I have the following definition of complemented subspaces in a normed vector space ( taken from Constantinescu, 2001 )
Let $X$ be a normed space, $M, N$ be algebraically complemented subspaces of $X$ (i.e. $M+N = X$ and $M \cap N = \{0\}$). Let $\phi: M\times N \rightarrow X$ be the natural isomorphism $(x,y) \mapsto x+y$. Then the subspaces $M, N$ are said to be topologically complemented (or simply complemented) if $\phi$ is an homeomorphism.
The questions are:
- Given the canonical injections (inclusions): \begin{align*} i_M: M &\longrightarrow X \\ i_N: N &\longrightarrow X \end{align*} Can I show that an equivalent requirement for two algebraically complemented subspaces $M,N$ in order to be topologically complemented is the topology on $X$ to be the finest topology such that $i_M,i_N$ are continuous (together with summation and scalar product, of course)?
- Can the first definition (and the second if they are equivalent) be generalized to topological vector spaces?
Thanks
EDIT: As pointed out by mjez, on $M,N$ I consider the topology induced by $X$.
This is my attempt: My goal is to prove that $\phi$ is continuous iff $i_M,i_N,+$ are continuous. To this end I notice that: \begin{align*} i_M &= \phi \circ \pi_M\\ i_N &= \phi \circ \pi_N\\ x+y &= \phi(\phi^{-1}(x)+\phi^{-1}(y))\\ \phi &= i_M \circ \pi_M + i_N \circ \pi_N \end{align*} Where I'm making an obvious distinction between summation in $X$ (whose continuity is not necessarily guaranteed) and summation in $M\times N$ (which is continuous by definition of product space).
So I get that $\phi$ is continuous iff $i_M,i_N,+$ are continuous. In particular, since $\phi$ is bijective, I get that $\tau$ is the finest topology s.t. $i_M,i_N,+$ are continuous if and only if $\tau$ makes $\phi$ an homeomorphism.
The answer by mjez uses an additional hypothesis of local convexity to prove the same result, so I suspect that the proof I just presented may be flawed. Am I right?
I think that the answer to your first question is affirmative.
Let us prove first that if $M$ and $N$ are complemented then the topology of $X$ is indeed the finest topological vector space topology on $X$ that makes $i_M$ and $i_N$ continuous. We denote by $\tau$ the topology on $X$. By definition of the induced topology, $i_M$ and $i_N$ are continuous when $X$ is endowed with $\tau$.
Let $\tilde{\tau}$ denote an other topology of topological vector space on $X$ that makes $i_M$ and $i_N$ continuous. By composition of continuous function, we see that the map
$$ \begin{array}{ccccc} \tilde{\phi}& : & M \times N & \to & (X,\tilde{\tau}) \\ & & (x,y) & \mapsto & x+y \end{array} $$ is continuous. Since $M$ and $N$ are complemented, so is $\tilde{\phi} \circ \phi^{-1} : (X,\tau) \to (X,\tilde{\tau})$. But this map is just the identity, and hence $\tau$ is finer than $\tilde{\tau}$.
Reciprocally, let us assume that $\tau$ is the finest topology if topological vector space on $X$ that makes $i_M$ and $i_N$ continuous. Let us denote by $\tau_{\textrm{prod}}$ the topology on $X$ that makes $\phi$ an isomorphism. This is just the product topology under the algebraic identification $X \simeq M \times N$. We want to prove that $\tau = \tau_{\textrm{prod}}$. Recall that by definition, $\tau_{\textrm{prod}}$ is the coarsest topology on $X$ that makes the projection $\pi_M, \pi_N$ according to the decomposition $X = M +N$ continuous. Notice that the map $\phi$ is always continuous, as a consequence of the definition of the induced topology and by continuity of $+$. This fact rewrites as $\tau \subseteq \tau_{\textrm{prod}}$.
In order to prove the converse inclusion, we only need to prove that $\tau$ makes $\pi_M$ and $\pi_N$ continuous. Noticing that $\pi_M \circ i_M = Id_{M}, \pi_M \circ i_N = 0, \pi_N \circ i_M = 0$ and $\pi_N \circ i_N = Id_{N}$ are continuous, we only need to prove the following fact :
Fact: If $E$ is a locally convex topological vector space and $L : X \to E$ is a linear map, then $L : (X,\tau) \to E$ is continuous if and only if $L \circ i_M$ and $L \circ i_N$ are continuous.
In order to prove this fact, notice that we only need to prove that if $U$ is a convex neighbourhood of $0$ in $E$ then $L^{-1}(U)$ is a neighbourhood of $0$ in $(X,\tau)$. Hence, the proof of the fact reduces to:
Fact': If $U$ is a convex subset of $X$ that contains $0$ and such that $i_M^{-1}(U)$ and $i_N^{-1}(U)$ are open, then $U \in \tau$.
To do so, define a filter $\mathcal{F}$ on $X$ by saying that $V \in \mathcal{F}$ if there is $U$ as in Fact' such that $U \subseteq V$. Then apply Theorem 3.1 p.21 in Topological Vector Spaces, Distributions and Kernels by F.Treves to see that $\mathcal{F}$ is the filter of neighbourhoods of $0$ for some topolgy $\tau'$ of topological vector space on $X$. In order to check the hypotheses of this theorem you shall use that $M$ and $N$ are locally convex (which is true since they are normed). The tricky hypothesis to check is (3.2). It is clear that $\tau'$ makes $i_M$ and $i_N$ continuous and hence $\tau' \subseteq \tau$, proving Fact' and ending the proof.
Remark 1: Concerning your second question, I think that the good context is the context of locally convex vector spaces (the proof above should still work).
Remark 2: If $X$ is assumed to be complete, it seems to me that it is equivalent to say that $M$ and $N$ are complemented and that they are closed (apply Open Mapping Theorem or Closed Graph Theorem), but I guess that the point was to give a definition that encompasses the case when $X$ is not complete.
Remark 3: There was a small ambiguity in your question. When you say that ``the topology on $X$ is the finest topology that makes $i_M$ and $i_N$ continuous'', there are two possible interpretations. What I think that you meant (and what I assumed in the proof) is that the topologies on $M$ and $N$ were fixed (as the induced topologies by the original topology on $X$), but one could argue that a priori the topologies on $M$ and $N$ still depend on the topology on $X$.