Linear projections and topological direct sum in normed/Banach spaces

Question

When learning about the concept of complemented subspaces of a Banach space, I'm curious with the following question:

Let $X$ be a vector space over $\mathbb{C}$ or $\mathbb{R}$ and let $E$ and $F$ be algebraic complemented subspaces with trivial intersection, such that $X=\ E\oplus F$. If defining a norm $\|\cdot\|$ on $X$, and equipping $E$ and $F$ with their norm-induced subspace topologies, respectively, is it always true that, as topological spaces, $X\simeq E\times F$?

A closely related proposition in terms of linear projections can be stated as follows.

Let $(X, \|\cdot\|)$ be a normed space over $\mathbb{C}$ or $\mathbb{R}$, and $P: X\to X$ a linear projection such that $P^2=P$. Then, do we always have a topological direct sum result $X\simeq \text{ran}(P) \oplus \text{Ker}(P)$, where the range and kernel are equipped with their subspace topologies, respectively?

Here I'm not assuming continuity of $P$ or whether the range or kernel are closed subspaces.

Background:

I'm trying to understand, on Banach spaces, why there can be discontinuous projections with a closed range.

In a Banach space $X$, if a subspace $E\subset X$ is closed, then its quotient space $X/E$ is a Banach space (complete) under the quotient norm. Purely algebraically, we would then have $X \simeq E \oplus (X/E)$. But, for a subspace $F\subset X$ algebraically complemented to $E$, we would also have algebraic direct sum result $X=E\oplus F$. This seems to suggest, the natural algebraic isomorphism from $F$ to $(X/E)$ should also be a homeomorphism. But, since not every closed subspace $E$ is (alg + top) complemented, that shouldn't be true.

My question is to help me understand this confusion. If the propositions in my question are true, then I can understand, even when $\text{ran}(P)$ is closed, the algebraic isomorphism from $X/\text{ran}(P)$ to $\text{Ker}(P)$ can fail to be a homeomorphism. So the kernel can still be non-closed.

I already know the following two facts/propositions about linear projections on Banach spaces.

If a projection $P$ is continuous, then $\text{ran}(P)$ is closed, since it is the kernel of $I-P$.

If a projeciton $P$ has closed range and closed kernel, it is continuous. This can be proved since everything is Banach (complete) and so we can freely apply the Open Mapping Thm/Inverse Mapping Thm wherever possible. See here.

Thanks in advance.

Update:

The conclusive answer should be the following.

Let $X$ be a normed vector space and $E\subset X$ a closed subspace, with its purely algebraic complemented subspace $F\subset X$ such that $X=E\oplus F$. Then, the original norm topology $\tau_X$ of the whole space, the product topology of the subspaces $\tau|_E \times \tau|_F$, and the product topology with the quotient space $\tau|_E \times \tau_{X/E}$, are in general all different.

Answer 1

If $X=E\times F$, let $p:X\to E$ be the projection $p:(x,y)\longmapsto x$. Let $V\subset E$ be open. Then $$ p^{-1}(V)=V\times F. $$ Because the topology is the product topology, $V\times F$ is open. Hence $p$ is continuous.

This means that if your projection $P$ is not continuous, the direct sum $\operatorname{ran}P\oplus\ker P$ is not topological.

It is easy to produce unbounded projections. Indeed, given any infinite-dimensional Banach space $X$, let $g$ be an unbounded linear functional. Fix nonzero $z\in X$ and define $$ Px=g(x)z. $$ Then $P$ is an unbounded rank-one projection, and $\ker P=\ker g$ is dense.

Answer 2

Based on Martin's answer, I can conclude the original proposition in my question is false. Before showing that, let me list a few basic lemmata to make it clearer.

Lemma 1. A net (sequence) $(x_i)_{i\in I}$ converges in the product space $X=\prod_{\alpha\in A}X_\alpha$ iff the component nets $\left(\pi_{\alpha}(x_i)\right)_{i\in I}$ converges in $X_\alpha$ for all $\alpha\in A$, where $\pi_\alpha: X\to X_\alpha$ are the set-theoretic projections. (A proof can be found here.)

Lemma 2. Let $(X, \|\cdot\|)$ be a normed vector space over $\mathbb{R}$ or $\mathbb{C}$ and $E\subset X$ a subspace. Then the subspace topology of $E$ is induced by the restricted norm on $E$, defined as $\|\cdot\|\Big|_E: e\mapsto \|\iota(e)\|$, where $\iota: E\to X$ is the inclusion map.

Lemma 3. Let $(X, \|\cdot\|_X)$ and $(Y, \|\cdot\|_Y)$ be normed vector spaces over $\mathbb{R}$ or $\mathbb{C}$. Then the product topology on $X\times Y$ is induced by the norm $|||\cdot|||:=\|\cdot\|_X+\|\cdot\|_Y$. (Can be proved using Lemma 1.)

Now we can state and prove the corrected proposition.

Let $(X, \|\cdot\|)$ be a normed space with algebraic complemented subspaces $E,F\subset X$ such that $X=E\oplus F$. Denote $\tau_{\|\cdot\|}$ the norm topology on $X$ and equip $E,F$ with their subspace topologies inherited from $\tau_{\|\cdot\|}$, respectively. Also, let $\tau_\prod$ be the product topology on $X$. Then $\tau_{\|\cdot\|}\subset \tau_\prod$.

Proof. By Lemma 2, the subspace topologies of $E$ and $F$ are induced by the restricted norms $\|\cdot\|\Big|_E$ and $\|\cdot\|\Big|_F$, respectively. Therefore, by Lemma 3, the product topology $\tau_\prod$ on $X$ is induced by the norm $|||\cdot|||:= \|\cdot\|\Big|_E + \|\cdot\|\Big|_F$. Note $\|\cdot\|\leq \|\cdot\|\Big|_E + \|\cdot\|\Big|_F$. Hence, the identity map, $\text{Id}: (X, \tau_\prod) \to (X, \tau_{\|\cdot\|})$ is a bounded operator between normed spaces. Hence $\tau_{\|\cdot\|}\subset \tau_\prod$. ${ \rm \square}$

As Martin has pointed out in his answer, however, we cannot say anything about whether it is true $\tau_\prod \subset \tau_{\|\cdot\|}$ without applying further conditions.

Especially, the following is true!!!

If a normed space $(X, \|\cdot\|)$ is Banach and the algebraic complemented subspaces $E$ and $F$ are both closed, then $\tau_{\|\cdot\|}= \tau_\prod$.

Proof. Since the identity map from $(X, |||\cdot|||)$ to $(X, \|\cdot\|)$ is bounded and bijective, by Open Mapping Thm, its inverse is also bounded. $\square$