $X/N$ from $X$. For example, normability, metrizability, local convexity. In particular, if $X$ is Banach, then $X/N$ is Banach as well. To do this, it suffices to define the quotient norm by $\lVert \pi(x) \rVert = \inf\{\lVert x-z \rVert:z \in N\}.$

# An Introduction to Quotient Space

I'm assuming the reader has some abstract algebra and functional analysis background. You may have learned this already in your linear algebra class, but we are making our way to functional analysis problems.

## Motivation

### The trouble with $L^p$ spaces

Fix $p$ with $1 \leq p \leq \infty$. It's easy to see that $L^p(\mu)$ is a topological vector space. But it is not a metric space if we define $d(f,g)=\lVert f-g \rVert_p.$ The reason is, if $d(f,g)=0$, we can only get $f=g$ a.e., but they are not strictly equal. With that being said, this function $d$ is actually a pseudo metric. This is unnatural. However, the relation $\sim$ by $f \sim g \mathbb{R}ightarrow d(f,g)=0$ is a equivalence relation. This inspires us to take the quotient set into consideration.

### Vector spaces are groups anyway

For a vector space $V$, every subspace of $V$ is a normal subgroup. There is no reason to prevent ourselves from considering the quotient group and looking for some interesting properties. Further, a vector space is an abelian group, therefore any subspace is automatically normal.

## Definition

Let $N$ be a subspace of a vector space $X$. For every $x \in X$, let $\pi(x)$ be the coset of $N$ that contains $x$, that is $\pi(x)=x+N.$ Trivially, $\pi(x)=\pi(y)$ if and only if $x-y \in N$ (say, $\pi$ is well-defined since $N$ is a vector space). This is a linear function since we also have the addition and multiplication by $\pi(x)+\pi(y)=\pi(x+y) \quad \alpha\pi(x)=\pi(\alpha{x}).$ These cosets are the elements of a vector space $X/N$, which reads, the quotient space of $X$ modulo $N$. The map $\pi$ is called the canonical map as we all know.

## Examples

First, we shall treat $\mathbb{R}^2$ as a vector space, and the subspace $\mathbb{R}$, which is graphically represented by $x$-axis, as a subspace (we will write it as $X$). For a vector $v=(2,3)$, which is represented by $AB$, we see the coset $v+X$ has something special. Pick any $u \in X$, for example, $AE$, $AC$, or $AG$. We see $v+u$ has the same $y$ value. The reason is simple since we have $v+u=(2+x,3)$, where the $y$ value remains fixed however $u$ may vary.

With that being said, the set $v+X$, which is not a vector space, can be represented by $\overrightarrow{AD}$. This proceed can be generalized to $\mathbb{R}^n$ with $\mathbb{R}^m$ as a subspace with ease.

We now consider a fancy example. Consider all rational Cauchy sequences, that is $(a_n)=(a_1,a_2,\cdots)$ where $a_k\in\mathbb{Q}$ for all $k$. In analysis class, we learned two facts.

1. Any Cauchy sequence is bounded.
2. If $(a_n)$ converges, then $(a_n)$ is Cauchy.

However, the reverse of 2 does not hold in $\mathbb{Q}$. For example, if we put $a_k=(1+\frac{1}{k})^k$, we should have the limit to be $e$, but $e \notin \mathbb{Q}$.

If we define the addition and multiplication term by term, namely $(a_n)+(b_n)=(a_1+b_1,a_2+b_2,\cdots)$ and $(\alpha a_n)=(\alpha a_1,\alpha a_2,\cdots)$ where $\alpha \in \mathbb{Q}$, we get a vector space (the verification is easy). The zero vector is defined by $(0)=(0,0,\cdots).$ This vector space is denoted by $\overline{\mathbb{Q}}$. The subspace containing all sequences converges to $0$ will be denoted by $\overline{\mathbb{O}}$. Again, $(a_n)+\overline{\mathbb{O}}=(b_n)+\overline{\mathbb{O}}$ if and only if $(a_n-b_n) \in \overline{\mathbb{O}}$. Using the language of equivalence relation, we also say $(a_n)$ and $(b_n)$ are equivalent if $(a_n-b_n) \in \overline{\mathbb{O}}$. For example, the two following sequences are equivalent: $(1,1,1,\cdots,1,\cdots)\quad\quad (0.9,0.99,0.999,\cdots).$ Actually, we will get $\mathbb{R} \simeq \overline{\mathbb{Q}}/\overline{\mathbb{O}}$ in the end. But to make sure that this quotient space is exactly the one we meet in our analysis class, there are a lot of verifications should be done.

We shall give more definitions for calculation. The multiplication of two Cauchy sequences is defined term by term à la the addition. For $\overline{\mathbb{Q}}/\overline{\mathbb{O}}$ we have $((a_n)+\overline{\mathbb{O}})+((b_n)+\overline{\mathbb{O}})=(a_n+b_n) + \overline{\mathbb{O}}$ and $((a_n)+\overline{\mathbb{O}})((b_n)+\overline{\mathbb{O}})=(a_nb_n)+\overline{\mathbb{O}}.$ As for inequality, a partial order has to be defined. We say $(a_n) > (0)$ if there exists some $N>0$ such that $a_n>0$ for all $n \geq N$. By $(a_n) > (b_n)$ we mean $(a_n-b_n)>(0)$ of course. For cosets, we say $(a_n)+\overline{\mathbb{O}}>\overline{\mathbb{O}}$ if $(x_n) > (0)$ for some $(x_n) \in (a_n)+\overline{\mathbb{O}}$. This is well defined. That is, if $(x_n)>(0)$, then $(y_n)>(0)$ for all $(y_n) \in (a_n)+\overline{\mathbb{O}}$.

With these operations being defined, it can be verified that $\overline{\mathbb{Q}}/\overline{\mathbb{O}}$ has the desired properties, for example, the least-upper-bound property. But this goes too far from the topic, we are not proving it here. If you are interested, you may visit here for more details.

Finally, we are trying to make $L^p$ a Banach space. Fix $p$ with $1 \leq p < \infty$. There is a seminorm defined for all Lebesgue measurable functions on $[0,1]$ by $p(f)=\lVert f \rVert_p=\left\{\int_{0}^{1}|f(t)|^pdt\right\}^{1/p}$ $L^p$ is a vector space containing all functions $f$ with $p(f)<\infty$. But it's not a normed space by $p$, since $p(f)=0$ only implies $f=0$ almost everywhere. However, the set $N$ which contains all functions that equal $0$ is also a vector space. Now consider the quotient space by $\tilde{p}(\pi(f))=p(f),$ where $\pi$ is the canonical map of $L^p$ into $L^p/N$. We shall prove that $\tilde{p}$ is well-defined here. If $\pi(f)=\pi(g)$, we have $f-g \in N$, therefore $0=p(f-g)\geq |p(f)-p(g)|,$ which forces $p(f)=p(g)$. Therefore in this case we also have $\tilde{p}(\pi(f))=\tilde{p}(\pi(g))$. This indeed ensures that $\tilde{p}$ is a norm, and $L^p/N$ a Banach space. There are some topological facts required to prove this, we are going to cover a few of them.

## Topology of quotient space

### Definition

We know if $X$ is a topological vector space with a topology $\tau$, then the addition and scalar multiplication are continuous. Suppose now $N$ is a closed subspace of $X$. Define $\tau_N$ by $\tau_N=\{E \subset X/N:\pi^{-1}(E)\in \tau\}.$ We are expecting $\tau_N$ to be properly-defined. And fortunately, it is. Some interesting techniques will be used in the following section.

### $\tau_N$ is a vector topology

There will be two steps to get this done.

$\tau_N$ is a topology.

It is trivial that $\varnothing$ and $X/N$ are elements of $\tau_N$. Other properties are immediate as well since we have $\pi^{-1}(A \cap B) = \pi^{-1}(A) \cap \pi^{-1}(B)$ and $\pi^{-1}(\cup A_\alpha)=\cup\pi^{-1}( A_{\alpha}).$ That said, if we have $A,B\in \tau_N$, then $A \cap B \in \tau_N$ since $\pi^{-1}(A \cap B)=\pi^{-1}(A) \cap \pi^{-1}(B) \in \tau$.

Similarly, if $A_\alpha \in \tau_N$ for all $\alpha$, we have $\cup A_\alpha \in \tau_N$. Also, by definition of $\tau_N$, $\pi$ is continuous.

$\tau_N$ is a vector topology.

First, we show that a point in $X/N$, which can be written as $\pi(x)$, is closed. Notice that $N$ is assumed to be closed, and $\pi^{-1}(\pi(x))=x+N$ therefore has to be closed.

In fact, $F \subset X/N$ is $\tau_N$-closed if and only if $\pi^{-1}(F)$ is $\tau$-closed. To prove this, one needs to notice that $\pi^{-1}(F^c)=(\pi^{-1}(F))^{c}$.

Suppose $V$ is open, then $\pi^{-1}(\pi(V))=N+V$ is open. By definition of $\tau_N$, we have $\pi(V) \in \tau_N$. Therefore $\pi$ is an open mapping.

If now $W$ is a neighbourhood of $0$ in $X/N$, there exists a neighbourhood $V$ of $0$ in $X$ such that $V + V \subset \pi^{-1}(W).$ Hence $\pi(V)+\pi(V) \subset W$. Since $\pi$ is open, $\pi(V)$ is a neighbourhood of $0$ in $X/N$, this shows that the addition is continuous.

The continuity of scalar multiplication will be shown in a direct way (so can the addition, but the proof above is intended to offer some special technique). We already know, the scalar multiplication on $X$ by \begin{aligned} \varphi:\Phi \times X &\to X \\ (\alpha,x) &\mapsto \alpha{x} \end{aligned} is continuous, where $\Phi$ is the scalar field (usually $\mathbb{R}$ or $\mathbb{C}$. Now the scalar multiplication on $X/N$ is by \begin{aligned} \psi: \Phi \times X/N &\to X/N \\ (\alpha,x+N) &\mapsto \alpha{x}+N. \end{aligned} We see $\psi(\alpha,x+N)=\pi(\varphi(\alpha,x))$. But the composition of two continuous functions is continuous, therefore $\psi$ is continuous.

## A commutative diagram by quotient space

We are going to talk about a classic commutative diagram that you already see in algebra class.

There are some assumptions.

1. $X$ and $Y$ are topological vector spaces.
2. $\Lambda$ is linear.
3. $\pi$ is the canonical map.
4. $N$ is a closed subspace of $X$ and $N \subset \ker\Lambda$.

Algebraically, there exists a unique map $f: X/N \to Y$ by $x+N \mapsto \Lambda(x)$. Namely, the diagram above is commutative. But now we are interested in some analysis facts.

$f$ is linear.

This is obvious. Since $\pi$ is surjective, for $u,v \in X/N$, we are able to find some $x,y \in X$ such that $\pi(x)=u$ and $\pi(y)=v$. Therefore we have \begin{aligned} f(u+v)=f(\pi(x)+\pi(y))&=f(\pi(x+y)) \\ &=\Lambda(x+y) \\ &=\Lambda(x)+\Lambda(y) \\ &= f(\pi(x))+f(\pi(y)) \\ &=f(u)+f(v) \end{aligned} and \begin{aligned} f(\alpha{u})=f(\alpha\pi(x))&=f(\pi(\alpha{x})) \\ &= \Lambda(\alpha{x}) \\ &= \alpha\Lambda(x) \\ &= \alpha{f(\pi(x))} \\ &= \alpha{f(u)}. \end{aligned}

$\Lambda$ is open if and only if $f$ is open.

If $f$ is open, then for any open set $U \subset X$, we have $\Lambda(U)=f(\pi(U))$ to be an open set since $\pi$ is open, and $\pi(U)$ is an open set.

If $f$ is not open, then there exists some $V \subset X/N$ such that $f(V)$ is closed. However, since $\pi$ is continuous, we have $\pi^{-1}(V)$ to be open. In this case, we have $f(\pi(\pi^{-1}(V)))=f(V)=\Lambda(\pi^{-1}(V))$ to be closed. $\Lambda$ is therefore not open. This shows that if $\Lambda$ is open, then $f$ is open.

$\Lambda$ is continuous if and only if $f$ is continuous.

If $f$ is continuous, for any open set $W \subset Y$, we have $\pi^{-1}(f^{-1}(W))=\Lambda^{-1}(W)$ to be open. Therefore $\Lambda$ is continuous.

Conversely, if $\Lambda$ is continuous, for any open set $W \subset Y$, we have $\Lambda^{-1}(W)$ to be open. Therefore $f^{-1}(W)=\pi(\Lambda^{-1}(W))$ has to be open since $\pi$ is open.

# A brief introduction to Fréchet derivative

Fréchet derivative is a generalisation to the ordinary derivative. Generally we are talking about Banach space, where $\mathbb{R}$​ is a special case. Indeed, the space discussed is not even required to be of finite dimension.

## Recall

A real-valued function $f(t)$ of a real variable, defined on some neighborhood of $0$, is said to be of $o(t)$ if $\lim_{t \to 0} \frac{f(t)}{t}=0.$ And its derivative at some point $a$ is defined by $f'(a)=\lim_{h \to 0}\frac{f(a+h)-f(a)}{h}.$ We also have this equivalent equation: $f(a+h)=f(a)+f'(a)h+o(h).$ Now suppose $f:U \subset \mathbb{R}^n \to \mathbb{R}^m$ where $U$ is an open set. The function $f$ is differentiable at $x_0 \in U$ if satisfying the following conditions.

1. All partial derivatives of $f$, i.e. $\frac{\partial f_i}{\partial x_j}$ exists for all $i=1,\cdots,m$ and $j = 1,\cdots,n$ at $f$. (Which ensures that the Jacobian matrix exists and is well-defined).

2. The Jacobian matrix $J(x_0)\in\mathbb{R}^{m\times n}$ satisfies $\lim_{|h| \to 0}\frac{|f(x_0+h)-f(x_0)-J(x_0)h|}{|h|}=0.$ In fact the Jacobian matrix has been the derivative of $f$ at $x_0$ although it's a matrix in lieu of number. But we should treat a number as a matrix in the general case. In the following definition of Fréchet derivative, you will see that we should treat something as linear functional.

## Definition

Let $f:U\to\mathbf{F}$ be a function where $U$ is an open subset of $\mathbf{E}$. We say $f$ is Fréchet differentiable at $x \in U$ if there is a bounded and linear operator $\lambda:\mathbf{E} \to \mathbf{F}$ such that $\lim_{\lVert y \rVert \to 0}\frac{\lVert f(x+y)-f(x)-\lambda y \rVert}{\lVert y \rVert}=0.$ We say that $\lambda$ is the derivative of $f$ at $x$, which will be denoted by $Df(x)$ or $f'(x)$. Notice that $\lambda \in L(\mathbf{E},\mathbf{F})$. If $f$ is differentiable at every point of $f$, then $f'$ is a map by $f':U \to L(\mathbf{E},\mathbf{F}).$

The definition above doesn't go too far from real functions defined on the real axis. Now we are assuming that both $\mathbf{E}$ and $\mathbf{F}$ are merely topological vector spaces, and still we can get the definition of Fréchet derivative (generalized).

Let $\varphi$ be a mapping of a neighborhood of $0$ of $\mathbf{E}$ into $\mathbf{F}$. We say that $\varphi$ is tangent to $0$ if given a neighborhood $W$ of $0$ in $\mathbf{F}$, there exists a neighborhood $V$ of $0$ in $\mathbf{E}$ such that $\varphi(tV) \subset o(t)W$ for some function of $o(t)$. For example, if both $\mathbf{E}$ and $\mathbf{F}$ are normed (not have to be Banach), then we get a usual condition by $\lVert \varphi(x) \rVert \leq \lVert x \rVert \psi(x)$ where $\lim_{\lVert x \rVert \to 0}\psi(x)=0$.

Still we assume that $\mathbf{E}$ and $\mathbf{F}$ are topological vector spaces. Let $f:U \to \mathbf{F}$ be a continuous map. We say that $f$ is differentiable at a point $x \in U$ if there exists some $\lambda \in L(\mathbf{E},\mathbf{F})$ such that for small $y$ we have $f(x+y)=f(x)+\lambda{y}+\varphi(y)$ where $\varphi$ is tangent to $0$. Notice that $\lambda$ is uniquely determined.

## Propositions

You must be familiar with some properties of derivative, but we are redoing these in Banach space.

### Chain rule

If $f: U \to V$ is differentiable at $x_0$, and $g:V \to W$ is differentiable at $f(x_0)$, then $g \circ f$ is differentiable at $x_0$, and $(g \circ f)'(x_0)=g'(f(x_0)) \circ f'(x_0)$

Proof. We are proving this in topological vector space. By definition, we already have some linear operator $\lambda$ and $\mu$ such that $f(x_0+y)=f(x_0)+\lambda{y}+\varphi(y) \\ g(f(x_0)+h)=g(f(x_0))+\mu{h}+\psi(h)$ where $\varphi$ and $\psi$ are tangent to $0$. Further, we got $f'(x_0)=\lambda \\ g'(f(x_0))=\mu$ To evaluate $g(f(x_0+y))$, notice that \begin{aligned} g(f(x_0+y))&=g[f(x_0)+(\lambda{y}+\varphi(y))] \\ &=g(f(x_0))+\mu(\lambda{y}+\varphi(y))+\psi(\lambda{y}+\varphi(y)) \\ &=g(f(x_0))+\mu\circ\lambda{y}+\mu\circ\varphi(y)+\psi(\lambda{y}+\varphi(y)) \end{aligned} It's clear that $\mu\circ\varphi(y)+\psi(\lambda{y}+\varphi(y))$ is tangent to $0$, and $\mu\circ\lambda$ is the linear map we are looking for. That is, $(g \circ f)'(x)=g'(f(x_0)) \circ f'(x_0).$

### Derivative of higher orders

From now on, we are dealing with Banach spaces. Let $U$ be an open subset of $\mathbf{E}$, and $f:U \to \mathbf{F}$ be differentiable at each point of $U$. If $f'$ is continuous, then we say that $f$ is of class $C^1$. The function of order $C^p$ where $p \geq 1$ is defined inductively. The $p$-th derivative $D^pf$ is defined as $D(D^{p-1}f)$ and is itself a map of $U$ into $L(\mathbf{E},L(\mathbf{E},\cdots,L(\mathbf{E},\mathbf{F})\cdots)))$ which is isomorphic to $L^p(\mathbf{E},\mathbf{F})$. A map $f$ is said to be of class $C^p$ if its $kth$ derivative $D^kf$ exists for $1 \leq k \leq p$, and is continuous. With the help of chain rule, and the fact that the composition of two continuous functions are continuous, we get

Let $U,V$ be open subsets of some Banach spaces. If $f:U \to V$ and $g: V \to \mathbf{F}$ are of class $C^p$, then so is $g \circ f$.

### Open subsets of Banach spaces as a category

We in fact get a category $\{(U,f_U)\}$ where $U$ is the object as an open subset of some Banach space, and $f_U$ is the morphism as a map of class $C^p$ mapping $U$ into another open set. To verify this, one only has to realize that the composition of two maps of class $C^p$ is still of class $C^p$ (as stated above).

We say that $f$ is of class $C^\infty$ if $f$ is of class $C^p$ for all integers $p \geq 1$. Meanwhile $C^0$ maps are the continuous maps.

## An example

We are going to evaluate the Fréchet derivative of a nonlinear functional. It is the derivative of a functional mapping an infinite dimensional space into $\mathbb{R}$ (instead of $\mathbb{R}$ to $\mathbb{R}$).

Consider the functional by \begin{aligned} \Gamma:C^0[0,1] &\to \mathbb{R} \\ u &\mapsto \int_{0}^{1}u^2(x)\sin\pi{x}dx. \end{aligned} where the norm is defined by $\lVert u \rVert = \sup_{x \in [0,1]}|u|.$

For $u\in C[0,1]$, we are going to find an linear operator $\lambda$ such that $\Gamma(u+\eta)=\Gamma(u)+\lambda{\eta}+\varphi(\eta),$ where $\varphi(\eta)$ is tangent to $0$.

Solution. By evaluating $\Gamma(u+\eta)$, we get \begin{aligned} \Gamma(u+\eta)&=\int_{0}^{1}(u+\eta)^2\sin\pi{x}dx \\ &= \Gamma(u)+2\int_{0}^{1}u\eta\sin\pi{x}dx+\int_{0}^{1}\eta^2\sin\pi{x}dx. \end{aligned} To prove that $\int_{0}^{1}\eta^2\sin{x}dx$ is the $\varphi(\eta)$ desired, notice that $\int_{0}^{1}\eta^2\sin\pi{x}dx \leq \lVert\eta\rVert^2\int_{0}^{1}\sin\pi{x}dx=2\lVert \eta \rVert^2.$ Therefore we have $0\leq\lim_{\lVert \eta \rVert \to 0}\frac{\int_{0}^{1}\eta^2\sin\pi{x}dx}{\lVert \eta \rVert} \leq \lim_{\lVert\eta\rVert\to0}2\lVert\eta\rVert=0$ as desired. The Fréchet derivative of $\Gamma$ at $u$ is defined by \begin{aligned} \Gamma'(u):C[0,1] &\to \mathbb{R} \\ \eta &\mapsto 2\int_{0}^{1}u\eta\sin\pi{x}dx. \end{aligned} It's hard to believe but, the derivative is not a number, nor a matrix, but a linear operator. But conversely, a real or complex number or matrix can be treated as a linear operator in the nature of things.