Reading:
Read sections 32, 33.
Exercises:
Consider the function \(f(x) = x^2 \sin (\frac 1x)\) when \(x \ne 0\) and \(f(0) = 0\) which has domain \(\mathbb R\) and is differentiable.
Calculate the derivative \(f’(x)\) for all \(x \in \mathbb R\).
Use your answer to part (a) to explain why \(f’(x)\) is discontinuous.
We can calculate that \(f’(-\frac 2\pi) = \frac 4\pi\) and \(f’(\frac 1{2\pi}) = -1\).
Which theorem guarantees that there exists a number \(c \in (-\frac 2\pi, \frac 1{2\pi})\) for which \(f’(c) = 0\)?
For \(x \ne 0\) we may use the product and chain rules to find that \(f’(x) = 2x\sin(\frac 1x) - \cos(\frac 1x)\). When \(x = 0\) we turn to the definition of the limit and compute
\[ f’(0) = \lim_{h\to0}\frac{f(h) - f(0)}{h} = \lim_{h\to0}\frac{h^2 \sin(\frac 1h)}{h} = 0. \]
Using our answer to part (a) we realize that while \(f’(0)\) exists, \(\lim_{x \to 0}f’(x)\) does not (it oscillates too rapidly). We thus conclude \(f’(x)\) is not continuous at \(0\) and hence is discontinuous.
The Intermediate Value Theorem for Derivatives still applies, even though the IVT for continuous functions does not (as \(f’\) is not continuous on \([-\frac 2\pi, \frac 1{2\pi}]\)).
Recall that a function \(f\) is Lipschitz continuous on a set \(S\) provided there exists \(K \ge 0\) so that for all \(x, y \in S\) with \(x \ne y\):
\[ \left\vert\frac {f(x)-f(y)}{x-y}\right\vert \le K.\]
Let \(f\) be a function which is differentiable on a closed interval \([a,b]\) and for which \(f’\) is continuous on \([a,b]\). Prove that \(f\) is Lipschitz continuous on \([a,b]\).
Hint: Use the Mean Value Theorem.
For any \(x, y \in [a,b]\) with \(x<y\), by the Mean Value Theorem, there exists \(c \in (x,y) \subseteq [a,b]\) so that
\[ \frac{f(x)-f(y)}{x-y} = f’(c). \]
As \(f’\) is continuous on the closed, bounded interval \([a,b]\) so too is \(\vert f’\vert \), so the set \(\left\{ \vert f’(c) \vert \;:\; c \in [a,b] \right\}\) is bounded above by some non-negative real number \(K \ge 0\).
Let \(x,y \in [a,b]\) with \(x < y\). Then there exists \(c \in [a,b]\) so that
\[ \left\vert \frac{f(x)-f(y)}{x-y} \right\vert = \vert f’(c) \vert \le K, \]
Notice that if on the other hand \(x,y \in [a,b]\) and \(x > y\) then we still have that
\[ \left\vert \frac{f(x)-f(y)}{x-y} \right\vert = \left\vert \frac{f(y)-f(x)}{y-x} \right\vert \le K. \]
So \(f\) is Lipschitz continuous with constant \(K\).
Show that \(ex \le e^x\) for all \(x \in \mathbb R\).
Notice that \(e x \le e^x\) is true if and only if \(0 \le e^x - ex\). Let \(g(x) = e^x - ex\). Then \(g(1) = 0\). Notice that \(g'(x) = e^x - e\). When \(x < 1\), \(g'(x) < 0\) and when \(x > 1\), \(g'(x) > 0\).
Let \(x < 1\). Then \(g(x) \ge 0\), as it must be decreasing always until \(x = 1\), when it is its minimum value of \(0\). On the other hand, when \(x > 1\), the function is always increasing from its minimum value of \(0\), so \(g(x) \ge 0\) as well. So in all cases \(g(x) \ge 0\).
Let \(f\) be real-valued and differentiable on \(\mathbb R\), and let \(g(x) = f(x+1) - f(x)\). Suppose additionally that \(\lim_{x\to \infty} f'(x) = 0\).
Prove that \(\lim_{x\to\infty} g(x) = 0\).
Hint: Use the Mean Value Theorem.
Let \(\epsilon > 0\), so there exists \(M\) so that for all \(c > M\) we have that \(|f'(c)| < \epsilon\). Let \(x > M\), and consider the interval \([x, x+1]\). By the Mean Value Theorem, there exists an input \(c\) between \(x\) and \(x+1\) for which \(f'(c) = (f(x+1) - f(x))/1 = g(x)\). But then \(|g(x)| = |f'(c)| < \epsilon\) as \(c > x > M\). This holds for all \(x > M\) and all \(\epsilon > 0\), so \(\lim_{x \to \infty} g(x) = 0\).