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Math 6710 • Applied Linear Operators and Spectral Methods

Fall 2022 • Question 2

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.

  1. Let \(x_0 \in X\) be such that \(f(x_0) = 1\). Explain why there is an \(\epsilon > 0\) for which the ball

    \[B(x_0, \epsilon ) \equiv \{x \in X \mid \|x - x_0\| < \epsilon \}\]

    satisfies \(B(x_0, \epsilon ) \subset X - \mathcal {N}(f)\).

  2. 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 \(f(y) \le 0\).
  3. 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 \(|f(u)| < 1/\epsilon \).
  4. To conclude, give an upper bound for \(\|f\|\).

Raw LaTeX

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.
\begin{enumerate}

\item
Let \(x_0 \in X\) be such that \(f(x_0) = 1\). Explain why there is an \(\epsilon > 0\) for which the
ball

\[B(x_0, \epsilon) \equiv \{x \in X \mid \|x - x_0\| < \epsilon\}\]

satisfies \(B(x_0, \epsilon) \subset X - \mathcal{N}(f)\).



\item
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 \(f(y) \le 0\).


\item
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 \(|f(u)| < 1/\epsilon\).


\item
To conclude, give an upper bound for \(\|f\|\).

\end{enumerate}