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In short, I am trying to find sufficient conditions for an approximation to the identity function $K_h$ so that, for $h$ small enough and fixed, the asymptotics at the origin of an $L^1 \cap L^2$ function $f$ with a singularity at the origin remain unchanged when convolved with $K_h$.

More precisely, let $f \in L^1 \cap L^2$ such that for some constant $0<c<1/2$ and some $\delta>0$

\begin{align} f(x) &= x^{-c} &\forall x \in (0,\delta] \, \, \, \, \, \\ f(x) &= 0 &\forall x \in [-\delta,0], \end{align}

and $K_h(x):=h^{-1}K(x/h)$ where $K$ is a band-limited kernel function (i.e. whose Fourier transform is compactly supported without loss of generality on $[-1,1]$, so $K$ cannot be compactly supported!), \begin{equation*} %\label{EqKProp} \int_{\mathbb{R}} K(x)dx=1 \qquad and \qquad \left|K(x)\right|+\left|K'(x)\right|\leq C (1+\left|x\right|)^{-\beta} \end{equation*} for some real constants $C>0$ and $\beta>2$.

The question is, is there an $h$ small enough such that (possibly under stronger conditions on $K$, respecting those above) for some constants $\varepsilon, \delta'>0$ and $0<c_1\leq c_2$

\begin{align} c_1 x^{-c} \leq &(f \ast K_h) (x) \leq c_2 x^{-c} &\forall x \in(0,\delta'] \, \, \, \, \, \, \\ -\varepsilon \leq &(f \ast K_h) (x) \leq \varepsilon &\forall x \in [-\delta', 0] ? \end{align}

Obviously the constants may depend on $h$, but if they do we would like $\delta'$ and $c_1$ to stay bounded below by some positive constant independent of $h$, and $\varepsilon $ and $c_2$ to stay bounded above by some positive constant independent of $h$.

The standard results on convolutions tell us that $\| (f\ast K_h)(x) - x^{-c} 1_{(0,\delta]} \| \to 0$ as $h\to 0$ on $[-\delta, \delta ]$ in the $L^1$ and $L^2$ norm and also uniformly on compact subsets of $(0,\delta)$ and $(-\delta,0)$, but I cannot seem to prove the above result.

In fact, I am trying to prove it for more general asymptotics, but proving it for the case above would be a good starting point. Any help or references will be greatly appreciated.

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1 Answer 1

As I argue in the next paragraph, the result I was looking for is too strong to be true. However, I do have a proof for a weaker result which essentially allows us to treat $(f\ast K_h)(x)$ as $x^{-c}1_{(0,\delta]}$ around the origin, and this is what I was looking for. I will include the weaker result and its proof when I have it written up in LaTex.

To see why the result above cannot be true note that, under its assumptions, $f\in L^1$ and $K_h\in L^{\infty}$ for any $h>0$ fixed, so $(f\ast K_h)$ is bounded and uniformly continuous (see Proposition 8.8 in Folland, Real analysis, if not familiarised with the area). Hence for any $\delta'>0$ it is impossible to find an $h>0$ and some constants $0<c_1\leq c_2$ such that $$ c_1 x^{-c} \leq (f\ast K_h)(x) \leq c_2 x^{-c} \qquad \forall x\in (0,\delta'].$$

Apologies.

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