(a) Consider a unit-step function with a step at origin. That is, f(x)=0 in (-infinity,0) and f(x)=1 in [0,infinity). It's obvious that f(x) is in F(-infinity,infinity) but is not in C(-infinity,infinity) since it's discontinuous.
(b) Consider a simple absolute value function f(x)=|x|. It's continuous on R but is not differentiable at origin, then it's not in C1.
(c) By the hint, indefinitely integrate |x| from origin to t, we have g(t)=t^2/2 while t>=0 and g(t)=-t^2/2 while t<0. It's easy to check g(t) is in C1 but is not in C2 since g'(t)=|t|.
(d) Just let f(x)=x or any polynomial with arbitrary degree.
(e) It's trivial. (Hint: Any function in C1 is in C, so in F.)

哇哦，全英的，没有办法哦

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