Discussion
Consider the following two spaces:
\(X = (C[-1,1], \|\cdot\|_{\infty})\), the space of all real-valued continuous functions defined on [-1, 1] equipped with the norm \(\|f\|_{\infty} = \sup_{t \in [-1,1]} |f(t)|\).
\(Y = (C[-1,1], \|\cdot\|_{2})\), the space of all real-valued continuous functions defined on [-1, 1] equipped with the norm \(\|f\|_{2} = \left(\int_{-1}^{1} |f(t)|^{2} dt\right)^{1/2}\).
Let W be the linear span over \(\mathbb{R}\) of all the Legendre polynomials. Then, which one of the following is correct?
Answer: C
W is dense in both X and Y.
The Legendre polynomials \(P_{0}, P_{1}, P_{2}, \ldots\) have degrees 0, 1, 2, ..., so their span W is the set of all polynomials. By the Weierstrass approximation theorem, polynomials are dense in \(C[-1, 1]\) under the sup norm, so W is dense in X.
For Y, note that \(\|f\|_{2} \le \sqrt{2}\|f\|_{\infty}\) on [-1, 1]. Any polynomial within \(\varepsilon\) of f in the sup norm is within \(\sqrt{2}\varepsilon\) in the 2-norm, so density in X carries over to Y. The weaker norm can only make it easier to approximate, never harder.
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