Let $\displaystyle f$ be differentiable, $\displaystyle f(x)=0$ for $|x| \geq 10 $ and $g(x)=\sum_{k \in \mathbb Z}f(x+k).$

Question

I came across the following problem that says:

Let $\displaystyle f \colon \mathbb R \rightarrow \mathbb R$ be differentiable function and $\displaystyle f(x)=0$ for $|x| \geq 10.$ Let $g(x)=\sum_{k \in \mathbb Z}f(x+k).$ Then which of the following is true?
$1.g$ is differentiable and $g'$ has infinitely many zeros
$2.g$ is continuous and $g'$ has no zeros
$3.g$ is differentiable and $g'$ has no zeros
$4.g$ is differentiable and $g'$ has only finitely many zeros.

I am not sure about how to progress with it.Can someone point me in the right direction? Thanks in advance for your time.

Answer 1

Differentiability: for sure $g$ is differentiable as this sum is locally finite and each $x\longmapsto f(x+k)$ is differentiable. Fix $n\in\mathbb{Z}$. Then observe $$ x>n-1\;\mbox{and}\; k\geq 11-n\quad\Rightarrow \quad x+k>n-1+11-n=10\quad\Rightarrow \quad f(x+k)=0 $$ and $$ x<n+1\;\mbox{and}\; k\leq -11-n\quad\Rightarrow \quad x+k<n+1-11-n=-10\quad\Rightarrow \quad f(x+k)=0. $$ So if $x$ belongs to $(n-1,n+1)$, the terms $f(x+k)$ vanish for all $k\geq 11-n$ and $k\leq -11-n$. Thus $$ g(x)=\sum_{k=-10-n}^{10-n}f(x+k)\qquad\forall x\in (n-1,n+1). $$ First, this shows that $g$ is well-defined. And since this is a finite sum of differentiable functions on $(n-1,n+1)$, $g$ is differentiable on $(n-1,n+1)$ and, simply, $$ g'(x)=\sum_{k=-10-n}^{10-n}f'(x+k)\qquad\forall x\in (n-1,n+1). $$ Now $\mathbb{R}=\bigcup_{n\in\mathbb{Z}}(n-1,n+1)$. Hence $g$ is differentiable on $\mathbb{R}$.

Note: observe that $g$ is $1$-periodic as the index set $\mathbb{Z}$ is invariant under the shift $x\longmapsto x+1$. So it suffices to prove the differentiability of $g$ on $(0,2)$. But it does not really simplify the argument.

Zeros of the derivative: by periodicity, as soon as $g'$ has a zero $x_0$, it must have infinitely many zeros as it vanishes on $x_0+\mathbb{Z}$. Now since $g$ is differentiable on $\mathbb{R}$ and $1$-periodic, it suffices to apply Rolle on the interval $[0,1]$, as $g(0)=g(1)$, to see that there exists $x_0\in (0,1)$ such that $g'(x_0)=0$.

Conclusion: 1 is the only correct option.

Answer 2

Try it with $f(x)=0$ for all $x\in \mathbb{R}$, what is $g(x)$ and $g'(x)$ than?

If you don't want to cheat use that $g(x)=g(x+1)$ for all $x$.

Answer 3

Let $x_0 \in \Bbb R$ and $\epsilon > 0$. Put $a = \lfloor x_0 - \epsilon \rfloor$ and $b = \lceil x_0 + \epsilon \rceil$. We have: $$ \forall x \in (x_0 - \epsilon, x_0 + \epsilon), \forall k \in \Bbb Z : \begin{cases} k > 10 - a & \implies x + k > a + 10 - a = 10\\ k < -10 - b & \implies x + k < b -10 - b = -10 \end{cases} $$

Hence, $g(x)$ is locally the sum of a finite number of shifts of $f$:

$$ \forall x \in (x_0 - \epsilon, x_0 + \epsilon) : g(x) = \sum_{k=-10-b}^{10-a} f(x + k) $$

Since the sum is finite and each $f(x + k)$ is differentiable, it follows that $g$ is differentiable at $x_0$. Since $x_0$ is arbitrary, it follows that $g$ is differentiable everywhere.

$g$ is a periodic function because: $$ g(x + 1) = \sum_{k \in \Bbb Z} f(x + k + 1) = \sum_{k' \in \Bbb Z} f(x + k') = g(x) $$ (where $k' = k + 1$)

By Rolle's theorem applied to $[a, a+1]$ for any $a \in \Bbb R$, $g'$ has a zero in $(a, a+1)$. Since $a$ is arbitrary, $g'$ has infinitely many zeros.

Answer 4

Put $10=:N$ and let an $R\geq1$ be given. When $|x|<R$ and $|k|>N+R$ then $$|x+k|\geq |k|-|x|>N+R-R=N\ .$$ It follows that $$f(x+k)=0 \quad\bigl(|k|>N+R)\ ,$$ so that for such $x$ we have $$g(x)=\sum_{k=-N-R}^{N+R} f(x+k)\ .\tag{1}$$ Since we now have a finite sum the value $g(x)$ is well defined; furthermore it easy to see that replacing $R$ by an $R'>R$ does not change the value of $g(x)$. As $R$ can be arbitrarily large it follows that $g$ is well defined and differentiable on all of ${\mathbb R}$.

When $|x|<R$ and $|x+1|<R$ we may use $(1)$ to compute both $g(x)$ and $g(x+1)$, and we obtain $$\eqalign{g(x+1)-g(x)&=\sum_{k=-N-R}^{N+R} f(x+1+k)-\sum_{k=-N-R}^{N+R} f(x+k) \cr &=\sum_{k'=-N-R+1}^{N+R+1} f(x+k')-\sum_{k=-N-R}^{N+R} f(x+k)\cr &=f\bigl((x+1)+N+R\bigr)-f(x-N-R)\cr &=0\ .\cr}$$ It follows that $g:\ {\mathbb R}\to{\mathbb R}$ is periodic with period $1$. Since $g$ is continuous it takes at least one $\max$ and one $\min$ per period, and this assures an infinity of zeros for $g'$.

It follows that (i) is true, and that the other proposed statements are false.