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I'm confused by the questions Discontinuous linear functional and Example of an unbounded operator which ask about unbounded linear functionals/operators on Banach spaces.

I don't understand how these can even exist.

Let $X$ be a Banach space. If $T$ is unbounded, then there exists a sequence $x_i \in X$ such that $\|x_i\|=1$ but $T(x_i)>i^3.$ Then we can let $x= \sum_i \frac{x_i}{i^2}.$ This is an element of $X$ by completeness but $T(x)$ is infinite. Hence $T$ isn't defined on $x.$

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user142700
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    Unbounded operators are not continuous. – Daniel Fischer Jul 15 '15 at 20:43
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    In case you didn't catch the point to Daniel's comment: When you say $Tx$ is infinite you're assuming that $T$ of that sum is the sum of $T$ of the individual terms. There's no reason that should be if $T$ is not continuous. – David C. Ullrich Jul 15 '15 at 20:46

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As the commenters said, your argument is flawed in the part where you conclude that (using notation $s_n=x_1+\dots+x_n$) $$s_n\to x \text{ and } Ts_n\to \infty \overset{?}{\implies} Tx=\infty $$

You did prove something, however: your argument shows that an unbounded operator must be discontinuous. (This isn't obvious; in fact, on an incomplete normed space one can have an unbounded continuous operator, such as $(x_n)\mapsto (nx_n)$ on the space of sequences that are eventually zero, equipped with any $\ell^p$ norm)

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    I think the last sentence is not true: A linear operator is bounded if and only if it is continuous. – gerw Feb 28 '18 at 07:55