(28.8) Let when is rational and otherwise.
Show that is continuous at .
Show that is discontinuous at all .
Prove that is differentiable at . (It is insufficient to simply claim )
Proof. Let \((x_n)\) be any sequence that converges to 0. Let \(\epsilon > 0\) be arbitrary. Convergence of \((x_n)\) to 0 means that there exists \(N\) so that for all natural numbers \(n > N\), \(\vert x_n \vert < \sqrt \epsilon \) Then \(\vert f(x_n) \vert \le x_n^2 < \epsilon \). So \((f(x_n))\) converges to \(f(0) = 0\), and so \(f\) is continuous at 0.
Proof. First, suppose is rational. Let be a sequence of irrationals converging to , which must exist by density of irrationals (or, if you like, by example, we have ). Then for all , so converges to 0, but .
On the other hand, if is irrational, let be a sequence of rationals converging to (if you like, is the sequence of decimal approximations to ), then but so converges to . So is not continuous at any .
Proof. The definition of the derivative of at 0 is . Let . Notice that for , we have that , proving that .
(28.14) Suppose is differentiable at .
Prove that .
Prove that .
Proof. Let , so that there exists for which implies that . But with , this means that implies that . This holds for all , so .
Proof. Notice that (a) actually also proves that . We have that
(29.5) Let be a real-valued function which is differentiable on . Prove that if for all , then is a constant function.
Proof. If , then for all , . But taking limits as of both sides, we realize that . But this means that is constant.
(29.9) Show that for all .
Proof. Notice that is true if and only if . Let . Then . Notice that . When , and when , .
Let . Then , as it must be decreasing always until , when it is its minimum value of . On the other hand, when , the function is always increasing from its minimum value of , so as well. So in all cases .
Let be real-valued and differentiable on , and let . Suppose additionally that .
Prove that .
Hint. Use the Mean Value Theorem.
Proof. Let , so there exists so that for all we have that . Let , and consider the interval . By the Mean Value Theorem, there exists an input between and for which . But then as . This holds for all and all , so .