# Understanding a step in Stein's proof that the Schwartz space is closed under convolution.

I am looking at Stein's proof that if $f,g \in \mathcal{S}(\mathbb{R})$, then so is $f*g$. The definition of the Schwartz space on $\mathbb{R}$ given is, the set of all indefinitely differentiable functions $f$ so that $f$ and all its derivatives are rapidly decreasing, in the sense that $$\sup_{x\in \mathbb{R}}|x|^k |f^{(l)}(x)| < \infty \; \text{for every}\; k,l\ge 0.$$ In the proof below, I have four questions.

First, how can I check the assertion that we have $\sup_x |x|^l |g(x-y)|\le A_l (1+|y|)^l$? I tried considering the two cases, but I cannot figure out how to get this bound. In fact, I don't really understand why we need to consider two separate cases. Shouldn't we be able to get $|x|^l |g(x-y)|\le |x-y|^l|g(x-y)|+|y|^l|g(x-y)|\le (1+|y|^l)A_l\le(1+|y|)^l A_l$, where $A_l=\sup_x |x|^l |g(x)|$?

Next, how can we guarantee that the integral $\int |f(y)|(1+|y|)^l dy$ is bounded for every $l\ge 0$?

Finally, in the last paragraph, how is the interchange of differentation and integration justified in the case by the rapid decrease of $dg/dx$? Also, why do we only take the differentiation on $g$ and not $f$? I am not aware of this kind of theorem.

I would greatly appreciate any hints or solutions.

1. For your proof $$|x|^l |g(x-y)|≤|x-y|^l |g(x-y)| + |y|^l |g(x-y)|$$ is incorrect, the triangle inequality doesn't hold for $|\cdot|^l$. Consider ie $x=2,y=1$, $l=2$ you have $|2|^2=4 \not≤|2-1|^2+|1|^2=2$.

I'm not sure how the hint should be used, but you can also do it like this: $$|x-y|^l|g(x)|≤\sum_kC_k |x|^{l-k}|y|^k|g(x)|≤A_l(1+|y|)^l$$ Where the Schwartz property of $g$ was used to bound $|x|^{l-k}|g(x)|$ in each summand. Then $$\sup_{x}|x|^l |g(x-y)|=\sup_{x} |x+y|^l|g(x)|$$ and the result follows.

2. The integral is bounded because $$|f(y)|(1+|y|^l)(1+|y|^{2})≤C$$ follows from the Schwartz condition. Then you have since $|f|(1+|y|^l)$ is measurable and pointwise dominated by $\frac C{1+|y|^2}$ it is integrable.

3. $$\frac{d}{dx}\int f(y)g(x-y)dy=\frac{d}{dx}\lim_{n\to\infty}\int_{-n}^n f(y)g(x-y)dy$$ We want to swap the limit with the derivative, this is possible if $\int_{-n}^n f(y)g(x-y) dy$ and $\int_{-n}^n f(y) g'(x-y) dy$ converge uniformly on compacta (wrt $x$). Here: $$\left|\int_{\mathbb R} f(y)g(x-y) dy -\int_{-n}^n f(y) g(x-y)dy\right|≤\left(\int_{-\infty}^{-n}+\int_n^\infty\right)|f(y)g(x-y)|dy$$ From the Schwartz condition bound $g$ by some constant and bound $f(y)$ by $\frac c{1+y^2}$ to see that the expression goes to zero uniformly. One does essentially the same for the derivative integral. This gives that $$\frac{d}{dx}\int f(y) g(x-y) dy = \lim_{n\to\infty}\frac{d}{dx}\int_{-n}^n f(y) g(x-y) dy = \lim_{n\to\infty}\int_{-n}^n f(y) g'(x-y) dy\\ =\int f(y) g'(x-y)dy$$ From the rule of derivation under the integral. Proceed via induction to get it for $\frac{d^k}{dx^k}$.
4. The expression $\frac{d^k}{dx^k}(f*g)=(f*g^{(k)})$ is useful because we know already that for any Schwartz $f,g$: $$\sup_{x}|x|^l |f*g|(x)<\infty$$ Thus, since $g^{(k)}$ is Schwartz, you have: $$\sup_{x}|x|^l|(f*g)^{(k)}|(x)=\sup_{x}|x|^l |f*g^{(k)}|(x)<\infty$$
• Does the binomial expansion hold on the absolute power as well? I mean, I know $(a+b)^l=\sum a^k b^{l-k}$, but I'm not sure how I can use it in this case $|a+b|^l$? Sep 25, 2016 at 16:50
• @takecare $|a+b|^l=\left|\sum_k C_k a^kb^{l-k}\right|≤\sum_k C_k |a|^k|b|^{l-k}$ from the triangle inequality. Sep 25, 2016 at 17:10
• Why do we have the first equality? The right side is the $|(a+b)^l|$ while the left side is $|a+b|^l$. I'm not sure why the two are the same. Shouldn't we argue, $|a+b|^l\le (|a|+|b|)^l=\sum_k C_k |a|^k|b|^{l-k}$ instead? Sep 25, 2016 at 17:20
• @takecare the absolute value is multiplicative, meaning $|a|\cdot |b|=|a\cdot b|$, so $|(a+b)^l|=|a+b|^l$ :) Sep 25, 2016 at 17:24

The other answers depend on the assumption that $$l$$ is an integer. There is nothing wrong with this assumption per se, since at the end of the day, we only need to consider integer values of $$l$$ for $$\mathcal{S}(\mathbb{R})$$.

But we can also easily prove the relationship $$\sup_x |x|^l |g(x-y)| \le A_l (1+|y|)^l$$ for any $$l \ge 0$$ in the following way.

If $$|x| \le 2 |y|$$, we have $$\sup_x |x|^l |g(x-y)| \le |2y|^l \sup_x |g(x-y)| \le M \frac{|2y|^l}{(1+|y|)^l} (1+|y|)^l = A_l (1+|y|)^l$$ where $$A_l \equiv M \frac{|2y|^l}{(1+|y|)^l}$$.

If $$|x| \ge 2 |y|$$, we have $$|x-y| \ge |y|$$, and hence $$|𝑥| \le (|𝑥−𝑦|+|𝑦|)≤2|𝑥−𝑦|$$. This gives us $$\sup_x |x|^l |g(x-y)| \le 2^l \sup_x |x-y|^l |g(x-y)| \le 2^l A'_l = A_l$$ where $$A_l \equiv 2^l A'_l$$.

• (I'm impressed that an Axolotl can type!) Jan 22, 2022 at 1:29

Since this case is about $\mathbb{R}$, follow your idea, we consider three cases, $l=0,l=1,l>1$.

$l=0$ case is trivial. $l=1$ case is indeed the triangle inequality, i.e. $$|x|=|x-y+y|\leq |x-y|+|y|$$.

For case $l>=2$, we can show that $$\left|\frac{x-y+y}{2}\right|^l\leq \frac{|x-y|^l+|y|^l}{2}$$, by the convexity of $x^l$.

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Directly, we can apply Minkowski inequality. (which can extend your proof to $\mathbb{R}^n$)