Fréchet space

A topological vector space ${\displaystyle X}$ is a Fréchet space if and only if it satisfies the following three properties:

1. It is locally convex.[note 2]
2. Its topology can be induced by a translation-invariant metric, that is, a metric ${\displaystyle d:X\times X\to \mathbb {R} }$ such that ${\displaystyle d(x,y)=d(x+z,y+z)}$ for all ${\displaystyle x,y,z\in X.}$ This means that a subset ${\displaystyle U}$ of ${\displaystyle X}$ is open if and only if for every ${\displaystyle u\in U}$ there exists an ${\displaystyle r>0}$ such that ${\displaystyle \{v:d(v,u)<r\}}$ is a subset of ${\displaystyle U.}$
3. Some (or equivalently, every) translation-invariant metric on ${\displaystyle X}$ inducing the topology of ${\displaystyle X}$ is complete.
   • Assuming that the other two conditions are satisfied, this condition is equivalent to ${\displaystyle X}$ being a complete topological vector space, meaning that ${\displaystyle X}$ is a complete uniform space when it is endowed with its canonical uniformity (this canonical uniformity is independent of any metric on ${\displaystyle X}$ and is defined entirely in terms of vector subtraction and ${\displaystyle X}$'s neighborhoods of the origin; moreover, the uniformity induced by any (topology-defining) translation invariant metric on ${\displaystyle X}$ is identical to this canonical uniformity).

Note there is no natural notion of distance between two points of a Fréchet space: many different translation-invariant metrics may induce the same topology.

The alternative and somewhat more practical definition is the following: a topological vector space ${\displaystyle X}$ is a Fréchet space if and only if it satisfies the following three properties:

1. It is a Hausdorff space,
2. Its topology may be induced by a countable family of seminorms ${\displaystyle ||\cdot ||_{k}}$ ${\displaystyle k=0,1,2,\ldots }$ This means that a subset ${\displaystyle U\subseteq X}$ is open if and only if for every ${\displaystyle u\in U}$ there exists ${\displaystyle K\geq 0}$ and ${\displaystyle r>0}$ such that ${\displaystyle \{v\in X:\|v-u\|_{k}<r{\text{ for all }}k\leq K\}}$ is a subset of ${\displaystyle U,}$
3. it is complete with respect to the family of seminorms.

A family ${\displaystyle {\mathcal {P}}}$ of seminorms on ${\displaystyle X}$ yields a Hausdorff topology if and only if[2]

$${\displaystyle \bigcap _{\|\,\cdot \,\|\in {\mathcal {P}}}\{x\in X:\|x\|=0\}=\{0\}.}$$

A sequence ${\displaystyle x_{\bullet }=\left(x_{n}\right)_{n=1}^{\infty }}$ in ${\displaystyle X}$ converges to ${\displaystyle x}$ in the Fréchet space defined by a family of seminorms if and only if it converges to ${\displaystyle x}$ with respect to each of the given seminorms.

In contrast to Banach spaces, the complete translation-invariant metric need not arise from a norm. The topology of a Fréchet space does, however, arise from both a total paranorm and an F-norm (the F stands for Fréchet).

Even though the topological structure of Fréchet spaces is more complicated than that of Banach spaces due to the potential lack of a norm, many important results in functional analysis, like the open mapping theorem, the closed graph theorem, and the Banach–Steinhaus theorem, still hold.

• Every Banach space is a Fréchet space, as the norm induces a translation-invariant metric and the space is complete with respect to this metric.
• The space ${\displaystyle \mathbb {R} ^{\omega }}$ of all real valued sequences (also denoted ${\displaystyle \mathbb {R} ^{\mathbb {N} }}$) becomes a Fréchet space if we define the ${\displaystyle k}$-th seminorm of a sequence to be the absolute value of the ${\displaystyle k}$-th element of the sequence. Convergence in this Fréchet space is equivalent to element-wise convergence.

• The vector space ${\displaystyle C^{\infty }([0,1])}$ of all infinitely differentiable functions ${\displaystyle f:[0,1]\to \mathbb {R} }$ becomes a Fréchet space with the seminorms
  $${\displaystyle \|f\|_{k}=\sup\{|f^{(k)}(x)|:x\in [0,1]\}}$$

  
  for every non-negative integer ${\displaystyle k.}$ Here, ${\displaystyle f^{(k)}}$ denotes the ${\displaystyle k}$-th derivative of ${\displaystyle f,}$ and ${\displaystyle f^{(0)}=f.}$ In this Fréchet space, a sequence ${\displaystyle \left(f_{n}\right)\to f}$ of functions converges towards the element ${\displaystyle f\in C^{\infty }([0,1])}$ if and only if for every non-negative integer ${\displaystyle k\geq 0,}$ the sequence ${\displaystyle \left(f_{n}^{(k)}\right)\to f^{(k)}}$ converges uniformly.
• The vector space ${\displaystyle C^{\infty }(\mathbb {R} )}$ of all infinitely differentiable functions ${\displaystyle f:\mathbb {R} \to \mathbb {R} }$ becomes a Fréchet space with the seminorms
  $${\displaystyle \|f\|_{k,n}=\sup\{|f^{(k)}(x)|:x\in [-n,n]\}}$$

  
  for all integers ${\displaystyle k,n\geq 0.}$ Then, a sequence of functions ${\displaystyle \left(f_{n}\right)\to f}$ converges if and only if for every ${\displaystyle k,n\geq 0,}$ the sequences ${\displaystyle \left(f_{n}^{(k)}\right)\to f^{(k)}}$ converge compactly.
• The vector space ${\displaystyle C^{m}(\mathbb {R} )}$ of all ${\displaystyle m}$-times continuously differentiable functions ${\displaystyle f:\mathbb {R} \to \mathbb {R} }$ becomes a Fréchet space with the seminorms
  $${\displaystyle \|f\|_{k,n}=\sup\{|f^{(k)}(x)|:x\in [-n,n]\}}$$

  
  for all integers ${\displaystyle n\geq 0}$ and ${\displaystyle k=0,\ldots ,m.}$
• If ${\displaystyle M}$ is a compact ${\displaystyle C^{\infty }}$-manifold and ${\displaystyle B}$ is a Banach space, then the set ${\displaystyle C^{\infty }(M,B)}$ of all infinitely-often differentiable functions ${\displaystyle f:M\to B}$ can be turned into a Fréchet space by using as seminorms the suprema of the norms of all partial derivatives. If ${\displaystyle M}$ is a (not necessarily compact) ${\displaystyle C^{\infty }}$-manifold which admits a countable sequence ${\displaystyle K^{n}}$ of compact subsets, so that every compact subset of ${\displaystyle M}$ is contained in at least one ${\displaystyle K^{n},}$ then the spaces ${\displaystyle C^{m}(M,B)}$ and ${\displaystyle C^{\infty }(M,B)}$ are also Fréchet space in a natural manner. As a special case, every smooth finite-dimensional complete manifold ${\displaystyle M}$ can be made into such a nested union of compact subsets: equip it with a Riemannian metric ${\displaystyle g}$ which induces a metric ${\displaystyle d(x,y),}$ choose ${\displaystyle x\in M,}$ and let
  $${\displaystyle K_{n}=\{y\in M:d(x,y)\leq n\}\ .}$$

  

  Let ${\displaystyle X}$ be a compact ${\displaystyle C^{\infty }}$-manifold and${\displaystyle V}$ a vector bundle over ${\displaystyle X.}$ Let ${\displaystyle C^{\infty }(X,V)}$ denote the space of smooth sections of ${\displaystyle V}$ over ${\displaystyle X.}$ Choose Riemannian metrics and connections, which are guaranteed to exist, on the bundles ${\displaystyle TX}$ and ${\displaystyle V.}$ If ${\displaystyle s}$ is a section, denote its j^th covariant derivative by ${\displaystyle D^{j}s.}$ Then

  $${\displaystyle \|s\|_{n}=\sum _{j=0}^{n}\sup _{x\in M}|D^{j}s|}$$

  
  (where ${\displaystyle \|\,\cdot \,\|}$ is the norm induced by the Riemannian metric) is a family of seminorms making ${\displaystyle C^{\infty }(M,V)}$ into a Fréchet space.

• Let ${\displaystyle H}$ be the space of entire (everywhere holomorphic) functions on the complex plane. Then the family of seminorms
  $${\displaystyle \|f\|_{n}=\sup\{|f(z)|:|z|\leq n\}}$$

  
  makes ${\displaystyle H}$ into a Fréchet space.
• Let ${\displaystyle H}$ be the space of entire (everywhere holomorphic) functions of exponential type ${\displaystyle \tau .}$ Then the family of seminorms
  $${\displaystyle \|f\|_{n}=\sup _{z\in \mathbb {C} }\exp \left[-\left(\tau +{\frac {1}{n}}\right)|z|\right]|f(z)|}$$

  
  makes ${\displaystyle H}$ into a Fréchet space.

Not all vector spaces with complete translation-invariant metrics are Fréchet spaces. An example is the space ${\displaystyle L^{p}([0,1])}$ with ${\displaystyle p<1.}$ Although this space fails to be locally convex, it is an F-space.

If ${\displaystyle X}$ is a locally convex space then the topology of ${\displaystyle X}$ can be a defined by a family of continuous norms on ${\displaystyle X}$ (a norm is a positive-definite seminorm) if and only if there exists at least one continuous norm on ${\displaystyle X.}$[12] Even if a Fréchet space has a topology that is defined by a (countable) family of norms (all norms are also seminorms), then it may nevertheless still fail to be normable space (meaning that its topology can not be defined by any single norm). The space of all sequences ${\displaystyle \mathbb {K} ^{\mathbb {N} }}$ (with the product topology) is a Fréchet space. There does not exist any Hausdorff locally convex topology on ${\displaystyle \mathbb {K} ^{\mathbb {N} }}$ that is strictly coarser than this product topology.[13] The space ${\displaystyle \mathbb {K} ^{\mathbb {N} }}$ is not normable, which means that its topology can not be defined by any norm.[13] Also, there does not exist any continuous norm on ${\displaystyle \mathbb {K} ^{\mathbb {N} }.}$ In fact, as the following theorem shows, whenever ${\displaystyle X}$ is a Fréchet space on which there does not exist any continuous norm, then this is due entirely to the presence of ${\displaystyle \mathbb {K} ^{\mathbb {N} }}$ as a subspace.

If ${\displaystyle X}$ is a non-normable Fréchet space on which there exists a continuous norm, then ${\displaystyle X}$ contains a closed vector subspace that has no topological complement.[14]

A metrizable locally convex space is normable if and only if its strong dual space is a Fréchet–Urysohn locally convex space.[9] In particular, if a locally convex metrizable space ${\displaystyle X}$ (such as a Fréchet space) is not normable (which can only happen if ${\displaystyle X}$ is infinite dimensional) then its strong dual space ${\displaystyle X_{b}^{\prime }}$ is not a Fréchet–Urysohn space and consequently, this complete Hausdorff locally convex space ${\displaystyle X_{b}^{\prime }}$ is also neither metrizable nor normable.

The strong dual space of a Fréchet space (and more generally, of bornological spaces such as metrizable TVSs) is always a complete TVS and so like any complete TVS, it is normable if and only if its topology can be induced by a complete norm (that is, if and only if it can be made into a Banach space that has the same topology). If ${\displaystyle X}$ is a Fréchet space then ${\displaystyle X}$ is normable if (and only if) there exists a complete norm on its continuous dual space ${\displaystyle X'}$ such that the norm induced topology on ${\displaystyle X'}$ is finer than the weak-* topology.[15] Consequently, if a Fréchet space is not normable (which can only happen if it is infinite dimensional) then neither is its strong dual space.

Note that the homeomorphism described in the Anderson–Kadec theorem is not necessarily linear.