Practice final exam and Answers

- If $f$ is continuous on $[a, b]$, then so is $|f|$.

True: this is $g \circ f(x)$ where $g(y)=|y|$ is also continuous. - If $f$ is differentiable on $[a, b]$, then so is $|f|$.

False: not even true for $f(x)=x$ on $[-1,1]$. - If $\lim _{n→0}\left|a_n\right|=0$, then $\sum_{n=0}^{\infty} a_n$ converges.

False: the harmonic series $a_n=1 / n$ is a counterexample. - The series $\sum_{n=0}^{\infty} e^{-n^2}$ converges.

True: $e^{-(n+1)^2} / e^{-n^2}=e^{-2 n-1} \rightarrow 0$ as $n \rightarrow \infty$, so converges by ratio test. - The series $\sum_{n=0}^{\infty} \sin (n x) / 2^n$ converges to a continuous function of $x$.

True: $\left|\sin (n x) / 2^n\right| \leq 1 / 2^n$, and $\sum 1 / 2^n$ converges, so this converges uniformly by Weierstrass M-test. And uniform limits of continuous functions are continuous. - If $A \in M_{m \times n}$ is a matrix such that $T_A=0$ (where $T_A$ is the corresponding linear map $\mathbb{R}^n→\mathbb{R}^m$ ), then $A=0$

True: we proved in class that the map taking $A$ to $T_A$ is an isomorphism, hence linear and injective, hence has kernel $=\{\overrightarrow{0}\}$ by A12#4.

- State the small-span theorem.(Notes)

If $f:[a, b] \rightarrow \mathbb{R}$ is continuous, then for all $\epsilon>0$ there exists a partition $a=x_0< x_1<\cdots< x_n=b$ such that for each $i \in\{1, \ldots, n\}, M\left(\left.f\right|_{\left[x_{i-1}, x_i\right]}\right)-m\left(\left.f\right|_{\left[x_{i-1}, x_i\right]}\right)<\epsilon$. Here $M$ and $m$ denote the absolute maximum and absolute minimum, respectively. - State the theorem on uniform convergence and integration.

If $f_n:[a, b] \rightarrow \mathbb{R}$ are continuous and converge uniformly to $f:[a, b] \rightarrow \mathbb{R}$, then the indefinite integrals $\int_a^x f_n(t) d t$ converge uniformly, as a function of $x$, to $\int_a^x f(t) d t$. - Define (a) convergence of a series; (b) absolute convergence of a series.

A series $\sum_{n=0}^{\infty} a_n$ converges if the sequence $\left\{\sum_{n=0}^m a_n\right\}$ of partial sums converges, that is, if $\lim _{m \rightarrow \infty} \sum_{n=0}^m a_n$ exists. It converges absolutely if the series $\sum_{n=0}^{\infty}\left|a_n\right|$ converges. - Define the standard basis vectors in $\mathbb{R}^n$.

The standard basis vectors $\vec{e}_1, \ldots, \vec{e}_n \in \mathbb{R}^n$ are the vectors whose components are $\left(\vec{e}_i\right)_j=\delta_{i, j}$. Here, for $i, j \in\{1, \ldots, n\}$, we define $\delta_{i, j}=1$ if $i=j$ and 0 if $i \neq j$.

- Let $f: S→T$ be a bijection of sets, and let $g: T→S$ be a function such that $g \circ f=\text{id}_S$. Prove that $f \circ g=\text{id}_T$.

Since $f$ is a bijection, it has an inverse, that is, a function $h: T \rightarrow S$ such that $f \circ h=\text{id}_T$ and $h \circ f=\text{id}_S$. Then $f \circ g=(f \circ g) \circ \text{id}_T=(f \circ g) \circ(f \circ h)=f \circ(g \circ f) \circ h=f \circ\left(\text{id}_S\right) \circ h=f \circ h=\text{id}_T$. - Suppose that $f: \mathbb{R}→\mathbb{R}$ is a differentiable function such that $f(0)$ and $f(2)$ are positive, but $f(1)$ is negative. Show that $f'(x)=0$ for some $x$.

Differentiable implies continuous, so by the intermediate value theorem, there exist $a \in(0,1)$ and $b \in(1,2)$ such that $f(a)=f(b)=0$. Then by Rolle's theorem (or the mean-value theorem) there exists $x \in(a, b)$ such that $f'(x)=0$. - Let $f(x)=\sin x+2 x$. Prove that $f: \mathbb{R}→\mathbb{R}$ is a bijection.
Since $\left|\sin x\right|≤1$, $f(x / 2-1)< x< f(x / 2+1)$. Since $f$ is continuous, by the intermediate value theorem there exists $y \in(x / 2-1, x / 2+1)$ such that $f(y)=x⇒f$ surjective.

If there were two distinct $y_1< y_2$ such that $f\left(y_1\right)=f\left(y_2\right)$, since $f$ is differentiable there would exist $c∈\left(y_1, y_2\right)$ such that $f'(c)=0$ by Rolle's theorem. But $f'(x)=\cos x+2$ is never zero, contradiction$⇒f$ injective. - Prove that if $\lim _{n→\infty} a_n=L$, then $\lim _{n→\infty}\left|a_n\right|=|L|$.

The hypothesis means that for all $\epsilon>0$, there exists $N \in \mathbb{N}$ such that $n \geq N$ implies $\left|a_n-L\right|<\epsilon$. By the triangle inequality, \[ |L|=\left|L-a_n+a_n\right| \leq\left|L-a_n\right|+\left|a_n\right| \] and \[ \left|a_n\right|=\left|a_n-L+L\right| \leq\left|a_n-L\right|+|L| . \] Subtracting $\left|L-a_n\right|+|L|$ from the first inequality and $|L|$ from the second gives \[ -\left|a_n-L\right| \leq\left|a_n\right|-|L| \leq\left|a_n-L\right|, \] that is, \[ \left|| a_n|-| L|\right| \leq\left|a_n-L\right| . \] Hence for $\epsilon, N$, and $n$ as above, $\left||a_n|-| L |\right| \leq\left|a_n-L\right|<\epsilon$, which implies the conclusion.

Alternative: By A9#8, a continuous function such as $f(x)=|x|$ takes convergent sequences to convergent sequences; more precisely, if $\lim _{n \rightarrow \infty} a_n=L$, then $\lim _{n \rightarrow \infty} f\left(a_n\right)=f(L)$, hence $\lim _{n \rightarrow \infty}\left|a_n\right|=|L|$. - Let $f_n:[0,1]→\mathbb{R}$ be given by $f_n(x)=n$ if $0< x< 1 / n$ and $f_n(x)=0$ otherwise. Show that $f_n→0$ pointwise but not uniformly.

Pointwise: Take $x \in[0,1]$. If $x=0$, then $f_n(x)=0$ for all $n$, so obviously $f_n(x) \rightarrow 0$. Otherwise, for any $\epsilon>0$, let $N$ be an integer $>1 / x$; then $n \geq N$ implies $1 / n \leq 1 / N< x$, so $f_n(x)=0$ and $\left|f_n(x)-0\right|=|0-0|<\epsilon$.

Uniform: Take $\epsilon=1$; then for all $n$ there exists $x \in[0,1]$ (say $x=1 / 2 n)$ such that $\left|f_n(x)-0\right|=n \geq \epsilon$, so convergence isn't uniform.

Alternative: $f_n$ is a step function, and $\int_0^1 f_n(x) d x=n / n=1$ doesn't converge to $\int_0^1 0 d x=0$, contradicting A10#3. - Prove that for every linear map $f: \mathbb{R}^n→\mathbb{R}$, there exists $B \in \mathbb{R}^n$ such that for all $X \in \mathbb{R}^n, f(X)=B⋅X$

Let $E_i$ be the $i$ th standard basis vector and let $B$ be the vector with $i$ th component $b_i=f\left(E_i\right)$. Then for any $X=\sum_{i=1}^n x_i E_i$, we have $f(X)=f\left(\sum_{i=1}^n x_i E_i\right)=\sum_{i=1}^n x_i f\left(E_i\right)=\sum_{i=1}^n x_i b_i=\sum_{i=1}^n b_i x_i=B⋅X$. - Let $V$ be the set consisting of all step functions $[0,1]→\mathbb{R}$.
(a) Prove that $V$ is a vector space.(a) We know that $\mathcal{F}([0,1], \mathbb{R})$ is a vector space, so it suffices to show that the set of step function is a subspace. But we know that if $s$ and $t$ are step functions, then so is $s+t$ (by passing to a common refinement). And if $s$ is a step function and $c \in \mathbb{R}$, then $c s$ is a step function (by using the same refinement as for $s$ ). Hence the set of step functions is closed under addition and scalar multiplication, so it is a subspace.

(b) Prove that the indefinite integral gives a linear map $V→\mathcal{F}([a, b], \mathbb{R})$.

(c) Is this map is injective? Why or why not?

(b) This is an immediate consequence of linearity (and homogeneity) for integrals of step functions: $\int_0^x(s+t)(y) d y=\int_0^x s(y) d y+\int_0^x t(y) d y$ and $\int_0^x c s(y) d y=c \int_0^x s(y) d y$. So if we let $\phi(s)=\int_0^x s(y) d y$, then $\phi(s+t)=\phi(s)+\phi(t)$ and $\phi(c s)=c \phi(s)$.

(c) This is not injective: both $s(x)=0$ and \[ t(x)=\left\{\begin{array}{l} 1 \text { if } x=0 \\ 0 \text { if } x \neq 0 \end{array}\right. \] have $\phi(s)=\phi(t)=0$