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For a large parameter $r>0$, consider the indicator function $1_{[-r,r]}$ and its convolution with the (normalized) Gaussian $\frac{1}{\sqrt{\pi}}e^{-x^2}$, that is, $$f_r(x) = \frac{1}{\sqrt{\pi}}\int_{-r}^r e^{-(x-y)^2} dy. $$ It is clear that $f_r(x) \to 1$ pointwise as $r\to \infty$. Is it true that $$\lVert f_r - 1_{[-r,r]} \rVert_{L^1} \leq C$$ for some constant $C>0$ independent of $r$? In the regions where $x\ll r$ and $x\gg r$ the above norm becomes exponentially small, but the critical area is where $x \approx r$. For example, let $0\leq a_r \leq r$ be such that $\lim_{r\to \infty} a_r \to \infty$. Then $$ \int_{\lvert x \rvert > r + a_r} \lvert f_r(x) - 1_{[-r,r]}(x) \rvert dx = \frac{1}{\sqrt{\pi}} \int_{\lvert x \rvert > r+a_r} \int_{-r}^r e^{-(x-y)^2} dy \leq C r e^{-a_r^2}, $$ and $$ \int_{r-a_r \leq \lvert x \rvert \leq r + a_r} \lvert f_r(x) - 1_{[-r,r]}(x) \rvert dx \leq 4a_r$$ and similarly for $\lvert x \rvert < r-a_r$.

Using this strategy, one can now optimize the choice of the function $a_r$ to obtain something growing slowly with $r$, but I was not able to obtain a bound independent of $r$. I am interested whether this is possible and, if not, whether there exists an optimal upper bound.

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Yes, this works, and the only ingredient we need is the estimate $\int_r^{\infty} e^{-t^2}\, dt\lesssim e^{-r^2}$.

We then have (for example) \begin{align*} \int_r^{\infty} |f_r(x)|\, dx &=\frac{1}{\sqrt{\pi}}\int_{-r}^r dy\int_r^{\infty} dx\, e^{-(x-y)^2} =\frac{1}{\sqrt{\pi}}\int_{-r}^r dy\int_{r-y}^{\infty}dt\, e^{-t^2} \\ & \lesssim \int_{-r}^r e^{-(r-y)^2}\, dy \lesssim 1 . \end{align*} Similarly, $$ \int_{-r}^r |1-f_r(x)|\, dx = \frac{1}{\sqrt{\pi}}\int_{-r}^r dx \left( \int_{-\infty}^{\infty}dy\, e^{-(x-y)^2} - \int_{-r}^r dy\, e^{-(x-y)^2} \right) , $$ and now we're back in the same situation as above.

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  • $\begingroup$ That is actually easier than what I did... somehow I overcomplicated things by introducing $a_r$. Thanks! $\endgroup$
    – Staki42
    Commented Sep 20, 2023 at 23:04

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