Let $T:X \to Y$ linear, $(X,||\cdot||_X), (Y, ||\cdot||_Y)$ being banach spaces. Is it still wrong that $\ker T$ closed $\Rightarrow T$ is continuous? Supposing $(X,|| \cdot||_X)$ is not a banach space, I can think of $$id: (C([0,1]),|| \cdot||_1) \to (C([0,1]),|| \cdot||_{\infty})$$ for real valued functions on the compact intervall $[0,1]$ but supposing $X$ is a banach space I can't think of a counter example. Can anyone help me here? Thank you in advance.
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1You can always pass to the quotient $X/\mathrm{ker}T$, so you are asking whether an injective linear map between Banach spaces is automatically continuous. This is true if the target is finite-dimensional, but I doubt it is true in general. However, I cannot think of a counter-example, which means that I don’t know how to answer your question. Still, I hope this comment and the link can help you somewhat. – Giuseppe Negro Jan 05 '21 at 18:05
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1Oh, here it is: https://math.stackexchange.com/a/170098/8157 – Giuseppe Negro Jan 05 '21 at 18:05
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oh thank you for the link will look into it! – goalgetter666 Jan 05 '21 at 18:26
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Yes it does since it fits the prerequisites of the assertion. Ty for providing the link. – goalgetter666 Jan 05 '21 at 19:14