Normed and Banach Spaces Questions and Answers

Answer: C

G is a closed subset of C[0, 1].

Suppose constant functions \(c_{n}\) converge uniformly to f. Then \(c_{n} = c_{n}(0)\) is a convergent sequence of numbers with limit \(c = f(0)\), and \(|f(x) - c| = \lim |c_{n}(x) - c_{n}| = 0\) for each x, so f is the constant c. The limit stays in G, so G is closed.

G is not bounded, since the constants n have sup norm n. It is not dense: the function \(x\) is at distance at least \(\frac{1}{2}\) from every constant. And it is not compact, since it is unbounded.

Answer: C

f is coercive if f(x) → +∞ as ||x|| → ∞.

Coercivity means the function grows without bound as one moves far out in the space. It is the condition that, together with lower semicontinuity and convexity, guarantees that a functional on a reflexive Banach space attains its minimum, because minimising sequences cannot escape to infinity.

Growth as \(\|x\| \to 0\) says nothing about behaviour at infinity. Tending to 0 or to \(-\infty\) at infinity is the opposite of coercive and would allow minimising sequences to run away.

Answer: B

M is an open subset of C[0, 1].

The evaluation map \(f \mapsto f\left(\frac{1}{2}\right)\) is continuous for the sup norm, since \(|f(\tfrac{1}{2}) - g(\tfrac{1}{2})| \le \|f - g\|_{\infty}\). M is the inverse image of the open set \(\mathbf{R} \setminus \{0\}\), so it is open. Concretely, if \(f\left(\frac{1}{2}\right) = c \neq 0\), every g within sup distance \(|c|/2\) of f also has \(g\left(\frac{1}{2}\right) \neq 0\).

M is not bounded (it contains every constant function n), not closed (the constants \(\frac{1}{n}\) lie in M but converge to 0, which does not), and so not compact.