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Let $T : X \rightarrow Y$ be a map between two normed spaces, where $Y$ is finite dimensional. Let $U$ be the kernel of $T$. Prove that $T$ is bounded if $U$ is closed.

I saw here the answer:

If $\ker(T)$ is closed then $X/\ker(T)$ is a normed vector space. Observe that the map $\overline{T}:X/\ker(T)\to Y$ given by $\overline{T}(x+\ker(T))=T(x)$ is a well-defined linear map by the first part $\overline T$ is continuous since $X/\ker(T)$ is a finite dimensional vector space (as it is isomorphic to a subspace of $Y$). Let $\pi: X \to X/\ker(T)$ denote the quotient map. Note that $T=\overline{T}\circ \pi$ hence $T$ is continuous since it is a composition of continuous functions.

I have a couple questions.

1) Is it possible to show this if $T$ is not linear? (My problem doesn't explicitly give this linearity; however, I kind of think that it is just assumed because we are talking about linear operators.)

2) I don't understand why $X/ker(T)$ is finite.

Thank you very much for any help you can provide.

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if $T$ is not linear the result is false even when $X=Y=\mathbb R$. (try to construct a function $f:\mathbb R \to \mathbb R$ which vanishes only at $0$ but not bounded on $[-1,1]$. Since there is no continuity assumption this is very easy.

For 2) define $S:X|ker(T) \to T(X)$ by $S(x+Ker(T))=Tx$. Verify that this linear map is a bijection. Hence the dimension of $X|ker(T)$ is equal to that of $T(X)$, hence less than or equal to the dimension of $Y$.

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  • $\begingroup$ Thank you very much. I appreciate the help. :) $\endgroup$
    – MathIsHard
    Nov 7 '18 at 23:58

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