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 \[ B(x_0,\epsilon )\equiv \{x\in X\mid \lVert x-x_0\rVert <\epsilon \} \] 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 \(f(y)<0\).
Any \(x\in B(x_0,\epsilon )\) can be written as \(x=x_0+\epsilon u\) for some \(u\) with \(\lVert u\rVert <1\). Use the result of part (b) to show
that \(\lvert f(u)\rvert <1/\epsilon \).
To conclude, give an upper bound for \(\lVert f\rVert \).