# Need some explanation to this Problem.

I was reading a theorem which says :

If $f \in L^p(\mathbb{R}^n)$ and $\Phi\in C^\infty(\mathbb{R}^n)$ then $f_\epsilon= \Phi_\epsilon \star f$ tends to $f$ as $\epsilon \to 0$.

(here, page number 2). But I am not able to understand the steps. Can anyone explain it to me? I understood the procedure but I am not able to understand the steps. Thanks.

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Which step is problematic to you? How far do you understand the proof? – abatkai May 21 '12 at 20:06
I am not able to understand how the smooth function is convolved ,the first step itself i am not able to understand . I can corelate it with the standard convolving procedure. Can u help me? :) – Theorem May 21 '12 at 20:11
Page 3, first line. Is it ok? – abatkai May 21 '12 at 20:20
Prof, No. How did that $a$ term disappear inside the integral ? – Theorem May 21 '12 at 20:24
Think of it as $a=1$ for now, but – abatkai May 21 '12 at 20:26

## 1 Answer

Using the definition of the constant $a$ and of the convolution product, we have that \begin{align*} (f \ast \varphi_t)(x) - af(x) & = (f \ast \varphi_t)(x) - f(x)\int_{\mathbb{R}^n} \varphi(y) ~dy \\ & = \int_{\mathbb{R}^n} f(x - y)\varphi_t(y) ~dy - \int_{\mathbb{R}^n} f(x)\varphi(y) ~dy. \end{align*} Now in the first integral, make the change of variables $y \mapsto tz$. Then $dy \mapsto t^n ~dz$ (because $tz = (tz_1, \dots, tz_n)$ and $dz$ is really shorthand for $dz_1 \cdots dz_n$ as we are working in $\mathbb{R}^n$) and $$\varphi_t(tz) = t^{-n} \varphi\left( \frac{tz}{t} \right) = t^{-n}\varphi(z).$$ Hence the first integral becomes \begin{align*} \int_{\mathbb{R}^n} f(x - y)\varphi_t(y) ~dy & = \int_{\mathbb{R}^n} f(x - tz)t^{-n}\varphi(z)t^n ~dz \\ & = \int_{\mathbb{R}^n} f(x - tz)\varphi(z) ~dz. \end{align*} So, returning to our original calculation, we have \begin{align*} (f \ast \varphi_t)(x) - af(x) & = \int_{\mathbb{R}^n} f(x - y)\varphi_t(y) ~dy - \int_{\mathbb{R}^n} f(x)\varphi(y) ~dy \\ & = \int_{\mathbb{R}^n} f(x - tz)\varphi(z) ~dz - \int_{\mathbb{R}^n} f(x)\varphi(z) ~dz \\ & = \int_{\mathbb{R}^n} (f(x - tz) - f(x))\varphi(z) ~dz \\ & = \int_{\mathbb{R}^n} ((\tau_{tz}f)(x) - f(x))\varphi(z) ~dz \\ & = \int_{\mathbb{R}^n} (\tau_{tz}f - f)(x)\varphi(z) ~dz. \\ \end{align*}

Now $\|\tau_{tz}f - f\|_p \longrightarrow 0$ by "Exercise I.6" in the notes you linked. The rough idea of the proof is to use the density of $C_c(\mathbb{R}^n)$ (compactly supported continuous real-valued functions on $\mathbb{R}^n$) in $L^p(\mathbb{R}^n)$. Given $\varepsilon > 0$, let $g \in C_c(\mathbb{R}^n)$ be such that $$\|f - g\|_p < \varepsilon.$$ Clearly since $g$ is continuous, for $t$ small enough $$\|\tau_{tz}g - g\|_p < \varepsilon.$$ Now we have that \begin{align*} \|\tau_{tz}f - f\|_p & = \|\tau_{tz}f - \tau_{tz}g + \tau_{tz}g - g + g - f\|_p \\ & \leq \|\tau_{tz}f - \tau_{tz}g\|_p + \|\tau_{tz}g - g\|_p + \|g - f\|_p \\ & \leq \|\tau_{tz}\|_{\text{op}}\|f - g\|_p + \|\tau_{tz}g - g\|_p + \|f - g\|_p \\ & = \|f - g\|_p + \|\tau_{tz}g - g\|_p + \|f - g\|_p \\ & < 3\varepsilon \end{align*} for $t$ small enough. Therefore $$\|\tau_{tz}f - f\|_p \longrightarrow 0 \text{ as } t \longrightarrow 0.$$ Since $$\|\tau_{tz}f - f\|_p \leq \|\tau_{tz}f\|_p + \|f\|_p = 2\|f\|_p$$ and $$\|\tau_{tz}f - f\|_p \longrightarrow 0 \text{ as } t \longrightarrow 0,$$ the hypotheses of the Dominated Convergence Theorem are satisfied and hence $$\|f \ast \varphi_t - af\|_p \longrightarrow 0 \text{ as } t \longrightarrow 0,$$ or in other words $f \ast \varphi_t \longrightarrow af$ in $L^p(\mathbb{R}^n)$ as $t \longrightarrow 0$.

Addendum: $\tau_{tz}$ is the translation operator $$\tau_{tz}: L^p(\mathbb{R}^n) \longrightarrow L^p(\mathbb{R}^n),$$ $$(\tau_{tz}f)(x) = f(x - tz).$$ To see why it has norm $1$, note that \begin{align*} \|\tau_{tz}\|_{\text{op}} & = \sup_{\|f\|_p = 1} \|\tau_{tz}f\|_p \\ & = \sup_{\|f\|_p = 1} \left(\int_{\mathbb{R}^n} |(\tau_{tz}f)(x)|^p ~dx\right)^{1/p} \\ & = \sup_{\|f\|_p = 1} \left(\int_{\mathbb{R}^n} |f(x - tz)|^p ~dx\right)^{1/p} \\ & = \sup_{\|f\|_p = 1} \left(\int_{\mathbb{R}^n} |f(u)|^p ~du\right)^{1/p} \\ & = \sup_{\|f\|_p = 1} \|f\|_p \\ & = 1. \end{align*} Above I used the change of variables $x - tz \mapsto u$. So $\|\tau_{tz}\|_{\text{op}} = 1$ is really just a consequence of the translation invariance of the Lebesgue measure.

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Thanks a TOOOOOOON !! – Theorem May 21 '12 at 20:35
@ Henry I was looking at the proof of denseness of $C^\infty$ in $L^p$ ( page 4) . i couldn't understand . Can u explain? – Theorem May 21 '12 at 21:17
@Ananda: I think that perhaps you should post your question about page 4 as a separate question, and try to specify what exactly you don't understand about it. In any case, I can help explain that proof. – Henry T. Horton May 21 '12 at 21:40
Thank you for suggestion :) – Theorem May 21 '12 at 21:42
T.Horton what is actually $||\tau_{tz}||$ and why is it's norm $1$. – Theorem May 22 '12 at 10:50