# Calculus Preliminaries

## Derivatives

If $$(X, {\left\lvert {{-}} \right\rvert})$$ is a metric space and $$f: X\to X$$ with \begin{align*} {\left\lvert {f(x) - f(y)} \right\rvert} \leq c {\left\lvert {x-y} \right\rvert} \text{ for some }c < 1, \forall x, y\in X ,\end{align*} then $$f$$ is a contraction. If $$X$$ is complete, then $$f$$ has a unique fixed point $$x_0$$ such that $$f(x_0) = x_0$$.

Uniqueness: if $$x, y$$ are two fixed points, then \begin{align*} 0 \leq {\left\lvert {x-y} \right\rvert} = {\left\lvert {f(x) - f(y)} \right\rvert}\leq c {\left\lvert {x-y} \right\rvert}\leq {\left\lvert {x-y} \right\rvert} ,\end{align*} forcing $${\left\lvert {x-y} \right\rvert} = 0$$

Existence: Define a sequence by picking $$x_0$$ arbitrarily and setting $$x_k \coloneqq f(x_{k-1})$$. Then \begin{align*} {\left\lvert {x_{k+1}-x_k} \right\rvert} = {\left\lvert {f(x_k) - f(x_{k-1}) } \right\rvert} \leq c{\left\lvert {x_k - x_{k-1}} \right\rvert} ,\end{align*} so inductively \begin{align*} {\left\lvert {x_{k+1}- x_k} \right\rvert}\leq c^k {\left\lvert {x_1 - x_0} \right\rvert} .\end{align*} The claim is that this makes $$\left\{{x_k}\right\}$$ a Cauchy sequence, this follows from the fact that if $$n < m$$ then \begin{align*} {\left\lvert {x_n - x_m} \right\rvert} \leq \sum_{n+1\leq k \leq m} {\left\lvert {x_k - x_{k-1}} \right\rvert} \leq \sum_{n \leq k \leq m-1} c^k {\left\lvert {x_1 - x_0} \right\rvert} \leq c^n {\left\lvert {x_1 - x_0 \over 1-c} \right\rvert} \to 0 .\end{align*}

## Implicit Function Theorem

Suppose $$f\in C^1({\mathbf{R}}^{n+m}, {\mathbf{R}}^n)$$, that $$f(a, b) = 0$$, and the derivative $$D_f(a, b)$$ at $$(a, b)$$ is an invertible linear map. Then there exists a neighborhood $$U\subseteq {\mathbf{R}}^n$$ containing $$a$$ and a unique $$g\in C^1(U, {\mathbf{R}}^m)$$ such that $$g(a) = b$$ and $$f(a, g(a)) = 0$$ for all $$x\in U$$.

A relation is locally the graph of a function wherever the derivative is nonsingular.

### Inverse Function Theorem

For $$f \in C^1({\mathbf{R}}; {\mathbf{R}})$$ with $$f'(a) \neq 0$$, then $$f$$ is invertible in a neighborhood $$U \ni a$$, $$g\coloneqq f^{-1}\in C^1(U; {\mathbf{R}})$$, and at $$b\coloneqq f(a)$$ the derivative of $$g$$ is given by \begin{align*} g'(b) = {1 \over f'(a)} .\end{align*} For $$F \in C^1({\mathbf{R}}^n, {\mathbf{R}}^n)$$ with $$D_f$$ invertible in a neighborhood of $$a$$, so $$\operatorname{det}(J_f)\neq 0$$, then setting $$b\coloneqq F(a)$$, \begin{align*} J_{F^{-1}}(q) = \qty{J_F(p)}^{-1} .\end{align*}

The version for holomorphic functions: if $$f\in \mathop{\mathrm{Hol}}({\mathbf{C}}; {\mathbf{C}})$$ with $$f'(p)\neq 0$$ then there is a neighborhood $$V\ni p$$ with that $$f\in \mathop{\mathrm{BiHol}}(V, f(V))$$.

A $$C^1$$ function is invertible in any neighborhood in which its derivative $$f'$$ is invertible.

Recall that absolutely convergent implies convergent, but not conversely: $$\sum k^{-1}= \infty$$ but $$\sum (-1)^k k^{-1}< \infty$$. This converges because the even (odd) partial sums are monotone increasing/decreasing respectively and in $$(0, 1)$$, so they converge to a finite number. Their difference converges to 0, and their common limit is the limit of the sum.

## Integrals

### Green’s Theorem

If $$\Omega \subseteq {\mathbf{C}}$$ is bounded with $${{\partial}}\Omega$$ piecewise smooth and $$f, g\in C^1(\overline{\Omega})$$, then \begin{align*}\int_{{{\partial}}\Omega} f\, dx + g\, dy = \iint_{\Omega} \qty{ {\frac{\partial g}{\partial x}\,} - {\frac{\partial f}{\partial y}\,} } \, \,dA.\end{align*} In vector form, \begin{align*} \int_\gamma F\cdot \,dr= \iint_R \operatorname{curl}F \,dA .\end{align*} As a consequence, areas can be computed as \begin{align*} \mu(\Omega) = {1\over 2}\oint_{{{\partial}}\Omega} \qty{y\,dx- x\,dy} = \oint_{{{\partial}}\Omega} x\,dy= -\oint_{{{\partial}}\Omega} y\,dx .\end{align*}

In general, $$\mu(\Omega) = \int_{\Omega} {\left\lvert {f'(z)} \right\rvert} \,dz$$.

Some basic facts needed for line integrals in the plane:

• Green’s theorem requires $$C^1$$ partial derivatives.
• $$\operatorname{grad}f = {\left[ { {\frac{\partial f}{\partial x}\,}, {\frac{\partial f}{\partial y}\,} } \right]}$$.
• If $$F = \operatorname{grad}f$$ for some $$f$$, $$F$$ is a vector field.
• Given $$f(x, y)$$ and $$\gamma(t)$$, the chain rule yields $${\frac{\partial }{\partial t}\,} (f\circ \gamma)(t) = {\left\langle { \operatorname{grad}f\circ \gamma)(t)},~{\gamma'(t)} \right\rangle}$$.
• For $$F(x, y) = {\left[ {M(x, y), N(x, y)} \right]}$$, $$\operatorname{curl}F = {\frac{\partial N}{\partial x}\,} - {\frac{\partial M}{\partial y}\,}$$ and $$\operatorname{Div}F = {\frac{\partial M}{\partial x}\,} + {\frac{\partial N}{\partial y}\,}$$.
• $$\int_\gamma F\cdot \,dr= \int_a^b F(\gamma(t))\cdot \gamma'(t) \,dt$$.

### Stokes Theorem

Suppose $$\omega \coloneqq f(z)\,dz$$ is a differential 1-form on an orientable manifold $$\Omega$$, then \begin{align*}\int_{{{\partial}}\Omega}\omega = \int_\Omega d\omega \qquad \text{i.e.} \qquad \int_{{{\partial}}\Omega}f(z)\,dz= \int_\Omega d(f(z)\,dz)\end{align*}

## Series and Sequences

• For every root $$r_i$$ of multiplicity 1, include a term $$A/(x-r_i)$$.
• For any factors $$g(x)$$ of multiplicity $$k$$, include terms $$A_1/g(x), A_2/g(x)^2, \cdots, A_k / g(x)^k$$.
• For irreducible quadratic factors $$h_i(x)$$, include terms of the form $${Ax+B \over h_i(x)}$$.

A series of functions $$\sum_{n=1}^\infty f_n(x)$$ converges uniformly iff \begin{align*} \lim_{n\to \infty} {\left\lVert { \sum_{k\geq n} f_k } \right\rVert}_\infty = 0 .\end{align*}

If $$\left\{{f_n}\right\}$$ with $$f_n: \Omega \to {\mathbf{C}}$$ and there exists a sequence $$\left\{{M_n}\right\}$$ with $${\left\lVert {f_n} \right\rVert}_\infty \leq M_n$$ and $$\sum_{n\in {\mathbb{N}}} M_n < \infty$$, then $$f(x) \coloneqq\sum_{n\in {\mathbb{N}}} f_n(x)$$ converges absolutely and uniformly on $$\Omega$$. Moreover, if the $$f_n$$ are continuous, by the uniform limit theorem, $$f$$ is again continuous.

Note that if a power series converges uniformly, then summing commutes with integrating or differentiating.

Consider $$\sum c_k z^k$$, set $$R = \lim {\left\lvert {c_{k+1} \over c_k} \right\rvert}$$, and recall the ratio test:

• $$R\in (0, 1) \implies$$ convergence.
• $$R\in (1, \infty] \implies$$ divergence.
• $$R=1$$ yields no information.

For $$f(z) = \sum_{k\in {\mathbb{N}}} c_k z^k$$, defining \begin{align*} {1\over R} \coloneqq\limsup_{k} {\left\lvert {a_k} \right\rvert}^{1\over k} ,\end{align*} then $$f$$ converges absolutely and uniformly for $$D_R \coloneqq{\left\lvert {z} \right\rvert} < R$$ and diverges for $${\left\lvert {z} \right\rvert} > R$$. So the radius of convergence is given by \begin{align*} R = {1\over \limsup_n {\left\lvert {a_n} \right\rvert}^{1\over n}} .\end{align*}

Moreover $$f$$ is holomorphic in $$D_R$$, can be differentiated term-by-term, and $$f' = \sum_{k\in {\mathbb{N}}} n c_k z^k$$.

Recall the $$p{\hbox{-}}$$test: \begin{align*} \sum n^{-p} < \infty \iff p \in (1, \infty) .\end{align*}

## Function Convergence

A sequence of functions $$f_n$$ is said to converge locally uniformly on $$\Omega \subseteq {\mathbf{C}}$$ iff $$f_n\to f$$ uniformly on every compact subset $$K \subseteq \Omega$$.

## Exercises

Compute $$\int_\Gamma \Re(z) \,dz$$ for $$\Gamma$$ the unit square.

Write $$\Gamma = \sum_{1\leq k \leq 4}\gamma_k$$, starting at zero and traversing clockwise:

Compute:

• $$\gamma_1$$: parameterize to get $$\int_0^1t1\,dt= 1/2$$.
• $$\gamma_2$$: $$\int_0^1 i \,dt= i$$
• $$\gamma_2$$: $$-\int_0^1 (1-t)\,dt= -1/2$$
• $$\gamma_2$$: $$- \int_0^1 0 \,dt= 0$$

So $$\int_\Gamma \Re(z) \,dz= i$$.

Show that if $$f$$ is a differentiable contraction, $$f$$ is uniformly continuous.

Show that a continuous function on a compact set is uniformly continuous.

Show that a uniform limit of continuous functions is continuous, and a uniform limit of uniformly continuous functions is uniformly continuous. Show that this is not true if uniform convergence is weakened to pointwise convergence.

Suppose $${\left\lVert {f_n - f} \right\rVert}_\infty\to 0$$, fix $${\varepsilon}$$, we then need to produce a $$\delta$$ so that \begin{align*} {\left\lvert {z-w} \right\rvert}\leq \delta \implies {\left\lvert {f(z) - f(w) } \right\rvert} < {\varepsilon} .\end{align*}

Write \begin{align*} {\left\lvert {f(z) - f(w)} \right\rvert} \leq {\left\lvert {f(z) - f_n(z)} \right\rvert} + {\left\lvert {f_n(z) - f_n(w)} \right\rvert} + {\left\lvert {f_n(w) - f(w)} \right\rvert} .\end{align*}

• Bound the first term by $${\varepsilon}/3$$ using that $$f_n\to f$$
• Bound the second term by $${\varepsilon}/3$$ using that $$f_n$$ is continuous
• Bound the third term by $${\varepsilon}/3$$ using that $$f_n\to f$$
• Pick $$\delta$$ to be the minimum $$\delta$$ supplied by these three bounds.

Why uniform convergence is necessary: need these bounds to holds for all $$z, w$$ where $${\left\lvert {z-w} \right\rvert} < \delta$$. Why pointwise convergence doesn’t work: $$f_n(z) \coloneqq z^n \overset{n\to\infty}\longrightarrow\chi_{z=1}$$

For uniform continuity: take $$\sup_{z, w}$$ on both sides: \begin{align*} \sup_{z, w} {\left\lvert {f(z) - f(w)} \right\rvert} &\leq \sup_{z} {\left\lvert {f(z) - f_n(z)} \right\rvert} + \sup_{z, w} {\left\lvert {f_n(z) - f_n(w)} \right\rvert} + \sup_{w} {\left\lvert {f_n(w) - f(w)} \right\rvert} \\ &\leq {\left\lVert {f - f_n} \right\rVert}_\infty + \sup_{z, w} {\left\lvert {f_n(z) - f_n(w)} \right\rvert} + {\left\lVert {f-f_n} \right\rVert}_\infty ,\end{align*} where now the middle term is bounded by uniform continuity of $$f_n$$.

Determine where the following real-valued function is or is not uniformly convergent: \begin{align*} f_n(x) \coloneqq{\sin(nx)\over 1+nx} .\end{align*}

This converges uniformly on $$[a, \infty)$$ for $$a$$ any constant: \begin{align*} {\left\lvert {\sin(nx) \over 1+nx} \right\rvert} \leq {1\over 1 + na} < {\varepsilon}= {\varepsilon}(n, a) .\end{align*}

This does not converge uniformly on $$(0, \infty)$$: \begin{align*} x_n \coloneqq{1\over n} \implies {\left\lvert {f_n(x_n)} \right\rvert} = {\left\lvert {\sin(1) \over 2} \right\rvert} > {\varepsilon} .\end{align*}

Show that $$\sum_{k\geq 0}z^k/k!$$ converges locally uniformly to $$e^z$$.

Apply the $$M{\hbox{-}}$$test on a compact set $$K$$ with $$z\in K \implies {\left\lvert {z} \right\rvert} \leq M$$: \begin{align*} {\left\lVert {e^z - \sum_{0\leq k \leq n} z^k/k!} \right\rVert}_\infty &= {\left\lVert {\sum_{k\geq n+1}z^k/k! } \right\rVert}_{\infty} \\ &\leq \sum_{k\geq n+1} {\left\lVert {z} \right\rVert}_\infty^k /k! \\ &\leq \sum_{k\geq 0} {\left\lVert {z} \right\rVert}_\infty^k /k! \\ &= e^{{\left\lVert {z} \right\rVert}_\infty} \\ &\leq e^{{\left\lvert {M} \right\rvert}} \\ &< \infty .\end{align*}

Show that if $$f_n\to f$$ uniformly then $$\int_\gamma f_n\to \int_\gamma f$$.

\begin{align*} {\left\lvert { \int_\gamma f_n(z) \,dz- \int_\gamma f(z) \,dz } \right\rvert} &= {\left\lvert {\int_\gamma f_n(z) - f(z) \,dz} \right\rvert} \\ &\leq \int_\gamma {\left\lvert {f_n - f} \right\rvert} {\left\lvert {\,dz} \right\rvert} \\ &\leq \int_\gamma {\left\lVert {f_n - f} \right\rVert}_{\infty, \gamma} \cdot {\left\lvert {\,dz} \right\rvert} \\ &= {\varepsilon}\cdot \mathop{\mathrm{length}}(\gamma) \\ &\overset{n\to\infty}\longrightarrow 0 .\end{align*}