the question why Frechet space is topological vector space.

Question

According to the following lecture (or Wikipedia),

Let $F(\omega)$ be the set of all real sequences and let $x=(x_1,x_2,...,x_k...) ,y=(y_1,y_2,...,y_k,...) \in F(\omega)$. Define the function $d : F(\omega) \times F(\omega) \to \mathbb{R} $, $~$ $d(x,y)= \sum_{i=1}^{\infty} \frac{1}{2^i}\frac{|x_i-y_i|}{1+|x_i-y_i|} $. $~~$(Clearly, the function is well-defined) Then, $(F(\omega),d)$ is a metric space which is well known as Fréchet space.

And according to the Wikipedia again on the Fréchet space, they are special topological vector space (TVS). I wonder that they really satisfy the TVS condition. More precisely, if $(F(\omega),d)$ is a TVS, following conditions

$~~$ (i) The vector addition $+: F(\omega) \times F(\omega) \to F(\omega) $ is continuous.

$~~$ (ii) The scalar multiplication $\cdot : \mathbb{R} \times F(\omega) \to F(\omega) $ is continuous.

would be satisfied.

Since $(F(\omega),d)$ is a metric space, I thought that it would be OK by using the continuous functions between metric spaces, but is it right idea? In order to check the condition, $F(\omega) \times F(\omega)$ must have the metric space, and I thought that the natural metric space on the (product) space is product metric space.

And how to approach in case (ii)? $\mathbb{R} \times F(\omega)$ also have a metric space. However, since the component of product is not coincidence, I don't imagine why scalar multiplication is continuous.

Answer 1

It really helps to just think of this as a question about convergent sequences. Note that a sequence $(x^n)_{n\in\mathbb{N}}\subseteq F(\omega)$ is convergent (with limit $x$) if and only if the sequence of real numbers $(x^n_k)_{n\in \mathbb{N}}$ is convergent with limit $x_k$ for every $k$.

To see this, note first of all that $f(t)=\frac{t}{1+t}=1-\frac{1}{1+t}$ is strictly increasing, continuous and maps $0$ to $0$. Thus,

$$ d(x,x^n)\geq 2^k f(|x_k-x_k^n|) $$ implies one direction.

For the other direction, note that $f(t)\leq 1$ for all $t\geq 0$, and thus, $$ d(x,x^n)\leq \max_{1\leq k\leq K} 2^{-k}|x_k-x^n_k|+\sum_{j=K+1}^{\infty} 2^{-j}=\max_{1\leq k\leq K} 2^{-k}|x_k-x^n_k|+2^{-K}, $$ which implies the other direction.

So to check continuity, we simply need to argue that convergent sequences are mapped to convergent sequences. Let $(x^n)_{n\in\mathbb{N}},(y^n)_{n\in\mathbb{N}}$ be convergent sequences in $F(\omega)$ with limits $x$ and $y$ respectively. Then, $x^n_k+y^n_k$ is convergent with limit $x_k+y_k$ for every $k$, since addition is continuous in $\mathbb{R}$. Thus, $x^n+y^n$ is convergent in $F(\omega)$ with limit $x+y$.

If, furthermore, $(\lambda_n)_{n\in\mathbb{N}}$ is convergent in $\mathbb{R}$ with limit $\lambda$, then $\lambda_n x^n_k$ is convergent with limit $\lambda x_k$, since multiplication is continuous on $\mathbb{R}$. Hence, $\lambda_nx_k^n$ is convergent in $F(\omega)$ with limit $\lambda x$.

Answer 2

$F(\omega)$ is just $\mathbb{R}^\mathbb{N}$ in the product topology (the said metric on $F(\omega)$ induces the product topology, as I showed here, in a more general setting, for a countable product of metric spaces). And a function $F: Y \to \mathbb{R}^\mathbb{N}$ in the product topology (for any function $F$ and any space $Y$) is continuous iff for all $n \in \mathbb{N}$ we have that $\pi_n \circ F$, the composition with the $n$-th projection, is continuous. That fact can even be seen as the defining property of the product topology, if we like (if you're interested in the theory behind that, you could read my answer here, e.g.).

So if we look at addition as a map $a: \mathbb{R}^\mathbb{N} \times \mathbb{R}^\mathbb{N} \to \mathbb{R}^\mathbb{N}$, with $a((x_n)_n, (y_n)_n)=(x_n + y_n)_n$ then for any $m$, $\pi_m \circ a$ sends the pair $((x_n)_n, (y_n)_n)$ to $x_m + y_m$ which is just $p \circ (\pi_m \times \pi_m)$, where $p: \mathbb{R} \times \mathbb{R} \to \mathbb{R}$ is addition on the reals (which I assume you know is continuous), so every composition with $\pi_m$ is itself continuous again as a composition of other continuous maps. So $a$ is continuous. In general, such pointwise operations on (topological) products of topological groups is again a continuous operation, in a similar way. The scalar multiplication map $m: F(\omega) \times \Bbb R \to F(\omega)=\mathbb{R}^\mathbb{N}$ is similarly continuous: $\pi_n \circ m = m' \circ (\pi_n \times \textrm{id}_{\Bbb R})$ where $m': \Bbb R \times \Bbb R \to \Bbb R$ is multiplication on $\Bbb R$ (hopefully you already know that that map is continuous).

So $F(\omega)$ is a TVS because $\Bbb R$ is and we have a topological product of copies of it. All we need is that the given metric just induces the product topology, which has the nice property that coordinate-wise operations are still continuous.

Answer 3

Note that what you wrote is not a definition of what Fréchet means. It is simply saying that $F(\omega)$ satisfies the conditions of being a Fréchet space. In particular, one has to check if the space is complete.

To see that $F(\omega)$ is a TVS, notice that the topology given by the metric $d$ is identical to the topology given by the (countable and separating) family of seminorms $$ p_k(x)=|x_k|,\qquad k=1,2,\ldots. $$ Then it is easy to show that this topology turns the underlying vector space into a TVS (in fact, it is a locally convex TVS).