# What goes up must come down

Let $f \in C^1(\mathbb{R})$ such that $f,f' \in L^1(\mathbb{R})$. Prove that $\int_\mathbb{R}f'=0$.

# One step forward, two steps back

Let $f:\mathbb{R}\to\mathbb{R}$ be an infinitely differentiable function. Does there existe $g:\mathbb R\to\mathbb R$ also infinitely differentiable such that $f(x) = g(x+1)-g(x)$ for all $x\in\mathbb R$?

Let $f,g:[a,b]\to\Bbb{R}$ be continuous functions. We call $\int_a^b x^kf(x)\,\mathrm{d}x$ the $k$-th moment of $f$. Prove that if $f$ and $g$ have identical $k$-th moments for all $k\geq 0$, then $f=g$.

# Constant Function

Let $f : \mathbb{R} \to \mathbb{R}$ be a continous and bounded function such that $f(x) = \int_x^{x+1} f(y) dy$ for all $x \in \mathbb{R}$

Then $f$ is constant.

# Wronskians and linear dependence

If $f_1,\dots,f_k:I\subseteq\Bbb{R}\to\Bbb{R}$ are sufficiently differentiable, their Wronskian is defined as

$W(f_1,\dots,f_k)=\mathrm{det}\begin{bmatrix} f_1&f_2&\dots&f_k\\f_1'&f_2'&\dots&f_k'\\ \vdots&\vdots&\ddots&\vdots\\ f_1^{(k-1)}&f_2^{(k-1)}&\dots&f_k^{(k-1)}\end{bmatrix}$

Obviously, if $f_1,\dots,f_k$ are linearly dependent, then their Wronskian vanishes identically on $I$, and the converse is true if the functions are analytic. Find non-analytic functions for which the converse fails.

# Balanced signs

Assume $(p_n)$ is a nonincreasing sequence of positive integers whose sum $p_1+p_2+\cdots$ diverges and assume signs $(\varepsilon_n)$ are chosen so that the modified series $\varepsilon_1p_1+\varepsilon_2p_2+\cdots$ converges. Then

$\liminf\limits_{n\to\infty}\dfrac{\varepsilon_1+\cdots+\varepsilon_n}n \leqslant 0\leqslant \limsup\limits_{n\to\infty}\dfrac{\varepsilon_1+\cdots+\varepsilon_n}n$

Assume now only the signed series converges, with no hypothesis on the unsigned sequence. Show

$\lim\limits_{n\to\infty} (\varepsilon_1+\cdots+\varepsilon_n)p_n=0$

# I know one when I see one

Suppose $f\in C^\infty([0,1])$ is such that for each $x\in[0,1]$, $f^{(n)}(x)$ is eventually zero. Then $f$ is a polynomial.