$\newcommand{\R}{\mathbb{R}}$$\newcommand{\diff}{\mathrm{d}}$Let $\rho\in C^\infty_c(\mathbb R^d;\mathbb R)$ be an even function, i.e. $\rho(-x)=\rho(x)$ for all $x\in\mathbb R^d$ that furthermore satisfies $\operatorname{supp}\rho\subset B(0,1)$ and
$$\int_{\mathbb R^d}\rho(x)\,\mathrm{d} x=1.$$
The function is not necessarily non-negative (which then implies $\int |\rho(x)|\, \diff x\geq 1$). We define
$$\rho^{(n)}(x)=2^{2d}\rho(2^nx).$$
Basically $\rho^{(n)}$ is somehow an approximation to the identity. We also define
$$\eta^{(n)}=\rho^{(0)}*\rho^{(1)}*\rho^{(2)}*\cdots * \rho^{(n)},$$
where $*$ is the convolution operator.
Problem. Show that $\eta^{(n)}$ converges in $C(\mathbb R^d)$ uniformly.
Attempt. The book (see Motivation section below) gives the hint to prove the limit exists in Fourier space. I imagine that they mean that I have to prove $\mathcal F\eta^{(n)}\to \hat\eta$ in the Schwartz space $\mathscr S(\mathbb R^d)$ and use continuity of the Fourier transform (-/inversion) from $\mathscr S(\mathbb R^d)$ to $\mathscr S(\mathbb R^d)$. So I did that and in particular I wanted to show that $\eta^{(n)}$ is Cauchy. So let $m\geq n$ and look at
$$(\mathcal F(\eta^{(m)}-\eta^{(n)}))(\xi)=\prod_{i=0}^m (\mathcal F\rho^{(i)})(\xi)-\prod_{i=0}^n (\mathcal F\rho^{(i)})(\xi)=\prod_{i=0}^n (\mathcal F\rho^{(i)})(\xi)\left(\left[\prod_{i=n+1}^m (\mathcal F\rho^{(i)})(\xi)\right]-1 \right).$$
It is not difficult to see that
$$(\mathcal F\rho^{(i)})(\xi)=(\mathcal F\rho)(2^{-i}\xi).$$
So in particular as $i\to\infty$ we easily see that $$\mathcal F\rho^{(i)}\to \mathcal F\rho(0)=\int\rho(x)\,dx= 1$$ in $\mathcal S'(\mathbb R^d)$, i.e. convergence as in tempered distribution. After this I was wishing the product written above also would converge to $1$, so I wrote
$$(\mathcal F(\eta^{(m)}-\eta^{(n)}))(\xi)=\prod_{i=0}^n(\mathcal F\rho)(2^{-i}\xi)\left( \left[\prod_{i=n+1}^m(\mathcal F\rho)(2^{-i}\xi)\right]-1\right).$$
I felt that this is just as stiff, because I do not know the rate for which the convergence happens. I was guessing through Taylor approximation $$(\mathcal F\rho^{(i)})(\xi)-1=(\mathcal F\rho)(2^{-i}\xi)-1\approx \xi 2^{-i},$$
but then $\xi$ is a polynomial so it does not belong in $\mathscr S(\mathbb R^d)$.
After this I tried showing it directly, something like
$$|\eta^{(n)}(x)-\eta^{(n+1)}(x)|\leq C 2^{-n}$$
uniformly, so that it is Cauchy in $C(\mathbb R^d)$. One thing one easily sees is that $\rho^{(n)}$ has support inside $B(0,2^{-n})$ and therefore $\eta^{(n)}$ has support inside
$$B(0,1)+B(0,2^{-1})+B(0,2^{-2})+\cdots + B(0,2^{-n})\subset B(0,3) $$ for all $n\in\mathbb N$ (by the support property of the convolution). In this case it is actually enough to show uniform convergence in say $C(\overline{B(0,3};\mathbb R)$. I could not show that either.
I appreciate any hint, or any methods to attack this problem.
Motivation. I am reading the book "A course on Rough Paths" by Friz and Hairer and I am struggeling in proving the previous result (Exercise 13.7, second edition) which is needed for the proof of the so-called reconstruction theorem (Theorem 13.26). This question is somehow independent of the context, because it serves as a useful tool from basic analysis.
Edit (Added attempt). Let us use the notation
$$\rho^{(n,m)}=\rho^{(n)}*\rho^{(n+1)}*\cdots *\rho^{(m)}.$$
Basically $\eta^{(n)}=\rho^{(0,n)}$. This notation is somehow easier for the things we will do soon.
We want to show that $(\eta^{(n)})_{n\in\mathbb N}$ is Cauchy in $C_b(\R^d)$. To that end we want to control
$$\eta^{(n+1)}-\eta^{(n)}=\rho^{(0,n+1)}-\rho^{(0,n)}=\rho^{(0,n)}*(\rho^{(n+1)}-\delta_0)=\rho^{(1,n)}*(\rho^{(n+1)}-\delta_0)*\rho.$$
Let us focus on the term coming after $\rho^{(1,n)}$, we have
$$((\rho^{(n+1)}-\delta_0)*\rho)(x)=\int \rho^{(n+1)}(y)(\rho(x-y)-\rho(x))\,\diff y.$$
We have by the Mean value theorem
$$|\rho(x-y)-\rho(x)|\leq \sup_{z\in \mathbb R^d} |D\rho(z)||y|$$
so that uniformly over $y\in\R^d$
$$|\rho(x-y)-\rho(x)|\leq C|y|.$$
Hence
$$|((\rho^{(n+1)}-\delta_0)*\rho)(x)|\leq C\int_{\R^d} |\rho^{(n+1)}(y)||y|\,\diff y=C2^{-n-1}\int_{\R^d} 2^{(n+1)d}|\rho(2^{(n+1)}y)||2^{(n+1)}y|\,\diff y. $$
We change variables to get
$$|((\rho^{(n+1)}-\delta)*\rho)(x)|\leq C2^{-n-1}\int_{\R^d} |\rho(y)||y|\,\diff y \leq \tilde C 2^{-n}.$$
Let us back to what we had
$$(\eta^{(n+1)}-\eta^{(n)})(z)=\int_{\R^d}\rho^{(1,n)}(z-x) \int_{\R^d} \rho^{(n+1)}(y)(\rho(x-y)-\rho(x))\,\diff y\,\diff x.$$
Bounding all this
$$|(\eta^{(n+1)}-\eta^{(n)})(z)|\leq \tilde C2^{-n} \int_{\R^d}|\rho^{(1,n)}(x)|\,\diff x.$$
Life would be much easier if we could bound the integral on the RHS. It is not difficult to see (through induction) that
$$\int_{\R^d}|\rho^{(1,n)}(x)|\,\diff x=\int_{\R^d}|\rho^{(0,n-1)}(x)|\,\diff x=\int_{\R^d}|\eta^{(n)}(x)|\,\diff x.$$
So it is enough to show that $\eta^{(n)}$ is bounded in $L^1(\R^d)$. I could not show that… According to the exercise it actually says that the integral is actually bounded. It is also enough to show that the Fourier transform $\mathcal F\eta^{(n)}$ is bounded in $L^1(\R^d)$. So here is where I am stuck now. Note. that $\rho$ is not necessarily non-negative, otherwise it was easy.
Edit 2. Since every little piece helps, let me add this. Define
$$A_n:=\int_{\R^d}|\rho^{(0,n)}(x)|\,\diff x=\int_{\R^d}|\eta^{(n)}(x)|\,\diff x.$$
Recall we ended the previous edit with the observation that it is enough to show $\sup_n A_n<\infty$. It is not difficult to find a recurrence relation for $A_n$, namely
$$A_{n+1}\leq \int_{\R^d}|\eta^{(n+1)}(x)-\eta^{(n)}(x)|\,\diff x+\int_{\R^d}|\eta^{(n)}(x)|\,\diff x\leq C' 2^{-n} A_{n-1}+A_n, $$
where $C'$ is some constant depending on $\tilde C$ above and the volume of $B(0,3)$. This recurrence is promising except that it doesn't lead to anywhere if we bound $A_n\leq A_0A_{n-1}$ through Young's inequality because $A_0=\int |\rho(x)|\,\diff x\geq 1$.
Remark. On the other hand if $\rho(x)\geq 0$ then necessarily $A_0=1$ and the above argument works. This is why the case $\rho(x)\geq 0$ is easy.
Best Answer
You have already done most of the work, well done!
I think it is possible to show that $A_n$ is bounded.
Let us simplify the recurrence relation. We define $B_n := \max_{k=0,\ldots,n} A_k$. Then one can obtain the relation $$ B_{n+1} \leq C' 2^{-n} B_{n-1} + B_n \leq C' 2^{-n} B_{n} + B_n = B_n (1+C' 2^{-n}). $$ It suffices to show that $B_n$ is bounded. From this recurrence relation we obtain $$ B_{n+1} \leq B_1 \prod_{k=1}^n (1 + C' 2^{-k}). $$ Taking logarithms yields $$ \log(B_{n+1}) \leq \log(B_1) + \sum_{k=1}^n \log(1+C' 2^{-k}) \leq \log(B_1) + \sum_{k=1}^n C' 2^{-k} \leq \log(B_1) + C'. $$ Thus, the $\log(B_{n+1})$ and therefore $B_n$ is bounded from above.