# If two functions are in Schwartz space,then their convolution is also in Schwartz space

For the proposition (i), I don't know how to show the inequality in the second line of the proof, can someone help me?

Notice that $$\sup_x|x|^l\cdot|g(x-y)|=\sup_t|t+y|^l\cdot|g(t)|.$$ Since $$\sup_{|t|\leqslant |y|}|t+y|^l\cdot|g(t)|\leqslant 2^l|y|^l\sup_{t}|g(t)|$$ and $$\sup_{|t|\gt |y|}|t+y|^l\cdot|g(t)|\leqslant 2^l\sup_{t}|t|^l|g(t)|,$$ we have $$\sup_t|t+y|^l\cdot|g(t)|\leqslant 2^l(1+|y|^l)\max\left\{\sup_{t}|g(t)|,\sup_{t}|t|^l|g(t)|\right\}.$$

I want to fill in the details for Davide's answer $\color{red}{\text{in red}}$.

Fix $y\in\mathbb R$.

$\color{red}{\text{Notice that}}$ $$\color{red}{\sup_{x}|x|^l\cdot|g(x-y)|=\sup_{t}|t+y|^l\cdot|g(t)|.}$$ This is true because the sets of numbers $\{|x|^l\cdot|g(x-y)|:x\in\mathbb R\}$ and $\{|t+y|^l\cdot|g(t)|:t\in\mathbb R\}$ are equal: for every element in the right hand set, we can find it in the left hand set, and vice-versa.

$\color{red}{\text{Since}}$ $$\color{red}{\sup_{|t|\leqslant |y|}|t+y|^l\cdot|g(t)|\leqslant 2^l|y|^l\sup_{t}|g(t)|}$$ We have some estimating to do. Again, $y$ is fixed. The left hand side $\sup_{|t|\leqslant |y|}|t+y|^l\cdot|g(t)|$ is the same as $\sup_{|t|\leqslant |y|}|t+y|^l\cdot\sup_{|t|\leqslant |y|}|g(t)|$ since everything is non-negative. We can estimate $\sup_{|t|\leqslant |y|}|g(t)|\leqslant \sup_{t\in\mathbb R}|g(t)|$ since the right hand sup is taken over a larger set. By the triangle inequality, $|t+y|\le |t| + |y|$, and since $l\ge 0$, the inequality $|t+y|^l\le (|t| + |y|)^l$ holds. Since we are taking the first sup over all $t$ satisfying $|t|\leqslant|y|$, we have the estimate $\sup_{|t|\leqslant|y|}(|t| + |y|)^l\leqslant\sup_{|t|\leqslant|y|}(2|y|)^l = 2^l|y|^l$ since this last expression doesn't depend on $t$. Putting these two estimates together gives the claimed estimate.

$\color{red}{\text{and}}$ $$\color{red}{\sup_{|t|\gt |y|}|t+y|^l\cdot|g(t)|\leqslant 2^l\sup_{t}|t|^l|g(t)|,}$$ This one is similar to the last one. This time, as the sup is taken over all $t>y$ [a fixed number], we have the estimate $|t+y|^l\leqslant |2t|^l=2^l|t|^l$ for every $t$. Thus we have the intermediate estimate $\sup_{|t|\gt |y|}|t+y|^l\cdot|g(t)|\leqslant 2^l\sup_{|t|>|y|}|t|^l|g(t)|$. Now, this latter expression is $\leqslant 2^l\sup_{t}|t|^l|g(t)|$, since this sup is taken over a larger set [all of $\mathbb R$. This gives us the claimed estimate.

$\color{red}{\text{we have}}$ $$\color{red}{\sup_t|t+y|^l\cdot|g(t)|\leqslant 2^l(1+|y|^l)\max\left\{\sup_{t}|g(t)|,\sup_{t}|t|^l|g(t)|\right\}.}$$ Since everything is non-negative, we can break up the left hand sup as $$\sup_t|t+y|^l\cdot|g(t)| \leqslant \sup_{|t|\leqslant |y|}|t+y|^l\cdot|g(t)|+\sup_{|t|\gt |y|}|t+y|^l\cdot|g(t)|.$$ Using our previous two estimates, we have $$\sup_{|t|\leqslant |y|}|t+y|^l\cdot|g(t)|+\sup_{|t|\gt |y|}|t+y|^l\cdot|g(t)|\leqslant 2^l|y|^l\sup_{t}|g(t)|+2^l\sup_{t}|t|^l|g(t)|.$$ The end's in sight now. Replace both $\sup_{t}|g(t)|$ and $\sup_{t}|t|^l|g(t)|$ by their maximum and factor it out to get the final claimed inequality: $$2^l|y|^l\sup_{t}|g(t)|+2^l\sup_{t}|t|^l|g(t)| \leqslant 2^l(1+|y|^l)\max\left\{\sup_{t}|g(t)|,\sup_{t}|t|^l|g(t)|\right\}.$$

• In the last part I do not catch why you can consider that the supremum of a set A is equal to the sum of the supremum of two sets B and C such that A = B U C ?. The other question y why it is necessay the explanation of the last red inequality, I am reasoning as follows: One we get the previous two inequalities we have two bounds for the sets B and C, now why isn't correct that the maximum of such bounds is a bound for A = B U C ? – Ale.B Jul 18 '18 at 20:56
• In the previous comment A is referring to the supremum taken all over the reals, B with $t >|y|$ and C for $t<|y|$ – Ale.B Jul 18 '18 at 20:59
• @Ale.B: Nice catch. You're right that it should be inequality, not equality. To your second point, I think you're right that we can just take the maximum instead of using the sum as an upper bound, but the goal is to get an upper bound of the form $A_ℓ(1+|y|)^ℓ$, which is not evidently the form that the maximum has unless I am missing something. – Alex Ortiz Jul 18 '18 at 21:28
• Respect to the second point, I think you are right because I was thinking something like this: Once you proved the first two inequalities the argument follows as: $sup |t+y|^l |g(t)| \leq 2^l max \{|y|^l sup |g(t)|,sup |t|^l |g(t)| \}$ and now, to finish, we need to use that this maximum is less than the sum of the two elements which is the same that you do. – Ale.B Jul 18 '18 at 22:11
• I think @Davide 's answer is neat and complete, i.e. there is no need to add anything to it. – Sam Wong Sep 27 '18 at 2:48

Consider the hints from the authors:

When $\left\vert x \right\vert \leq 2\left\vert y \right\vert$, then \begin{equation*} \left\vert x \right\vert^\ell\left\vert g(x-y) \right\vert\leq C_0\left\vert 2y \right\vert^\ell\leq2^\ell C_0 (1+\left\vert y \right\vert)^\ell, \end{equation*} where $C_0\geq \sup\left\vert g(x-y) \right\vert$.

When $\left\vert x \right\vert > 2\left\vert y \right\vert$, then \begin{align*} \left\vert x \right\vert^\ell\left\vert g(x-y) \right\vert &= \left\vert (x-y) + y \right\vert^\ell\left\vert g(x-y) \right\vert\\ &=\left\vert (x-y) \right\vert^\ell \left\vert 1 + \frac{\left\vert y \right\vert}{\left\vert x-y \right\vert} \right\vert^\ell\left\vert g(x-y) \right\vert\\ &\leq\left\vert (x-y) \right\vert^\ell \left\vert 1 + \frac{\left\vert y \right\vert}{\left\vert x \right\vert - \left\vert y \right\vert} \right\vert^\ell\left\vert g(x-y) \right\vert\\ &\leq\left\vert (x-y) \right\vert^\ell \left\vert 1 + \frac{\left\vert y \right\vert}{\left\vert 2y \right\vert - \left\vert y \right\vert} \right\vert^\ell\left\vert g(x-y) \right\vert\\ &=\left\vert (x-y) \right\vert^\ell 2^\ell\left\vert g(x-y) \right\vert\\ &\leq 2^\ell\left\vert (x-y) \right\vert^\ell C_{\ell}\left\vert x-y \right\vert^{-\ell}\\ &\leq 2^\ell C_{\ell}(1+\left\vert y\right\vert)^{\ell}, \end{align*} where $C_\ell\geq \sup \left\vert (x-y) \right\vert^\ell\left\vert g(x-y) \right\vert$.

Now let $A_\ell = 2^\ell\max\{C_0, C_\ell\}$.