$T$ is bounded iff $T^{-1}(\{0\})$ is closed [duplicate]

If $T$ is the null-functional, it's obviously continuous, so suppose it's not. We can find $z \in X$ with $T(z)=1$ (standard scaling).

Now if $T$ we're unbounded we'd have $v_n$ with $\|v_n\|\le 1$ and $|T(v_n)| > n$, say.

Then define $w_n = z - \frac{1}{T(v_n)} \cdot v_n$.

Then $T(w_n)=0$ by linearity and so $w_n \in T^{-1}[\{0\}]$ for all $n$. But $w_n \to z$ and $z \notin T^{-1}[\{0\}]$, contradicting the closedness of that set. So $T$ is not unbounded.

Hint: if $T$ is not bounded, for any $n$ there is $x_n$ with $\|x_n\| \le 1$ and $|T x_n| > n$. Show that $T^{-1}(\{0\})$ is dense: for any $u \in X$ and $\varepsilon > 0$, find $u + c x_n$ for some $n$ and $c$ with $T(u + c x_n) = 0$ and $\|c x_n\| < \varepsilon$.