- Main Idea: Exact same idea as 1, just a more complicated check.
Take \(H(x, t) = P(tf(x) + (1-t)g(x))\). This is well defined; the only case to check is when the denominator is zero. But \(\norm{x} = 0\) iff \(x =0\), which would imply \(tf(x) +(1-t)g(x) = 0\) and so \(tf(x) = -(1-t)g(x)\).
Taking norms and observing that since \(f,g \in S^n \implies \norm{f} = \norm{g} = 1\), this forces \(t = 1-t\) and thus \(t=1/2\). But this would force \((1/2)f(x) = (-1/2)g(x)\) and thus \(f(x) = -g(x)\), which we assumed was not the case.
Solution
1. Let \(P(v)=v/\|v\|\) for \(v\ne0\) and define\[H(x,t)=P(tf(x)+(1-t)g(x)).\]
We need only verify that the vector being normalized is never zero.
2. If \(tf(x)+(1-t)g(x)=0\), taking norms gives \(t=1-t\), hence \(t=1/2\).
Both \(f(x)\) and \(g(x)\) have norm \(1\), so \(t\|f(x)\|=(1-t)\|g(x)\|\).
3. At \(t=1/2\), the equality would imply \(f(x)=-g(x)\), contrary to hypothesis.
Multiply the zero equation by \(2\).
4. Therefore \(H\) is well-defined and continuous, with \(H(x,0)=g(x)\) and \(H(x,1)=f(x)\). Thus \(f\simeq g\).
At the endpoints normalization does nothing because \(f(x),g(x)\in S^n\).