We claim that \(f^{(n)}(0)=0\) for all \(n\in\N\).

The derivative at \(x_0\) is given by

\begin{equation*} \lim_{t\to 0} \frac{f(x_0+t)-f(x_0)}{t}. \end{equation*}

It is local (by that we mean that if \(f\) coincides with \(g\) on an open ball and \(g\) is differentiable at a point of that ball, then \(f\) is too and \(f'=g'\) at that point).

Let us compute the derivatives \(f^{n}(x)\) for \(x\gt 0\). Using the chain rule:

\begin{equation*} \begin{aligned}[t] & f(x)= e^{-1/x} \\ & f'(x)= x^{-2} e^{-1/x} \\ & f''(x)= (x^{-4}-2 x^{-3}) e^{-1/x} \end{aligned} \end{equation*}

We claim that for each \(n\in\N\) the \(n\)th derivative \(f^{(n)}(x)\) has the form

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

where \(P_n\) is some polynomial. Let us show this by induction. It is true for \(n=0\) with \(P_0(t)=1\). Supposing that it is true for \(n\) we have

\begin{equation*} \begin{aligned}[t] f^{(n+1)}(x)&=\frac{d }{dx}f^{(n)}(x)=\frac{d }{dx}\Big( P_{n}(1/x)e^{-1/x}\Big) = \\ &= -(1/x^2) P'_n(1/x)e^{-1/x} + P_n(1/x) (1/x^2) e^{-1/x} =(P_n(1/x) - P'_n(1/x))(1/x^2) e^{-1/x} = P_{n+1}(1/x) e^{-1/x} \end{aligned} \end{equation*}

where \(P_{n+1}(t)= t^2(P_n- P_n'(t))\).

Let us show by induction that \(f^{(n)}(0)=0\). This is true for \(n=0\) and \(n=1\) (by HW).

Supposing that \(f^{(n)}(0)=0\) let us compute \(f^{(n+1)}(0)\):

\begin{equation*} \begin{aligned}[t] f^{(n+1)}(0)=\lim_{t\to 0} \frac{f^{(n)}(t)-f^{(n)}(0)}{t}=\lim_{t\to 0} \frac{f^{(n)}(t)}{t} \end{aligned} \end{equation*}

Let us split the limit into \(t\to 0^-\) and \(t\to 0^+\). For \(t\to 0^-\) this is \(0\) since \(f^{(n)}(t)=0\) if \(t\lt 0\) (by locality of derivatives and by the fact that for \(t\lt 0\) one has \(f(t)=0\)).

For \(t\gt 0\) we use the inductive claim about the form of \(f^{(n)}(t)\) to obtain

\begin{equation*} \begin{aligned}[t] \lim_{t\to 0^+} \frac{f^{(n)}(t)}{t} = \lim_{t\to 0^+} \frac{P_n(1/t) e^{-1/t}}{t} \end{aligned} \end{equation*}

Changing variables like in HW06P1.5 we compute

\begin{equation*} \begin{aligned}[t] \lim_{t\to 0^+} \frac{P_n(1/t) e^{-1/t}}{t}= \lim_{s\to +\infty} sP_n(s) e^{-s} =0 \end{aligned} \end{equation*}

The last identity follows by the fact that \(\lim_{s\to +\infty} \frac{sP_n(s)}{s^d}=0\) for any \(d\gt \mathrm{deg}(P_n)\) and \(\lim_{s\to +\infty}s^de^{-s}=0\) as per solution to the HW06P1.5.

We need to show that \(\lim_n \|f_n\|_\sup =0\). Since \(f\) is either even or odd (depending on \(n\)) we can only study it for \(x\geq 0\)

Let us show that \(f_n\) is bounded and find bounds for \(|f_n(x)|\).

We have (by solution of HW06P1.5 or L'Hopital) that for any fixed \(n\) we have. L

\begin{equation*} \lim_{x\to +\infty} f_n(x)=0 \end{equation*}

So by midterm01 the function above is non-negative. Also, being continuous on \([0,+\infty)\), \(f_n\) is bounded and attains max (since \(0\) cannot be its \(\sup\)). Since \(f_n\) is differentiable (being a composition and product of differentiable functions). The point of maximum \(x\) must satisfy

\begin{equation*} f'_n(x)=0 \end{equation*}

i.e.

\begin{equation*} n\,2^{-n^2}x^{n-1} e^{-x^2} - 2\,2^{-n^2} x^{n+1}e^{-x^2}=0 \end{equation*}

i.e.

\begin{equation*} x^{n-1}(n - 2 x^2) =0. \end{equation*}

The maximum is not at \(x=0\) where \(f(x)=0\) so it is at \(x=\sqrt{n/2}\). There the value of the function is

\begin{equation*} f_n(\sqrt{n/2}) = 2^{-n^2}(n/2)^{n/2} e^{-n/2}\leq e^{-\log(2)n^2 + (\log(n/2)-1) n/2} \end{equation*}

For the numerator we have

\begin{equation*} \lim_{n\to \infty}\Big(-\log(2)n^2 + (\log(n/2)-1) n/2\Big)=\lim_{n\to \infty}n^2\Big(-\log(2) + (\log(n/2)-1)/( 2n)\Big)=-\infty \end{equation*}

since \((\log(n/2)-1)/( 2n)\to 0\). To see this last part apply either change variables \(n= e^t\) to obtain:

\begin{equation*} \lim_{n\to +\infty}(\log(n/2)-1)/( 2n)=-\lim_{t\to +\infty} \frac{t-\log(2)-1}{e^t}=0 \end{equation*}

or apply L'Hopital to obtain

\begin{equation*} \lim_{n\to +\infty}(\log(n/2)-1)/( 2n)=\lim_{n\to +\infty}(1/n)/(2) = 0. \end{equation*}

Thus we have shown that

\begin{equation*} 0\leq \|f_n\|_{\sup}\leq f_n(\sqrt{n/2})\leq e^{-\log(2)n^2 + (\log(n/2)-1) n/2}\to 0 \end{equation*}

concluding our claim.

Show that \(F\) has bounded first and second derivative on any set of the form \([-M;M]\).

Show that \(T\) is continuous as a map from \(\Big(C^0\big([0,1];\R\big),\, \|\cdot\|_{\sup}\Big)\) to \(\Big(C^0\big([0,1];\R\big),\, \|\cdot\|_{\sup}\Big)\).

Show that \(T\) is continuous as a map from \(\Big(C^1\big([0,1];\R\big),\, \|\cdot\|_{C^1}\Big)\) to \(\Big(C^1\big([0,1];\R\big),\, \|\cdot\|_{C^1}\Big)\).

To solve the last two points you may assume initially that \(F'\) \(F''\) are bounded.

Then use the first point to remove that assumption.

Suppose that \(f\) is differentiable at \(x_0\). Show that

\begin{equation*} \lim_{t\to 0} \frac{f(x_{0}+t)-f(x_{0}-t)}{2t} = f'(x_{0}) \end{equation*}

Careful, no information about the behavior of \(f\) outside of \(x_0\) is given, except for what follows from differentiability at \(x_0\).

Suppose that \(f\) is twice differentiable at \(x_0\). Show that

\begin{equation*} \lim_{t\to 0} \frac{f(x_{0}+t)+f(x_{0}-t)-2f(x_{0})}{t^{2}} = f''(x_{0}) \end{equation*}

Careful, no information about the behavior of \(f\) outside of \(x_0\) is given, except for what follows from differentiability at \(x_0\).

Let \(p_{1},p_{2},p_{3},p_{4}\) be four numbers in \(\R\). Show that a sufficient and necessary condition for

\begin{equation*} \lim_{t\to 0} \frac{\sum_{i=1}^{4}a_{i}f(x_{0}+p_{i}t)}{t^{3}}= f'''(x_{0}) \end{equation*}

for any \(f\colon(-1,1)\to \R\) three times differentiable at \(x_{0}\in(-1,1)\) is that

\begin{equation*} \begin{aligned} &\sum_{i=1}^{4}a_{i}p_{i}^{k}=0 \qquad \forall k\in\{0,1,2\} \\ & \sum_{i=1}^{4}a_{i}p_{i}^{3}=3! \end{aligned} \end{equation*}

Careful, no information about the behavior of \(f\) outside of \(x_0\) is given, except for what follows from differentiability at \(x_0\).

Hint: To check necessity test against \(f\) that are polynomials. To show that it is sufficient, Taylor expand.