Determine $\lim_{h\rightarrow0}\sup_{f\in X}\|f(x+h)-f(x)\|_{L^2(\mathbb R)}$ for $f_n=\sqrt{n}\,\mathbb 1_{(0,1/n)}$

Question

Let $f_n=\sqrt{n}\,\mathbb 1_{(0,1/n)}$ and $X=\left\{f_n|\ n\in\mathbb N\right\}$. How can you show that $$ \lim_{h\rightarrow0}\sup_{f\in X}\left\|f(x+h)-f(x)\right\|_{L^2(\mathbb R)}\neq0? $$ I've tried to calculate the $L^2$ norm but it didn't get me any further. How can you show this?

Answer 1

We have, for $0<h<1/n$, \begin{align} \|f_n(x+h)-f_n(x)\|_2^2&=\int_{-\infty}^\infty |\sqrt n\,1_{(0,1/n)}(t+h)-\sqrt n\,1_{(0,1/n)}(t)|^2\,dt\\ \ \\ &=n\,\int_{-\infty}^\infty|1_{(-h,-h+1/n)}(t)-1_{(0,1/n)}(t)|^2\\ \ \\ &=n\,\left(\int_{-h}^01\,dt+\int_{-h+1/n}^{1/n}1\,dt \right)\\ \ \\ &=2nh. \end{align} We need to show that we can find arbitrarily large $n$ and arbitrarily small $h>0$ such that $2nh$ is bigger than a fixed constant. If we take $h_n=\frac1{2n}$, then $$ \|f_n(x+h_n)-f_n(x)\|_2^2=\frac{2n}{2n}=1. $$ Thus $$ \lim_{h\rightarrow0}\sup_{f\in X}\|f(x+h)-f(x)\|_{L^2(\mathbb R)}\geq1. $$

Answer 2

OK, when $n>1/h$ then $f_n(x+h)$ and $f_n(x)$ have disjoint support (when one is non-zero the other is zero). So $\|f_n(x+h)-f_n(x)\|^2 = \|f_n(x+h)\|^2 + \|f_n(x)\|^2 = 1 + 1 = 2$. (The notation using $x$ inside the norms is not nice btw). So the difference has norm $\sqrt{2}$ and can not be larger. So it is the limit as well.