Let $T$ be a linear functional on a Banach space. If $T^{-1}(1)$ is closed, is $T$ bounded?

Question

Let $T$ be a linear functional on a Banach space. If $T^{-1}(1)$ is closed, is $T$ bounded?

This is a problem from a qualifying exam. I am not sure how to prove it. Reminds me of the closed graph theorem but i am not sure how to apply it here.

Answer 1

If $T=0$, we are done. Else, fix $x\in T^{-1}(1)$ and notice that $\ker(T)=(-x)+T^{-1}(1)$ is also closed. Now, every linear functional with closed kernel is bounded.

Answer 2

Prove the contrapositive: suppose that $T$ is unbounded. Then for each $n$, we can find $x_n$ with $\lVert x_n \rVert = 1$ and $| T(x_n) | > n$. Now $y_n := \frac{x_n}{T(x_n)}$ converges to 0, and each $y_n \in T^{-1}(1)$, so $T^{-1}(1)$ is not closed.