$V$ is finite dimensional if and only if all linear operators $T\colon V \rightarrow \mathbb{K}$ are continuous.

Question

Where $V$ is a a normed $\mathbb{K}$-vector space ($\mathbb{K}$ = $\mathbb{C}$ or $\mathbb{R}$ ). I have showed right implication. I believe I also showed ($\Leftarrow$) but it seems too easy, so I think I might be missing something. Here is the reasoning.

We assume $V$ is infinite dimensional, then we can easily construct a linear and non-bounded operator, and hence non continuous. Let $B = \{ e_1, e_2,\dots, \}$ be a basis of $V$. Let $z \in V$, so we have $z = \sum_i \alpha_i e_i$. We define

\begin{equation} T(z) = \sum_i \alpha_i \end{equation}

Then $T$ is linear, but for say $z = \sum_i e_i$, we have $\lVert z\rVert_V < \infty$ and $\lVert T(z)\rVert = \infty$

Answer 1

Actually, your statement is the opposite of the correct one: $V$ is finite-dimensional if and only if every linear map from $V$ into $\mathbb K$ is continuous.

Your proof is wrong. How can you possibly $\|T(z)\|=\infty$? The norm of a vector is always a non-negative real number and $\infty$ is not a number.

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Your proof is wrong for another reason: there is no such thing as $\sum_ie_i$. This sum diverges in general.