Topological direct sum of vector spaces

Question

Here is the theorem I want to prove (from wikipedia fr).

Let $V$ be a normed topological vector space and let $V_1$ and $V_2$ be two subspaces such that $V=V_1\oplus V_2$.

The sets $V_1$ and $V_2$ are equipped with the induced topology.

I suppose that the bijection $\psi: V_1\times V_2\to V$ given by $\psi(v_1,v_2)=v_1+v_2$ is homeomorphic (bijective, continuous wit continuous inverse).

Then:

$V_1$ and $V_2$ are closed in $V$
the projection $s:V_2\to V/V_1$ is homeomorphic.

I'm able to prove that $V_1$ and $V_2$ are closed and that $s$ is continuous.

I'm stuck on proving that $s^{-1}$ is continuous.

QUESTION: how to prove the continuity of $s^{-1}$ ?

Here is what I do. Let $O$ be open in $V_2$. There exists an open part $A$ of $V$ such that $O=A\cap V_2$.

Let $p:V\to V/V_1$ be the projection. I know that $s(O)$ is open iff the part $p^{-1}(s(O))$ is open in $V$. I also knwo that

$p^{-1}(s(O))=O+V_1$

So let $x\in O+V_1$. I have to found a neighborhood of $x$ in $V$ which is contained in $O+V_1$. For the sake of simplicity I suppose $x\in O$.

from here, I'm quite sure that I'm going the wrong way

Let $r>0$ be such that $B(x,r)\subset A$. I want to prove that the ball $B(x,r)$ is included in $O+V_1$.

Let $y\in B(x,r)$. I write $y=x+h$ and I decompose $h=h_1+h_2$ with $h_1\in V_1$ and $h_2\in V_2$.

I have $\|h\|=\|x-y\|\leq r$ and then $\|h_2\|\leq r$ in such a way that $x+h_2\in B(x,r)\subset O$. Then $y=x+h_2+h_1\in O+V_1$.

Where it's wrong : I cannot deduce $\|h_2\|\leq r$ because $V_1$ and $V_2$ are not othonormal. If the dimension is finite, I can create a scalar product adapted to the decomposition $V_1\oplus V_2$ and then invoke the fact that all the norms are equivalent.

EDIT -- QUESTION Is the statement true ? Bourbaki seems to assume completeness of the spaces (hence Banach)

$s$ is a linear and bijective mapping between the Banach spaces $V_2$ and $V/V_1$. Since $s$ is continuous it has a continuous inverse. — Gerd, Aug 20 '22 at 14:32
As stated on Wikipedia, the spaces are not assumed to be complete (hence, not Banach). However, Bourbaki(Bourbaki (fr)) seems to assume completeness. This may be a missing hypothesis on Wikipedia. I edit the question to make it clear. — Laurent Claessens, Aug 20 '22 at 14:46
Ah, yes. I overlooked that $V$ is not assumed to be complete. — Gerd, Aug 20 '22 at 20:40
full proof in french Search for the label "PropKZDqTR " (disclaimer: I'm the author). See also the discussion on Wikipedia fr — Laurent Claessens, Aug 23 '22 at 05:00

score 2 · Answer 1 · answered Aug 20 '22 at 15:48

We can consider $V_1 \times V_2$ as direct sum $V_1 \oplus V_2$ with norm $$ \|(v1,v2) \| = \max\{\|v_1\|,\|v_2\|\} $$ Looking this way, $\phi$ and $\phi^{-1}$ are bounded linear transformations. Hence, there exists some $C>0$ such that $$ \|\phi^{-1}(v_1 + v_2)\| = \max\{\|v_1 \|,\|v_2\|\} \leq C \|v_1 + v_2\|$$

Now, let $x_n + V_1$ be a sequence in $\frac{V}{V_1}$ ( $x_n \in V_2$) such that $$\| x_n + V_1\| = d(x_n, V_1) \rightarrow 0$$ where $d(x_n,V_1)$ is the distance between $x_n$ and $V_1$. Hence, there exists a sequence $v_n \in V_1$ such that $$ \|x_n + v_n\|\rightarrow 0$$ Hence $$ \| x_n \| \leq \max\{\|x_n \| , \|v_n \| \} \leq C\|x_n + v_n \|\rightarrow 0 $$

This proves that $s^{-1}$ is continuous at 0 and hence continuous.

For the sake of completeness, this answers uses the induced norm which is discussed here, here and here — Laurent Claessens, Aug 20 '22 at 17:36

score 1 · Accepted Answer · answered Aug 21 '22 at 21:41

1

The topology of $V$ does not need to be induced by a norm.

Since $\psi^{-1}:V\to V_1\times V_2$ continuous, its second component $q:V\to V_2$ is continuous.

But $q=s^{-1}\circ\pi$, where $\pi$ is the quotient map $V\to V/V_1$.

Hence $s^{-1}$ is continuous.

answered Aug 21 '22 at 21:41

Anne Bauval

49,005

Even better. Thanks. – Laurent Claessens Aug 23 '22 at 04:44

Topological direct sum of vector spaces

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