($\forall x\in E_1 \exists y\in E_2: f_1(x_1)=f_2(x_2))\ \Longrightarrow \ x \mapsto y$ continuous

41 Views Asked by At

I'm reading through some notes on functional analysis where on a couple of occasion we where in the following setting: $E_1$ and $E_2$ are two Frechet spaces and $f_i:E_i \rightarrow F$ two continuous linear mappings into a seminormed space $F$. For each $x \in E_1$ we know that there exists a unique $y\in E_2$ with $f_1(x)=f_2(y)$. (This seems hard to understand intuitively for me, not just on a technical level - can someone maybe provide some insight/concerete example where this occurs ?)

Then the mapping that maps such an $x$ to the corresponding $y$ came up and its continuity was used. Can someone help fill in this gap how to prove the continuity ?

1

There are 1 best solutions below

2
On BEST ANSWER

The continuity of the map $\varphi = f_2^{-1}\circ f_1 \colon x \mapsto y$ follows by the closed graph theorem, assuming that $F$ is a Hausdorff space.

The graph of $\varphi$,

$$\Gamma(\varphi) = \left\{(x,y) \in E_1 \times E_2 : f_1(x) = f_2(y)\right\}$$

is the preimage of the diagonal of $F$ under the continuous map $f_1\times f_2 \colon E_1\times E_2 \to F\times F$, hence closed, since in a Hausdorff space, the diagonal is closed. The closed graph theorem yields the continuity of $\varphi$.