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.
I know that for a Banach space $X$ and a linear functional $T:X\rightarrow\mathbb{R}$ in its dual $X'$ the following holds: \begin{align}T \text{ is continuous } \iff \text{Ker }T \text{ is closed}\end{align} which probably holds for general operators $T:X\rightarrow Y$ with finite-dimensional Banach space $Y$. I think the argument doesn't work for infinite-dimensional Banach spaces $Y$. Is the statement still correct? I.e. continuity of course still implies the closedness of the kernel for general Banach spaces $X,Y$ but is the converse still true?
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