$T:X\to Y$ a linear operator, $Y$ finite-dimensional, then $\ker T$ is closed iff $T$ is continuous

Question

I have to prove that if $T:X\to Y$ a linear operator on normed spaces, $Y$ finite-dimensional, then $\ker T$ is closed iff $T$ is continuous.

On a lot of places I see a proof that looks like this: it starts by assuming $Y=\mathbb{R}$ for simplicity. (I'm also aware of this that use a completely different construction). Anyway, I did not at all find it obvious that there was any kind of generalisation of the first linked proof to higher dimensions.

I will give my (quite long a tedious) generalisation here, but my question is: is there a more direct/easier generalisation of the construction given in the first linked proof? (I also wouldn't object to someone checking the actual correctness of this proof)

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We will take $Y=\mathbb{R}^n$.

Take $x_n\in X$ s.t. $||x_n||=1$, $||Tx_n||\to \infty$. w.l.o.g. we may assume that $\frac{Tx_n}{||Tx_n||}\to v_1\in \overline{B(0,1)}\subset \mathbb{R^n}$, by the compactness of the closed unit ball in finite dimensions. We now take $y_n=x_j$ for some $j$ s.t. $||\frac{Tx_n}{||Tx_n||}-v_1||<\frac{1}{n}$. Then we adjust the $y_n$ a little bit: if $||Ty_n||>n$ we set $y_n=\frac{y_n}{||Ty_n||}n$. We now obtain a sequence $y_n$ s.t. $$ \begin{align*} ||y_n||\leq 1\\ ||Ty_n||\leq n\\ ||Ty_n||\to \infty \end{align*} $$ In particular we have that $||Ty_n-||Ty_n||v_1||<1$ (this is the critical property that we needed).

We now take an orthonormal basis $\{v_i\}$ for $\mathcal{R}T$ containing $v_1$, and $v_i'$ s.t. $Tv_i'=v_i$. We conclude that for $j>1$ we must have $(v_j,Ty_n)<1$. We now define $z_n=y_n-\sum_{j>1}(v_j,Ty_n)v_j'$. Then $$||z_n||=||y_n-\sum_{j>1}(v_j,Ty_n)v_j'||\leq ||y_n||+\sum_{j>1}(v_j,Ty_n)||v_j'||\leq 1+\sum_{j>1}||v_j'||\leq C$$ Hence these $z_n$ are bounded. Furthermore we clearly still have that $$||Tz_n||\to\infty$$ because the norm of $T\sum_{j>1}(v_j,Ty_n)v_j'$ is bounded.

We can now finally define $$w_n=v'_1-\frac{z_n}{||Tz_n||}$$ Then $$Tw_n=v_1-\frac{1}{||Tz_n||}(Ty_n-\sum_{j>1}(v_j,Ty_n)v_j)=0$$ Since $Ty_n-\sum_{j>1}(v_j,Ty_n)v_j$ is by construction a vector pointing in the direction of $v_1$. But again $w_n\to v'_1$, which contradicts the closedness of the kernel.

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For the intuition: We take the unbounded sequence in $X$. The the project is onto the unit ball, which is compact, hence the sequence a convergent subsequence. We take the sequence, with limit $v_1$. Then we 'limit the growth' of our unbounded sequence to make sure that if $\frac{Tx}{||Tx||}$ is close to $v_1$, then $Tx$ is close to the line spanned by $v_1$. Then we remove any orthogonal components from the $Tx$, and because $Tx$ is now close to $<v_1>$ we can remove the orthogonal components by subtracting a vector of a certain bounded length. This finally gives a bounded sequence in $X$ with unbounded images in $Y$ all lying on the lines $<v_1>$, and from here the previous argument goes through.

Thanks for reading.

Answer 1

I think this is basically the same argument as you have written, maybe a bit cleaner. I don't think you can simplify it further too much without changing the nature of the argument (I like the topological one much more, myself).

Suppose we have a (real or complex, say) linear map $F\colon V\to W$ and $W$ is finite-dimensional, with both $V$ and $W$ being normed. Then we can assume without loss of generality that $W$ is an inner product space (since all finite-dimensional normed spaces of fixed dimension are isomorphic) and $F$ is onto.

Now, suppose towards contradiction that $v_n$ is a bounded sequence with $\lvert F(v_n)\rvert\to \infty$.

We may assume (by compactness) without loss of generality that $\frac{F(v_n)}{\lvert F(v_n)\rvert}\to w$ for some $w\in W$. Moreover, we can choose $z_1,z_2,\ldots, z_d\in V$ such that $F(z_j)$ form an orthonormal basis.

Let $z\in V$ be such that $F(z)=w$. Take $v_n'=z-\frac{v_n}{\lvert F(v_n)\rvert}$. Now, $v_n'\to z$ and $F(v_n')\to w-w=0$.

Then let $v_n''=v_n'-\sum_{j=1}^d \langle F(z_j),F(v_n')\rangle z_j$. It is easy to check that $v_n''\in \ker F$ and still $v_n''\to z\notin \ker F$. Thus, $\ker F$ is not closed.