Is this proof correct? (left inverse and topologically complementary subsets)

Question

I want to prove the following theorem:

Theorem. Assume $T \in \mathcal L ( X, Y )$ is injective. The following statements are equivalent:

1. $T$ admits a left inverse;
2. Im($T$) is closed and admits a complement in $Y$.

Notation. $\mathcal L(X,Y)$ is the space of all continuous linear transformation from $X$ to $Y$. Im($T$) is the image (or the range) of $T$. Ker($T$) shall denote the kernel of $T$. When $M$ and $N$, closed linear subset of a Banach space $X$, are complentary we write $X = M \bigoplus N$.

It is a well known theorem (for example, is the theorem 2.13 of Brezis's book on functional analysis) and should be simple to prove, but all proofs I saw seem to be too concise, therefore some doubts hold on correctness of my try.

Let's recall a Lemma (for example, Lemma 4.47 in Rynne and Youngson)

Lemma. Suppose $X$ is a Banach space, $Y$ is a normed space and $T \in \mathcal L (X, Y)$. If there exists $\alpha > 0$ such that $\lVert Tx \rVert \geq \alpha \lVert x \rVert$ for all $x \in X$, then Im($T$) is closed.

and the characterization of topological complementarity:

Lemma. Assume $M$ is a closed linear subspace of a Banach space $X$. $M$ admits a complement in $X$ iff there exits $P \in \mathcal L(X)$ projection such that Im($P$) = $M$.

and finally (Lemma 5.61(b) in Rynne and Youngson)

Lemma. Suppose that $P$ is a projection on $X$. Then the subspaces Im($P$), Im($I-P$) are complementary.

So here my try.

Proof. (1) $\Rightarrow$ (2). Let $R\in \mathcal L(Y,X)$ the left inverse of $T$. So $RT = Id_X$ and $\lVert x \rVert = \lVert RT x \rVert \leq \lVert R \rVert \lVert Tx \rVert$. Hence $\lVert Tx \rVert \geq \alpha \lVert x \rVert$, $\alpha = \lVert R \rVert^{-1}$ (that is finite because $R$ is bounded) and Im($T$) is closed. Let $P = TR$; then $P$ is a projection and Im($P$) = Im($T$). Furthermore, Im($I - P$) is a complement of Im($P$) and Im($I - P$) = Ker($P$) = Ker($TR$) = Ker($R$) (because $T$ is injective). [Is it correct?]

(2) $\Rightarrow$ (1). Assume that Im($T$) is closed and admits a complement, $N$, in $Y$. Then exists $P \in \mathcal L(Y)$, $P^2 = P$ such that Im($P$) = Im($T$). Since $Py \in$ Im($T$), there exists a unique $x \in X$ such that $T x = Py$. Set $Ry = x$. So $RT = Id_X$ and $R$ is continuous by theorem of inverse operator.

The second part is the same as Brezis's one but the first part is different. He only says that it is an obvious fact that Im($T$) is closed. So my questions are:

• Why does he state that it is so obvious?
• Which is the quickest way to verify that an operator has closed image?
• Are there other possible complements?

Sorry for such a long question, but since notations are not universal in this subject I preferred to be explicit. Thank you.

References:

• H. Brezis, Functional analysis, Sobolev spaces and partial differential equations
• B. P. Rynne, M.A. Youngson, Linear functional analysis

Answer 1

After the typo fix, your proof is correct.

In the first part, you could already have stopped after "$P$ is a projection", but that would have gone against

"but all proofs I saw seem to be too concise".

As for

• Why he states the it is so obvious?
• Which is the quickest way to verify that an operator has closed image?

I can only make an educated guess for the first: experience. If you work a lot in one area, the important facts in that area tend to become obvious to you.

In the case at hand, the two quickest ways to see that $T$ has closed range are

1. what you did, find a dilation constant for $T$ from its left inverse.
2. the closed range theorem. Since $RT = \operatorname{id}_X$, we have $T^\ast\circ R^\ast = \operatorname{id}_{X^\ast}$, so $T^\ast$ has closed range (it's surjective), hence $T$ has closed range (closed range theorem).

Both ways can be applied in many similar situations, and when that is the case, they tend to be the easiest ways of showing the closedness of $\operatorname{Im} T$.

There are other possible complements?

If the question is whether $\operatorname{Im} T$ can have other complements than $\ker R$, the answer is "in general yes". In the finite dimensional case, every subspace of complementary dimension that has trivial intersection with $E$ is a complement of $E$, and that carries over partly to the infinite dimensional case. If $E$ has finite codimension, every subspace of dimension $\operatorname{codim} E$ that has trivial intersection with $E$ is a complement. Generally, if you have $X = E \oplus F$ with neither summand trivial, choose a $\xi \in F\setminus \{0\}$, and extend the linear form

$$\lambda_0 \colon E \oplus \mathbb{K}\xi \to \mathbb{K};\qquad \lambda_0(c\cdot\xi + e) = c$$

to a linear form $\lambda \in X^\ast$. Then, for $e \in E\setminus\{0\}$, the operator

$$T_e \colon x \mapsto x + \lambda(x)\cdot e$$

is an isomorphism of $X$ that leaves $E$ fixed, hence $T_e(F)$ is also a complement of $E$, and $T_e(F) \neq f$ since $\xi + e \in T_e(F)\setminus F$.

If $E = \{0\}$ or $E = X$, the complement of $E$ in $X$ is of course unique.