Sequence of functions $f_{n} \in L_{1}(\mathbb{R}) \cap L_{2}(\mathbb{R})$ for $n=1,2,3,....$ s.t. $\frac{||f_n||_1}{||f_n||_2} \rightarrow 0$.

Question

Give an example of a sequence of functions $f_{n} \in L_{1}(\mathbb{R}) \cap L_{2}(\mathbb{R})$ for $n=1,2,3,....$ such that $\frac{||f_n||_1}{||f_n||_2} \rightarrow 0$.

Thoughts I can find functions such that $\frac{||f_n||_2}{||f_n||_1} \rightarrow 0$ but any sequence of functions I can think of that tend to zero in $L_{2}$ either tend to zero in $L_{1}$ aswell or aren't in $L_{1}$ at all.

Answer 1

For a given $f\in L^1(\mathbb{R}) \cap L^2(\mathbb{R})$ and $n\in \mathbb{N}$ consider the $L^1$ norm of the scaled function $f_n(x) := f(nx)$. Namely $$ ||f_n||_{L^1(\mathbb{R})} = \int\limits_{\mathbb{R}} f(nx) dx = \frac{1}{n} \int\limits_{\mathbb{R}} f(x) dx = \frac 1n ||f||_{L^1(\mathbb{R})}. $$ Hence, $$||f_n||_{L^2(\mathbb{R})} = \left( ||f_n^2||_{L^1{(\mathbb{R}})} \right)^{1/2} = \frac{1}{n^{1/2}} ||f||_{L^2(\mathbb{R})}.$$ It follows that $$ \frac{||f_n||_{L^1}}{||f_n||_{L^2}} = \frac{1}{n^{1/2}} \frac{||f||_{L^1(\mathbb{R})}}{||f||_{L^2(\mathbb{R})}} \to 0. $$

As an explicit example of $f$ take $f(x) = \frac{1}{ 1 +x^2}$ and set $f_n (x) = \frac{1}{1+n^2 x^2}$.

Answer 2

If you can find a function $f\in L^1(\mathbb{R})\setminus L^2(\mathbb{R})$ bounded, truncating $f$ at a compact interval will be in $L^1(\mathbb{R})\cap L^2(\mathbb{R})$. Using this you can find a desired sequence of funtions.