Let us write \(P(t)=\sum_{k=k_0}^{d} a_{k}t^{k} \) where \(a_k\in\R\), \(a_{k_0}\neq 0\), and \(k_0\) is the smallest degree such that \(P\) contains a monomial of that degree

We have that

\begin{equation*} \sum_{k=k_0}^{d} a_{k}t^{k}{t^d}= \Big(a_{k_0}t^{k_0-d} \big(1+\sum_{k=k_0}^{d} a_{k}/a_{k_0}t^{k-k_0}\big)\Big) \end{equation*}

The second factor convegres to \(1\) as \(t\to 0^+\). The term \(a_{k_0}t^{k_0-d}\) either goes to \(a_{k_0}\) if \(k_0=d\) or is unbounded if \(k_0\lt d\) (we always have \(k_0\leq d\)). thus we always will have that the expression above eigher converges to \(a_{k_0}\) or diverges, contradicting our assumption.

Take two Taylor polynomials \(T_1\) and \(T_2\) and write out the definitions

\begin{equation*} \begin{aligned}[t] & f(x_0+t)=T_1(t) + |t|^dr_{x_0,1}(t) \\ & f(x_0+t)=T_2(t) + |t|^dr_{x_0,2}(t) \end{aligned} \end{equation*}

Subtract the above two identities and obtain

\begin{equation*} T_1(t)-T_2(t)= - |t|^d\big(r_{x_0,1}(t) -r_{x_0,2}(t)\big) \end{equation*}

thus

\begin{equation*} \lim_{t\to 0}\frac{T_1(t)-T_2(t)}{|t|^d }= - \lim_{t\to 0} \big(r_{x_0,1}(t) -r_{x_0,2}(t)\big)=0 \end{equation*}

and the limit exists since it exists for for the terms on the RHS by definition. Since \(T_1\) and \(T_2\) are at most degree \(d\) it follows from the previous point that \(T_1-T_2\) is the zero polynomial i.e. \(T_1=T_2\).

Notice that if \(P(\cdot)\) is a Polynomial of degree \(d\) then \(P(x_0+\cdot)\) is a polynomial of degree \(d\). We clearly have

\begin{equation*} P(x_0+t) = T_{x_0}^d(t) +0 \end{equation*}

where \(+0\) emphasizes the constant \(0\) error term. This shows the claim

Take \(f(x)= x^{200}\sin(x^{-500})\) outside \(0\) and \(f(0)=0\). By the previous HW assignment we have that \(f'(x)\) exists on \((-1,1)\) but is discontinuous at (and unbounded around) \(x=0\). This shows that \(f''(0)\) does not exists because then \(f'\) would have to be continuous at \(x=0\)

However notice that \(0\) is a Taylor polynomial of order \(100\) for \(f\) at \(0\) since we have

\begin{equation*} \lim_{t\to 0 } \frac{|f(t)-f(0)|}{|t|^{100}} =0 \end{equation*}

that can be seen by sandwiching:

\begin{equation*} \frac{|f(t)-f(0)|}{|t|^{100}} = |t|^{100} |\sin(t^{-500})| \leq |t|^{100}. \end{equation*}

For any two point \(x,y\) we have

\begin{equation*} \begin{aligned}[t] f(y)-f(x)=+f'(x)(y-x)+ \mathrm{err}_x(y-x) f(x)-f(y)=+f'(y)(x-y)+ \mathrm{err}_{y}(x-y) \end{aligned} \end{equation*}

Adding the two gives

\begin{equation*} f'(x)-f'(y) = \frac{\mathrm{err}_{y}(x-y) -\mathrm{err}_{x}(y-x)}{y-x} \end{equation*}

The condition on the error with \(t=(x-y)\) and \(t=(y-x)\) respecively gives

\begin{equation*} |f'(x)-f'(y)| \leq \frac{|\mathrm{err}_{y}(x-y)| }{|y-x|} + \frac{|\mathrm{err}_{x}(y-x)| }{|y-x|}\leq 2 C |y-x|^{\alpha} \end{equation*}

showing that \(f'(x)\) is \(\alpha\)-Hölder continuous. Furthermore it also follows that \(f'(x)\) is bounded on \((a,b)\) (even though this was not asked by the problem). Fix some \(x_0\in(a,b)\) and he have

\begin{equation*} |f'(x)-f'(x_0)| \leq 2 C |x-x_0|^{\alpha}\leq 2C |b-a|^\alpha \end{equation*}

so

\begin{equation*} f'(x_0)+ 2C |b-a|^\alpha\leq f'(x) \leq f'(x_0)+ 2C |b-a|^\alpha \end{equation*}

We have that \(f_n'\) is Cauchy in \(\|\cdot\|_{\sup}\). If we show that \(f_n\) is Cauchy in \(\|\cdot\|_{\sup}\) then we have shown that \(f_n\) is Cauchy in \(\|\cdot\|_{C^1}\). Since \(C^1_b((-1,1);\R)\) is complete this proves the claim. To show that \(f_n\) is Cauchy we use the triangle inequality and Lagrange

\begin{equation*} \begin{aligned}[t] |f_m(x)-f_n(x)|& \leq |(f_m-f_n)(x) - (f_m-f_n)(0)| \;+ \; |(f_m-f_n)(0)| \\ & \leq \sup_{c\in(-1,1)} \|f_n'(c)-f_m'(c)\| |x-0| + |(f_m-f_n)(0)| \\ &\leq \|f_n'-f_m'\|_{\sup} + |(f_m-f_n)(0)| \end{aligned} \end{equation*}

By choosing \(N=N(\epsilon)\) large enough so that both \( |(f_m-f_n)(0)|\lt \epsilon\) and \( \|f_n'-f_m'\|_{\sup}\lt \epsilon\) for all \(n,m\ge N(\epsilon)\) we conclude that

\begin{equation*} |f_m(x)-f_n(x)|\lt 2\epsilon \end{equation*}

for all \(n,m\ge N(\epsilon)\) showing that \(f_n\) is Cauchy in \(\|\cdot\|_{\sup}\) as required.

Take

\begin{equation*} f_n(x) = e^{-(x/n)^2}-1 \end{equation*}

We have that \(f_n(0)=0\) for all \(n\). Pointwise \(f_n(x)\to 0\) since \(-x/n\to 0\) and \(e^{-t^2}\to 1\) as \(t\to 0\).

The functions \(f_n\) do NOT converge uniformly to \(0\) since \(f_n(n)=e^{-1}-1\) that is not \(0\).

The derivatives \(f_n'\) converge uniformly to \(0\). To see this compute

\begin{equation*} f'_n(x)= -2\frac{x}{n^2} e^{-(x/n)^2} = \frac{1}{n} \Big(-2\frac{x}{n} e^{-(x/n)^2}\Big) = \frac{1}{n} f_1'(x/n). \end{equation*}

Where in particular

\begin{equation*} f'_1(x)= -2x e^{-x^2} \end{equation*}

Notice that \(f'_1(x)\) is bounded since it \(\lim_{x\to +\infty }-2x e^{-x^2}=\lim_{x\to -\infty }-2x e^{-x^2}=0\) and \(-2xe^{-x^2}\) is continuous (a variant of a problem from Midterm 01). We have thus that \(\|f'_1\|_{\sup}\lt \infty\) so \(\|f'_n\|_{\sup}\leq \|f'_1\|_{\sup}\frac{1}{n}\) so \(f_n'\rightrightarrows 0\).

This finishes the counterexample

Let \(D\) be the dense set on which pointwise convergence holds.

Fix \(\epsilon \gt 0\) and let \(\mc{A}_\epsilon\) be a finite \(\epsilon\) net for \((-1,1)\) made of points in \(D\). This can be done density by choosing one point from \(D\) that is \(\epsilon/100\) close to each of the points \(k\epsilon/100\) with \(k\in\N\), \(- 100/ \epsilon\lt k\lt 100/ \epsilon \).

For each \(z\in\mc{A}_\epsilon\) let \(N(\epsilon,z)\) be an index such that \(|f_n(z)|\lt \epsilon\) for all \(n\gt N(\epsilon,z)\). Such an index exists for each such \(z\) since \(z\in D\) and thus \(f_n(z)\to 0\). Let \(N(\epsilon)=\max_{z\in\mc{A}_\epsilon} {N(\epsilon,z)}\) that is allowed since \(\mc{A}_\epsilon\) is finite. This way \(|f_n(z)|\lt \epsilon\) for all \(n\gt N(\epsilon)\) and for all \(z\in\mc{A}_\epsilon\).

Finally for each \(x\in(-1,1)\) choose \(z_x\in\mc{A}_\epsilon\) such that \(|x-z_x|\lt \epsilon\); this is possible since \(\mc{A}_\epsilon\) is an \(\epsilon\)-net.

We have by Lagrange:

\begin{equation*} \begin{aligned}[t] |f_n(x)| &\leq |f_n(x)-f_n(z_x)| +|f_n(z_x)| \\ & \leq \epsilon + |f_n(x) -f_n(z_x)| \leq \epsilon + |z_x-x| \sup_{c\in(-1,1)} |f'_n(c)| \\ & \leq \epsilon + \epsilon M \end{aligned} \end{equation*}

The RHS does not depend on \(x\) so

\begin{equation*} \|f_n\|_{\sup }\leq \epsilon(M+1) \end{equation*}

for all \(n\gt N(\epsilon)\). This shows that \(f_n\) is Cauchy in \(\|\cdot\|_{\sup}\).

Either copy the example from class or take

\begin{equation*} f_n(x) = \frac{1}{n}\sin(nx) \end{equation*}

Clearly \(\|f_n\|_{\sup}=1/n\to 0\) while \(f'_n(x) = \cos(nx)\) is bounded in absolute value by \(1\). Taking \(x= \pi\) we have

\begin{equation*} f'_n(\pi)= (-1)^n \end{equation*}

that does not converge even pointwise.