Discussion
Let C[0,1] denote the set of all real valued continuous functions defined on [0,1] and \(\|f\|_{\infty} = \sup\{|f(x)| : x \in [0,1]\}\) for all \(f \in C[0,1]\). Let
\[ X = \{f \in C[0,1] : f(0) = f(1) = 0\}. \]
Define \(F : (C[0,1], \|\cdot\|_{\infty}) \to \mathbb{R}\) by \(F(f) = \int_{0}^{1}f(t)\,dt\) for all \(f \in C[0,1]\).
Denote \(S_{X} = \{f \in X : \|f\|_{\infty} = 1\}\).
Then the set \(\{f \in X : F(f) = \|F\|\} \cap S_{X}\) has
Answer: A
The set has NO element.
F is integration over [0,1], and \(|F(f)| \le \|f\|_{\infty}\), with equality for the constant function 1, so \(\|F\| = 1\). Functions in X can come arbitrarily close to this value: a function equal to 1 except on short ramps near 0 and 1 has integral close to 1.
But no \(f \in S_{X}\) attains it. If \(\|f\|_{\infty} = 1\) and \(\int_{0}^{1}f = 1\), then \(\int_{0}^{1}(1-f) = 0\) with \(1 - f \ge 0\) continuous, forcing f = 1 everywhere, which contradicts f(0) = 0. So the set is empty.
This is the standard example of a functional on the closed subspace X that does not attain its norm.
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