I ws thinking that how a linear map $F$ from a normed space $X$ to a normed space $Y$ is continuous iff it sends open sets unit ball $U$ in $X$ to a bounded subset in $Y$ ?
It says refer sec 6.2
so i saw that $F$ is continuous if $F$ is bounded on $\bar{U}(0,r)$ for some $r>0$.
We denote $C(X,Y)$ the set of all compact operators from $X$ to $Y$
What is a compact operator?
So Let $X,Y$ re normed linear spaces then a linear operator $T:X \rightarrow Y$ is a compact if every bounded subset $B$ of $X$ , $\bar{T(B)}$ is a compact subset of $Y$.
We explore some theorems -
Theorem1 -
Let $X$ and $Y$ are NLS then $T: X \rightarrow Y$ is a compact linear map $\textbf{iff}$ for every bounded sequence $\{x_{n}\} \in X$ $T(x_{n})$ has a convergent subsequence.
$\rightarrow$ direction -
So we have $T$ is a compact linear map and we have a bounded sequence say $\{x_{n}\}$,
Now we consider the set $B =\{x_{n} , n=1,2,3,...\}$
the set $B$ is bounded in $X$
so as $T$ is compact so by defn. $\bar{T(B)}$ is a compact subset of $Y$
$T(x_{n})$ is a bounded sequence in $\bar{T(B)}$
but as $T(B)$ is compact so $\{T(x_{n})\}$ has a convergent subsequence
$\textbf{doubt} $- I was thinking why $\{T(x_{n})\}$ is a sequence in $\bar{T(B)}$ ?
@BAYMAX This theorem seems to be more-or-less direct consequence of the fact that, in metric spaces, compactnes and sequential compactness are equivalent.
@BAYMAX You show that every sequence in $\overline{T(B)}$ has a convergent convergence, which is sequential compactness. (And now we are working with metric spaces, so this is the same as compactness.)
By "from the definition of closure" I mean this: If $y_n\in\overline{T(B)}$, then every neighborhood of $U$ intersects $T(B)$.
Just take an open ball around $y_n$ with radius $1/n$ for $U$.
Let $X$ and $Y$ are NLS then $T: X \rightarrow Y$ is a compact linear map $\textbf{iff}$ for every bounded sequence $\{x_{n}\} \in X$ $T(x_{n})$ has a convergent subsequence.
You are now trying to show the direction where you assume that for every bounded sequence $\{x_{n}\} \in X$ $T(x_{n})$ has a convergent subsequence. And you want to show that $T$ is compact.
Let $k \in \Bbb{K}$ and $F,G \in C(X,Y)$ then $kF $ and $F+G \in C(X,Y)$
Proof -
Let $\{x_{n}\}$ be a bounded sequence in $X$ then we have that there existsa convergent subsequence of $F(x_{n})$
so let $F(x_{n})_{j}$
but why he has taken a subsequence of $\{x_{n}\}$ that is $x_{n_{j}}$ amn dmoreover also used the same subsequence in the convergent subsequence of $\{F(x_{n})\}$
It seems reasonable to use the characterization from the first part of the theorem, so probably you want to show that for each bounded sequence $(x_n)$ there is a convergent subsequence of $H(x_n)$.
@MartinSleziak perhaps we are planning to discuss separate cases like, when we take the same subsequence in botht the cases and next different subsequences
You know that there is a subsequence $(x_{n_k})$ such that $F(x_{n_k})$ converges. But you know nothing about $G(x_{n_k})$, so you can't say how $H$ behaves for terms of this sequence.
You know that there is a subsequence $(x_{n_k})$ such that $F(x_{n_k})$ converges. But you know nothing about $G(x_{n_k})$, so you can't say how $H$ behaves for terms of this sequence.
@BAYMAX No, just a reformulation of the same thing. (Mainly because you seemed uncomfortable with taking subsequence of $(F(x_n))$ and subsequence of $(x_n)$.
> “Your mind is like this water, my friend. When it is agitated, it becomes difficult to see. But if you allow it to settle, the answer becomes clear.”
Any subsequence is determined by the set of indices you take - i.e., by infinite subset of $\mathbb N$ or by increasing sequence of natural numbers - whichever way of looking at it you prefer.
> Show that a linear functional $\phi$ defined on a real linear topological space X is discontinuous if and only if the set $\{x\in X:\phi(x)\neq 0\}$ is connected.
The comments posted there seem to contradict each other:
