91Ó°ÊÓ

(Begin by considering any scalar \(a\), the function \(g\), and a vector \(v\) not in \(W\). Then, one needs to show the existence of the function \(f\), given by \(f(v) = a\) and \(f(x) = g(x)\) for all \(x \in W\). Firstly, extend \(W\) to a larger subspace \(U \subset V\) which includes \(v\) as well. This can be done by taking the set of linear combinations of vectors in \(W\) and \(v\), formally \(U = \{w + \lambda v \mid w \in W, \lambda \in F\}\), where \(F\) is the underlying field of the vector space. Now define a function \(f: U \rightarrow F\) by \(f(w + \lambda v) = g(w) + \lambda a\). This function is well-defined, and it's easy to see that it satisfies \(f(v) = a\) and \(f(x) = g(x)\) for \(x \in W\). But, we need \(f\) to be defined on \(V\), not just \(U\). For this, use the fact that any function defined on a subspace can be extended to the whole space (as per Exercise 35 of Section 2.1). So, find such an extension \(f': V \rightarrow F\). Now, this function satisfies \(f'(v) = a\) and \(f'(x) = g(x)\) for \(x \in W\), hence, it's the required function. This completes the proof for (a)).

(To prove the existence of a non-zero linear functional \(f \in V^{*}\) such that \(f(x)=0\) for all \(x \in W\), we can apply part (a) of the problem. Choose \(v \in V\) but \(v \notin W\) and let \(a=1\). By part (a), we have a functional \(f\) such that \(f(v) = 1\) and \(f(x) = 0\) for \(x \in W\). This function \(f\) is nonzero (since \(f(v) = 1\)), and \(f(x) = 0\) for all \(x \in W\), as required. This completes the proof for (b).