The Exponential Function: Examples and Counterexamples

For every $n>2$, the base is positive and $a_n=(1-2\mathrm{/}\iota (n){)}^n$, so [L1] applies at $x=-2$ and gives $a_n\to \exp (-2)$ on that tail.

By [L3], the whole sequence has the same limit as the tail in step 1.1, and [L2] identifies that limit as $e^{-2}>0$.

At the moving point $x=\iota (n)$, $h_n(x)=2^n$, whereas $\exp (x)=e^n$.

Since $e>2$, the difference $e^n-2^n$ is at least $e-2>0$ and in fact grows; therefore ${\sup }_x\mathrm{\mid }{h}_n(x)-\exp (x)\mathrm{\mid }\to ̸0$.

Hence the pointwise convergence is not uniform on $\mathbb{R}$.

On $x>0$, repeated product and chain rules give ${\varphi }^{(m)}(x)=P_m(1\mathrm{/}x)\exp (-1\mathrm{/}x)$ for a polynomial $P_m$; on $x<0$, every derivative is $0$.

If the formula holds at order $m$, differentiating produces another polynomial $P_{m+1}(1\mathrm{/}x)$ times $\exp (-1\mathrm{/}x)$. By [L1], this tends to $0$ as $x\to 0^{+}$.

The difference quotient for the $m$-th derivative at $0$ is again a polynomial in $1\mathrm{/}x$ times $\exp (-1\mathrm{/}x)$, so it tends to $0$. Induction therefore extends every derivative continuously across $0$, with value $0$.

The function $q(x)=x^2$ is a polynomial, so it is differentiable at every real with $q^{\mathrm{\prime }}(x)=2x$, and $\psi (x)=\varphi (q(x))$ for every $x\in \mathbb{R}$.

The derivative of any finite sum of functions of the form $p(x){\varphi }^{(j)}(q(x))$, with $p$ a polynomial, is again a finite sum of this form: each $p$ is differentiable with polynomial derivative $p^{\mathrm{\prime }}$ by [L5], each ${\varphi }^{(j)}$ is differentiable by [L1], and $q$ is differentiable with $q^{\mathrm{\prime }}(x)=2x$ by [L5], so [L2] gives $({\varphi }^{(j)}\circ q{)}^{\mathrm{\prime }}(x)=2x{\varphi }^{(j+1)}(q(x))$ and then [L3] gives $(p\cdot ({\varphi }^{(j)}\circ q){)}^{\mathrm{\prime }}(x)=p^{\mathrm{\prime }}(x){\varphi }^{(j)}(q(x))+2xp(x){\varphi }^{(j+1)}(q(x)),$ in which $p^{\mathrm{\prime }}$ and $2xp$ are again polynomials; [L3] then adds the finitely many summands.

If $x\mathrm{\ne }0$, then $q(x)=x^2>0$, and therefore $\psi (x)=\varphi (q(x))>0$.

Starting from $\psi =\varphi \circ q$ and applying step 1.2 repeatedly shows that every derivative of $\psi$ exists; moreover, for each $m\ge 0$, ${\psi }^{(m)}$ is a finite sum of functions $p(x){\varphi }^{(j)}(q(x))$.

For every $m\ge 0$, step 2.1 makes ${\psi }^{(m)}$ differentiable at every real, because ${\psi }^{(m+1)}$ exists there, and [L6] then makes ${\psi }^{(m)}$ continuous on $\mathbb{R}$. So every derivative of $\psi$ exists on $\mathbb{R}$ and is continuous there, which by [L7] is exactly $\psi \in C^{\mathrm{\infty }}(\mathbb{R})$.

At $x=0$, every summand in step 2.1 vanishes because $q(0)=0$ and ${\varphi }^{(j)}(0)=0$. Hence ${\psi }^{(m)}(0)=0$ for every $m\ge 0$.

By the definition of the Maclaurin series, every coefficient of the Maclaurin series of $\psi$ is zero, so the series converges everywhere to $0$.

Thus the everywhere-convergent Maclaurin series agrees with $\psi$ at $x=0$ but disagrees with it at every $x\mathrm{\ne }0$, refuting the stated claim.

On $\mathrm{\mid }x\mathrm{\mid }<1$, $1-x^2>0$, so $b(x)>0$; on $\mathrm{\mid }x\mathrm{\mid }\ge 1$, $1-x^2\le 0$, so $b(x)=0$.

Away from $\pm 1$, smoothness follows from [L2]. At $\pm 1$, every one-sided derivative from inside is a finite sum of derivatives of $\varphi$ at $0$, all zero by [L1], matching the zero function outside.

Thus $b$ is smooth and its nonzero set is $(-1,1)$, whose closure is $[-1,1]$; this is its support.

Extend $\{1,b\}$ to a Hamel basis. Define the $\mathbb{Q}$-linear map $A$ by $A(1)=1$, $A(b)=0$, and $A(v)=v$ on the remaining chosen basis elements. Then $A$ is additive but not the identity.

Put $F(x)=\exp (A(x))$. Positivity and additivity give $F(x+y)=F(x)F(y)$, and $F(1)=e$.

If $F$ were continuous, Regular normalized multiplicative Cauchy equations characterize the exponential would give $F=\exp$; injectivity of $\exp$ would then give $A(x)=x$, contradicting $A(b)=0\mathrm{\ne }b$.

Thus $F$ is a discontinuous positive multiplicative solution. The construction uses Choice exactly in the basis extension.

For $n\ge 1$, let $x_n=\iota (n)$ and $y_n=\iota (n)+1\mathrm{/}\iota (n)$. Then $\mathrm{\mid }{y}_n-x_n\mathrm{\mid }=1\mathrm{/}\iota (n)\to 0$.

By the mean value theorem, $\exp (y_n)-\exp (x_n)=\exp (c_n)\mathrm{/}\iota (n)$ for some $c_n\in (x_n,y_n)$. Since exponential is increasing, this is at least $\exp (\iota (n))\mathrm{/}\iota (n)$, which tends to $+\mathrm{\infty }$ by [L2].

Thus arbitrarily close pairs have image distances bounded away from $0$, contradicting the uniform-continuity condition [L1].