My Question is regarding the answer in the following question Deriving the Oversampling formula
To be more specific let $g(\xi) = \underbrace{\frac{1}{2a} \chi_{[-a,a]} \ast...\ast \frac{1}{2a} \chi_{[-a,a]}}_{k-\text{times}} \ast \chi_{[-B-ka,B+ka]}$
Let $B'=B+2ka$, I want to show that
$$g(\xi)=0\qquad \text{if }\ \ |\xi|>B',$$
and
$$g(\xi)=1\qquad \text{if }\ \ |\xi|\leq B.$$
In the Question mentioned above there was only an intuitive explaination given (and linked). I would like to do it "by calculation".
For the first part, I was able to show that for two functions $f$ and $g$ with $\operatorname{supp}(f)=I_1$ and $\operatorname{supp}(g)=I_2$ we get $\operatorname{supp}(f \ast g) \subseteq I_1+I_2$. By using this we get $\operatorname{supp}(g)\subseteq [-B',B'] $.
The part where I am struggling at the moment is to show that $g(\xi)=1$ if $|\xi|<B$.
I have tried raw calculation but this didn't lead anywhere.
Denote $g_1=\frac1{2a}\chi_{[-a,a]} \ast \chi_{[-B-ka,B+ka]}$, $g_2=\frac1{2a}\chi_{[-a,a]} \ast g_1$ and so on: $g_{j+1}=\frac1{2a}\chi_{[-a,a]} \ast g_j$ for $1\leq j\leq k-1$, then $g=g_k$.
We prove this by induction. For $|\xi|\leq B+(k-1)a$, we have $$g_1(\xi)=\int_{-a}^a\frac1{2a}\chi_{[-B-ka,B+ka]}(\xi-\eta)\,d\eta.$$ For $|\xi|\leq B+(k-1)a$ and $|\eta|\leq a$, we have $|\xi-\eta|\leq B+ka$, hence $g_1(\xi)=\int_{-a}^a\frac1{2a}\,d\eta=1$. This verifies the claim for $j=1$.
Assume that the claim holds for $j\geq1$, we consider the $j+1$ case. For $|\xi|\leq B+(k-j-1)a$, we have $$g_{j+1}(\xi)=\int_{-a}^a\frac1{2a}g_{j}(\xi-\eta)\,d\eta.$$ For $|\xi|\leq B+(k-j-1)a$ and $|\eta|\leq a$, we have $|\xi-\eta|\leq B+(k-j)a$, hence by induction hypothesis we have $g_j(\xi-\eta)=1$ and thus $g_{j+1}(\xi)=\int_{-a}^a\frac1{2a}\,d\eta=1$.
By induction, the claim holds. Taking $j=k$ in the claim gives the desired result.