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As already recalled, any kernel of a non-continuous linear form is a dense hyperplane, and non-continuous forms exist in infinite dimension by the existence of Hamel basis. That said, it's worth recalling a relevant fact in the affirmative direction, which is a corollary of the open mapping theorem:

A linear subspace in a Banach space, of finite codimension, and which is the image of a Banach space via a linear bounded operator is closed.

Btw, the property of being complemented has also a particular characterization for those subspaces that are images of operators: the image of $R:X\to Y$ is complemented if and only if $R$ is a right inverse, meaning that there is a bounded operator $L:Y\to X$ such that $LR=I$. A linear projector onto the subspace it is then $RL$.

As already recalled, a kernel of any non-continuous linear form is a dense hyperplane, and non-continuous forms exist in infinite dimension as a consequence of the existence of Hamel basis. That said, it's worth recalling a relevant fact in the affirmative direction, which is a corollary of the open mapping theorem:

A linear subspace in a Banach space, of finite codimension, and which is the image of a Banach space via a linear bounded operator, is closed.

Btw, the property of being complemented has also a particular characterization for those subspaces that are images of operators: the image of $R:X\to Y$ is complemented if and only if $R$ is a right inverse, meaning that there is a bounded operator $L:Y\to X$ such that $LR=I$. A linear projector onto the subspace it is then $RL$.

As already recalled, any kernel of a non-continuous linear form is a dense hyperplane, and non-continuous forms exist in infinite dimension by the existence of Hamel basis. That said, it's worth recalling a relevant fact in the affirmative direction, which is a corollary of the open mapping theorem:

A linear subspace in a Banach space, of finite codimension, and which is the image of a Banach space via a bounded operator is closed.

As already recalled, any kernel of a non-continuous linear form is a dense hyperplane, and non-continuous forms exist in infinite dimension by the existence of Hamel basis. That said, it's worth recalling a relevant fact in the affirmative direction, which is a corollary of the open mapping theorem:

A linear subspace in a Banach space, of finite codimension, and which is the image of a Banach space via a linear bounded operator is closed.

Btw, the property of being complemented has also a particular characterization for those subspaces that are images of operators: the image of $R:X\to Y$ is complemented if and only if $R$ is a right inverse, meaning that there is a bounded operator $L:Y\to X$ such that $LR=I$. A linear projector onto the subspace it is then $RL$.

As already recalled, any kernel of a non-continuous linear form is a dense hyperplane. That said, it's worth recalling a relevant fact in the affirmative direction, which is a corollary of the open mapping theorem:

A linear subspace in a Banach space, of finite codimension, and which is the image of a Banach space via a bounded operator is closed.

As already recalled, any kernel of a non-continuous linear form is a dense hyperplane, and non-continuous forms exist in infinite dimension by the existence of Hamel basis. That said, it's worth recalling a relevant fact in the affirmative direction, which is a corollary of the open mapping theorem:

A linear subspace in a Banach space, of finite codimension, and which is the image of a Banach space via a bounded operator is closed.