Discussion
Let \(V = C^{1}[0,1]\), \(X = (C[0,1], \|\ \|_{\infty})\) and \(Y = (C[0,1], \|\ \|_{2})\). Then \(V\) is
Answer: C
\(V\) is dense in both \(X\) and \(Y\).
By the Weierstrass approximation theorem, polynomials are dense in \(C[0,1]\) under the supremum norm. Polynomials are continuously differentiable, so \(C^{1}[0,1]\), which contains them, is dense in \(X\).
For \(Y\), note that \(\|f\|_{2} = \left(\int_{0}^{1} |f|^{2}\right)^{1/2} \le \|f\|_{\infty}\). So any approximation that is close in the supremum norm is at least as close in the \(L^{2}\) norm, and \(V\) is dense in \(Y\) as well.
That rules out every option that denies density in either space.
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