Can every closed subspace be realized as kernel of a bounded linear operator from a Banach space to itself? Let $X$ be a Banach space and $Y$ a closed subspace of $X$. 
Does there exist a bounded linear operator $T \colon X \to X$ such that $\ker T=Y$ ?
 A: Recently, there was written the following paper on this topic:
https://arxiv.org/abs/1811.02399
A: The answer is yes if $X$ is separable, but no in general. To cast the problem slightly differently, let us recall the fact that for any bounded linear operator $V:W\longrightarrow Z$ there is a factorisation $V=\tilde{V}Q$, where $\tilde{V}$ is one-to-one and $Q:W\longrightarrow W/\ker(V)$ is the quotient map (see, e.g., Theorem 1.7.13 of Megginson's book An Introduction to Banach Space Theory). It is therefore easy to see that OP's question is equivalent to the following question: Given a Banach space $X$ and $Y$ a closed subspace of $X$, does there exist a one-to-one bounded linear operator from $X/Y$ to $X$?
We first show the answer is affirmative when $X$ is separable. To this end let $X$ be a separable Banach space and $Y$ a closed subspace of $X$. To achieve the positive answer in the separable case we shall define one-to-one bounded linear operators $J_1$, $J_2$, $J_3$, $J_4$ and $J_5$ such that the composition $J_5J_4J_3J_2J_1$ is well-defined and a bounded linear operator from $X/Y$ to $X$. Since every separable Banach space is (isometrically) isomorphic to a subspace of $C[0,1]$ (see, e.g., Theorem 5.17 in the book Functional Analysis and Infinite Dimensional Geometry by Fabian et al), there exists a linear isometry $J_1$ from $X/Y$ into $C[0,1]$. Let $J_2: C[0,1]\longrightarrow L_2[0,1]$ be the formal identity map and let $J_3:L_2[0,1]\longrightarrow \ell_2$ be an isometric isomorphism (note that such an operator $J_3$ exists since $L_2[0,1]$ and $\ell_2$ are both infinite dimensional separable Hilbert spaces). Let $J_4$ be the operator from $\ell_2$ to $\ell_1$ that maps each element $(a_n)_{n=1}^\infty$ of $\ell_2$ to the element $(2^{-n}a_n)_{n=1}^\infty$ of $\ell_1$. Let $(x_n)_{n=1}^\infty$ be a bounded Schauder basic sequence in $X$ (see, e.g., Theorem 6.14 of Fabian et al) and let $J_5: \ell_1\longrightarrow X$ be the unique bounded linear operator from $\ell_1$ to $X$ satisfying $J_5((a_n)_{n=1}^\infty) = \sum_{n=1}^\infty a_nx_n$ for each $(a_n)_{n=1}^\infty\in\ell_1$, noting that $J_5$ is one-to-one by the defining property of Schauder basic sequences. As all of the operators $J_1$, $J_2$, $J_3$, $J_4$ and $J_5$ are bounded, linear and one-to-one, and since the composition $J_5J_4J_3J_2J_1$ is well-defined, the proof of the affirmative answer in the separable case is complete.
For a counterexample in the general case, let $\Gamma$ be an uncountable set and let $R:\ell_1(\Gamma)\longrightarrow c_0(\Gamma)$ be a surjective bounded linear operator (for the existence of such an operator $R$, see Exercise 5.46 in Fabian et al). Let $Q:\ell_1(\Gamma)\longrightarrow \ell_1(\Gamma)/\ker(R)$ denote the quotient map and let $\tilde{R}: \ell_1(\Gamma)/\ker(R)\longrightarrow c_0(\Gamma)$ be one-to-one and such that $R=\tilde{R}Q$ (see the reference in the first paragraph above). Let $\iota$ denote the formal identity map from $\ell_1(\Gamma)$ to $\ell_2(\Gamma)$, which is linear, one-to-one and bounded (with $\Vert\iota\Vert=1$).
Since $\tilde{R}$ is one-to-one and onto, the Open Mapping Theorem implies that $\tilde{R}^{-1}$ is a well-defined, linear, one-to-one and bounded operator from $c_0(\Gamma)$ onto $\ell_1(\Gamma)/\ker(R)$. If there were to exist a one-to-one bounded linear operator $S: \ell_1(\Gamma)/\ker(R)\longrightarrow \ell_1(\Gamma)$, then $\iota S\tilde{R}^{-1}$ would be a one-to-one bounded linear operator from $c_0(\Gamma)$ into the Hilbert space $\ell_2(\Gamma)$. However, as noted in this answer, an argument due to Olagunju shows that there is no one-to-one bounded linear operator from $c_0(\Gamma)$ into any Hilbert space, therefore no such operator $S$ can exist. In particular, we have that a counterexample exists when $X=\ell_1(\Gamma)$, for $\Gamma$ an uncountable set.
