Let \(X\) be a real Banach space. Assume \(f \in X^*\) has a closed nullspace \(\mathcal {N}(f)\). The goal of this
problem is to show that \(f\) must be a bounded linear functional.
Let \(x_0 \in X\) be such that \(f(x_0) = 1\). Explain why there is an \(\epsilon > 0\) for which the ball
satisfies \(B(x_0, \epsilon ) \subset X - \mathcal {N}(f)\).
Prove that \(f(x) > 0\) for all \(x \in B(x_0, \epsilon )\), where \(\epsilon \) is as in part (a). To prove this, you may assume
for contradiction that there is some \(y \in B(x_0, \epsilon )\) with \(f(y) \le 0\).
Any \(x \in B(x_0, \epsilon )\) can be written as \(x = x_0 + \epsilon u\) for some \(u\) with \(\|u\| < 1\). Use the result of part (b) to show
that \(|f(u)| < 1/\epsilon \).