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.