TT is continuous if and only if kerT\ker T is closed

Let X,Y be normed linear spaces. Let T:XY be linear. If X is finite dimensional, show
that T is continuous. If Y is finite dimensional, show that T is continuous if and only if kerT is closed.

I am able to show that X, finite dimensional T is bounded, hence continuous.

For the second part: This is what I have:

Suppose T is continuous. By definition kerT={xX:Tx=0} , and so kerT is the continuous inverse of a closed set. Hence kerT is closed.

First, is what I have attempted okay. How about the other direction?


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

Source : Link , Question Author : Jay , Answer Author : azarel

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