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I have a convolution operator $T$ in $L^1(\mathbb R)$ defined as $T:f \rightarrow f*g$ , for some $g \in L^1(R)$.
I need to prove that $||T||_{L^1}=||g||_{L^1}$.
I have found such problem here: Limit of convolution and i've tried to solve mine in a similar way, but i have some trouble with it.
As i understand it, this solution uses the fact, that $g$ is non-negative to approximate an integral.
Here is what @Martín-Blas Pérez Pinilla states:
"Namely, if $\epsilon>0$, then $\exists n_\varepsilon\in{\Bbb N} $ s.t. for $x\in[-n_\varepsilon,n_\varepsilon]$: $\|g\|_1-\epsilon/2\le\int_{-n_\epsilon}^{n_\epsilon}g" $
I understand how we can do this for positive function and i assume that a similar approach can be used for negative functions. But when $Im(g) \in R$ such approximation can't be done, if I'm not mistaken. It seems to me that in such case it still can be showed that $||f*g||_{L^1}\geq||g||_{L^1}-\epsilon$. I've tried to represent $g$ as difference of two positive functions i.e. $g^+$ and $g^-$ such that $g=g^+-g^-$, but when I try to estimate $||f*(g^+-g^-)||_{L^1}$, i can't come up with a desired inequality.
Would be very glad for any help!

Noob mathematician
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Антон Рябченко
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1 Answers1

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You don't need to worry about these decompositions. A standard mollification argument can be used here.

Let $\varphi:\mathbb{R}\rightarrow [0,\infty)$ be a smooth, compactly supported function such that $\int\varphi=1$. For $\epsilon>0$ define $$ \varphi_\epsilon(x)=\frac{1}\epsilon\varphi(x/\epsilon). $$ Note that $\int\varphi_\epsilon=1$ and we have \begin{align*} |T\varphi_\epsilon(x)-f(x)|&=|\int (f(x-y)-f(x))\varphi_\epsilon(y)\,dy|\le \int |f(x-y)-f(x)|\varphi_\epsilon(y)\,dy\\ &=\int |f(x-\epsilon y)-f(x)|\varphi(y)\,dy. \end{align*} Consequently, \begin{align*} \|T\varphi_\epsilon-f\|_{L^1}&\le \int \|f(\cdot-\epsilon y)-f(\cdot)\|_{L^1}\varphi(y)\,dy\longrightarrow 0, \quad\textrm{as}\quad\epsilon\rightarrow 0. \end{align*} In particular it implies \begin{align*} \|T\|\ge \lim_{\epsilon\rightarrow 0}\|T\varphi_\epsilon\|_{L^1}=\|f\|_{L^1}. \end{align*}

Tony419
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  • Thank you for your answer. I have just a couple of questions here: What means compactly supported function? Am i right, that there is a typo and $\epsilon$>0? Did you forget about the modulus in the first integral? And i don't understand why $||T||_{L^1} \geq limit$, why is it implied? – Антон Рябченко May 14 '20 at 18:59
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  • it should be $\epsilon>0$ and I forgot about a modulus, I corrected it. Thanks. 2) Observe that $|T|=\sup_{|g|{L^1}=1} |Tg|{L^1}$ so in particular $|T|\ge \sup_{\epsilon}|T\varphi_\epsilon|{L^1}$, since all $\varphi\epsilon$ have $L^1$-norm $1$, and $\sup$ is always greater than a limit.
    • – Tony419 May 14 '20 at 19:07
  • All right, now i get it. And the last question about compactly supported function. Am i right that this means, that function is non-zero only on some compact? – Антон Рябченко May 14 '20 at 19:10
  • And one question about that inequality for norm. Why is it less than integral? Am i right, that we used Fubini's theorem for it? – Антон Рябченко May 14 '20 at 19:13
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    A function is compactly supported if it's zero outside a compact set (compact means essentially closed and bounded; you can think that it is zero outside the interval $(0,1)$). As to inequality involving a norm, it is moving the absolute value inside the integral and then applying Fubini-Tonelli theorem. – Tony419 May 14 '20 at 19:31
  • Now i understand everything. Thank you very much for your answer! – Антон Рябченко May 14 '20 at 20:22
  • You're welcome :] – Tony419 May 14 '20 at 20:32