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I've just looked at this answer from another question regarding the proof that "a hyperplane is closed $\Leftrightarrow$ $f$ is continuous".

I understand everything except this part:

Taking $y \in E \setminus \ker f$ , if $f$ were not bounded then we should have a sequence of unit vectors $x_n$ such that $f(x_n) \geq n$ for each $n \in \mathbb{N}$. But then it is $$ f(y)f(x_n)^{-1}x_n -y \to y $$ with $f(y)f(x_n)^{-1}x_n -y \in \ker f$

I don't really get how $f(y)f(x_n)^{-1}x_n -y$ converges to $y$. And even if it does, why is it in $\ker(f)$ if $y\in E\setminus \ker(f)$?

Does anyone know?

Quotenbanane
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2 Answers2

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Since $y$ is fixed, $x_n$ is a unit vector, and $f(x_n) \to \infty$, we have $f(y) f(x_n)^{-1} x_n \to 0$ as $n \to \infty$. The right-hand side should be $-y$, not $y$. It would make more sense to consider $y - f(y) f(x_n)^{-1} x_n$ instead.

It is in $\ker f$ since $f(f(y) f(x_n)^{-1} x_n - y) = f(y) f(x_n)^{-1} f(x_n) - f(y) = f(y) - f(y) = 0$.

Joshua P. Swanson
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I think there is a typo. It must be $$ f(y)f(x_n)^{-1}x_n-y\to -y. $$ Indeed, $$ \|f(y)f(x_n)^{-1}x_n\|\le |f(y)|/n\to 0\Rightarrow f(y)f(x_n)x_n-y\to -y $$ as $n\to\infty$. On the other hand, $$ f(f(y)f(x_n)^{-1}x_n-y)=f(y)-f(y)=0 $$ so $f(y)f(x_n)^{-1}x_n-y\in ker(f)$. But this is a contradiction, since the kernel is assumed to be closed. Observe that $y\notin Ker(f)$ is equivalent to $-y\notin Ker(f)$ since $Ker(f)$ is a subspace.

GReyes
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