Show that $f_n$ is continuous at $0$

Question

The function in the Lemma $16.1$ is

$$f(x)=\begin{cases}e^{-1/x}&\text{for }x > 0\\0&\text{for }x \leq 0\end{cases}$$

(a) To show that $a<e^a$ I have tried to look at the Taylor series of $e^x$ and thus come to the conclusion that $e^a=1+a+a^2/2!+a^3/3!+...$ and therefore $a<e^a$, is this fine?

Using the help, we do $a=t/2n$ and so $\frac{t}{2n}<e^{t/2n}$ so $\frac{t^n}{(2n)^n}<e^{t/2}$ so $\frac{t^n}{e^t}<\frac{(2n)^n}{e^{t/2}}$, then, if $t=1/x$ then $\frac{1}{e^{1/x}x^n}< \frac{(2n)^n}{e^{1/2x}}$ and as $\frac{(2n)^n}{e^{1/2x}}\to 0$ if $x\to 0$ we conclude that $\lim_{x\to 0}f_n(x)=0=f_n(0)$.

(b) $\lim_{x\to 0}\frac{f_n(x)-f_n(0)}{x}=\lim_{x\to 0}\frac{1}{e^{1/x}x^{n+1}}$, but I do not know how to calculate this limit, could I use what I did in (a)? ($\frac{1}{e^{1/x}x^n}< \frac{(2n)^n}{e^{1/2x}}$)

(c) If $x\leq 0$ is clear, then consider $x>0$, from here we have that $f'_n(x)=\frac{x^{n-2}e^{-1/x}-e^{-1/x}nx^{n-1}}{x^{2n}}=x^{-n-2}e^{-1/x}-ne^{-1/x}x^{-n-1}=f_{n+2}(x)-nf_{n+1}(x)$

(d) How do I prove that $f$ is of class $C^{\infty}$? Can I use (c)? Thank you very much.

Answer 1

(a) Yes, that argument is fine for $a \geq 0$, which is really all that's necessary here. I'd phrase it as $e^a \geq 1 + a > a$ for clarity, but the argument is fine.

(b) You can use the result of (a)! You're trying to find $\lim_{x \to 0} \frac{1}{e^{1/x} x^{n+1}}$, but this is exactly $\lim_{x \to 0} f_{n+1}(x)$, which we know from (a) to be $0$.

(c) Yup.

(d) You can absolutely use (c). $f_n'$ can be written in terms of $f_{n+1}$ and $f_{n+2}$, which are both differentiable functions. Hence $f_n'$ is itself differentiable for all $n$. Then, we know $f_{n+1}$ and $f_{n+2}$ are both twice differentiable, so $f_n$ is thrice differentiable, and so on, giving you $C^\infty$ by easy induction.

Answer 2

For (a) you are fine as $a < S_N := \sum_{n=1}^N \frac{a^n}{n!}$ for all $N \in \mathbb{N}$, thus as $(S_N)$ is increasing with limit $e^a$ we have that $a< e^a$. The rest of your argument is good too.

For (b) you need to check the limit $\lim_{h \rightarrow 0}\frac{f_n(h) -f_n(0)}{h}$ not $\lim_{h \rightarrow 0}\frac{f_n(h) -f(0)}{h}$. This is considerably easier as $\frac{f_n(h) -f_n(0)}{h} = f_{n+1}(h) = 0$.

For (d) we note that by (c) we have $f_n'(x) = f_{n+2}(x) + n f_{n+1}(x)$ which is continuous, thus using an inductive argument we have that $f_n$ is smooth for each $n \in \mathbb{N}$ (with $f^{(k)}_n(0) =0$ for all $k \in \mathbb{N}$).

For showing that this proves that $f$ is smooth we define for each $k \in \mathbb{N}$ the Banach space $$X_k := \{ g \in C^k: \|g\|_{C^k} := \|g\|_\infty + \frac{1}{1}\|g'\|_\infty + \ldots + \frac{1}{k}\|g^{(k)}\|_\infty < \infty \}$$ and note that $f_n \in X_k$ for all $n,k \in \mathbb{N}$. We now show that $f_n \rightarrow f$ in $X_k$ as $n \rightarrow \infty$ and so $f \in X_k \subset C^k$ for each $k \in \mathbb{N}$, thus $f$ is smooth.

Answer 3

$(a)$ Since you need to show $a < e^a$ for all $a$ including negative $a$, using the Taylor Series of $e^a$ is clear in showing the strict inequality for non-negative $a$. For negative $a$, note that $f(x) = e^x$ is a non-negative function. Hence, the strict inequality follows for all $a$.

$(b)$ Indeed from the hint in part $(a)$, we have $\frac{1}{x^{n+1}e^{\frac{1}{x}}} < \frac{(2n)^n}{xe^{\frac{1}{2x}}}$. We want to prove that the right hand side goes to $0$ as $x \rightarrow 0^{+}$. By L'hospital's Rule, $\lim_{x \rightarrow 0^{+}} \frac{1}{xe^{\frac{1}{2x}}} = \lim_{x \rightarrow 0^{+}} \frac{x^{-1}}{e^{\frac{1}{2x}}} = \lim_{x \rightarrow 0^{+}} \frac{-x^{-2}}{e^{\frac{1}{2x}}(\frac{-1}{2}x^{-2})} = \lim_{x \rightarrow 0^{+}}\frac{2}{e^{\frac{1}{2x}}} = 0$. Verifying the definition of differentiability from the left hand side is easier, and shows that $f_n$ is differentiable at $0$ for all $n$.

$(c)$ Looks correct.

$(d)$ From part $(b)$, we showed that for all $n$, $f_n(x)$ is differentiable for all $x$. From part $(c)$ we can write $f_{n}^{'}(x)$ in a recursive way using $f_{n+2}$ and $f_{n+1}$. Thus, $f_{n}^{''}(x)$ exists since both $f_{n+2}^{'}$ and $f_{n+1}^{'}$ exist, etc. Applying this reasoning further shows that $f_n$ is $C^{\infty}$ for all $n$.