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Let $A,B$ be Banach spaces and let $T:A \to B$ be a surjective, bounded, linear operator. Let $A_1$ be a non-empty subset of $A$, then:

$T(A_1)$ is closed if and only if $A_1+ \textrm{ker}(T)$ is closed.

I have shown that if $A_1+ \textrm{ker} (T)$ is closed, then $T(A_1)$ is closed, but am unsure of how to proceed in the other direction. Note that we know that $T(A \setminus [A_1 + \textrm{ker}(T)]) = B \setminus T(A_1)$. Figured the reverse direction could be proved using some modification of the Open Mapping Theorem or Closed Graph theorem, but not sure how to tackle this. Thanks.

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Assume that $T(A_1)$ is closed. Let $\{x_n\} \subset A_1$, $\{y_n\} \subset ker(T)$ and $x_n +y_n \to z$. Since T is continuous we get $Tx_n =Tx_n +Ty_n \to Tz$. Thus $T(z)$ is the limit of the sequence $\{Tx_n\}$ which lies in $T(A_1)$. It follows that $Tz=Tw$ for some $w \in A_1$. But then $z=w+(z-w) \in A_1 +ker (T)$. This proves that $A_1 +ker (T)$ is closed.

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Let

$A_1' = T^{-1}(T(A_1)) = \{ y \in A \mid T(y) \in T(A_1) \} ; \tag 1$

clearly,

$A_1 \subset A_1', \tag 2$

now if $w \in A_1$ and $u \in \ker T$, then

$T(w + u) = T(w) + T(u) = T(w) + 0 = T(w) \in T(A_1), \tag 3$

or

$w + u \in A_1', \tag 4$

which shows that in fact

$A_1 + \ker T \subset A_1'; \tag 5$

and if $z \in A_1'$ then $T(z) \in T(A_1)$, and there is some $x \in A_1$ such that $T(z) = T(x)$, whence

$T(z - x) = T(z) - T(x) = 0; \tag 6$

thus

$z - x \in \ker T, \tag 7$

or

$z \in x + \ker T \subset A_1 + \ker T, \tag 8$

whence

$A_1' \subset A_1 + \ker T; \tag 9$

(5) and (9) together yield

$A_1' = A_1 + \ker T; \tag{10}$

we now conclude by observing that since

$A_1'= T^{-1}(T(A_1)) \tag{11}$

(as (1) asserts), and $T(A_1)$ is closed, the continuity of $T$ implies $T^{-1}(T(A_1))$ is closed as well; thus (10) and (11) together yield $A_1 + \ker T$ is closed.

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