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An approximate identity is a function $\phi_{\epsilon} \in L^1(\mathbb{R}^n)$ with the following properties:

(a) $\| \phi_{\epsilon} \|_{L^1(\mathbb{R}^n)} \leq c$ for some constant $c>0$.

(b) $\int_{\mathbb{R}^n} \phi_{\epsilon} dx =1$.

(c) for $\delta >0$ it holds that $\int_{|x| \geq \delta} |\phi_{\epsilon} (x) | dx \rightarrow 0$ as $\epsilon \rightarrow 0$.

I would like to find an approximate identity for any function $\phi$ with the property $\int_{\mathbb{R}} \phi(x) dx=1$. I therefore defined

$\phi_{\epsilon}(x)=\frac{1}{\epsilon} \phi(\frac{x}{\epsilon}).$

I would now like to show the three properties stated above for my newly defined function $\phi_{\epsilon}$. I’m only having troubles with (a).

For (a), I just computed the integral and with substitution I derived that $\|\phi_{\epsilon}\|_{L^1(\mathbb{R})} =\|\phi\|_{L^1(\mathbb{R})}$.Is that enough? I know that $\int_{\mathbb{R}} \phi(x) dx=1$ by assumption. But does this also imply that $\phi \in L^1(\mathbb{R})$?

Both (b) and (c) also follow easily by just computing the integrals and using the substitution $y=x/\epsilon$.

Rócherz
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Andreas804
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  • You may want to additionally suppose that $\phi\in L^1(\mathbb{R})$, then $\phi_\varepsilon$ is integrable as well due to the fact that $\Vert \phi_\varepsilon \Vert_{L^1(\mathbb{R})} = \Vert \phi \Vert_{L^1(\mathbb{R})}$. From only $\int_\mathbb{R} \phi_\varepsilon dx = 1$ you cannot infer integrability of $\phi_\epsilon$. – stange Nov 06 '23 at 17:48
  • That is exactly what I thought! So there may be a mistake in the exercise then? (Or we could also suppose that $\phi \geq 0$) – Andreas804 Nov 06 '23 at 17:50
  • You could require that $\phi \geq 0$, but usually one supposes that $\phi\in L^1(\mathbb{R})$. – stange Nov 06 '23 at 18:16
  • Depending on what your prevailing paradigm is, $\int_\mathbb{R} \phi = 1$ may actually require/imply $\phi \in L^1$. For instance, in the context of the Lebesgue integral, $\int_\mathbb{R} \phi$ doesn't exist as a real number unless $\phi \in L^1$. Alternatively, in the context of singular integrals, we no longer have that implication. In that sense, omitting $\phi \in L^1$ not necessarily technically a mistake, but it's a bit sloppy. – Brian Moehring Nov 06 '23 at 18:31
  • Thank you! Could you clarify why the existence of the integral $\int_\mathbb{R} \phi$ would imply $ \phi \in L^1$ the context of the Lebesgue integral? – Andreas804 Nov 07 '23 at 12:06
  • @Andreas804 Go back to the definition of the Lebesgue integral. You write $\phi = \phi^+ - \phi^-$ where $\phi^+,\phi^- \geq 0$ and $\phi^+\phi^-=0$. Then $\int\phi^+, \int\phi^- \in [0,\infty]$. If both are infinite, $\int\phi$ doesn't exist. Otherwise, $\int\phi := \int\phi^+ - \int\phi^-$. Assuming this difference exists and is finite implies both integrals we're subtracting exist and are finite. Then just notice $\int|\phi| = \int\phi^+ + \int\phi^-$ – Brian Moehring Nov 09 '23 at 20:09

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