# $L^1$ error between indicator function and smoothed out version

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.

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