Given arbitrary $a,b \in (0, \infty)$. Is it possible to find a function $f \in L^1(\mathbb{R}) \cap L^2(\mathbb{R})$ with $||f||_1 = a$ and $||f||_2 = b$?

The problem is easy to solve, if we just ask for a function $f \in L^1(\mathbb{R})$ with $||f||_1 = a$. Given $a \in \mathbb{R}$, we can find $f_1:\mathbb{R} \to \mathbb{R}, x \mapsto \chi_{[0,a]}(x)$, so $\int_\mathbb{R} |\chi_{[0,a]}(x) |\text{d}\lambda(x) = a$ or more exotic $f_2 : \mathbb{R} \to \mathbb{R}, x \mapsto \frac{a}{\sqrt{\pi}} e^{-x^2}$, so $\int_\mathbb{R}|\frac{a}{\sqrt{\pi}} e^{-x^2}| \text{d}\lambda(x) = \frac{a}{\sqrt{\pi}} \int_{\mathbb{R}} e^{-x^2} \text{d} \lambda(x) = a $.

But if I try to construct a funtion $f \in L^1(\mathbb{R}) \cap L^2(\mathbb{R})$ with $||f||_1 = a$ and $||f||_2 = b$, I can not cancel out the dependence on the first parameter $a$ in $||f||_2$. For example $f : \mathbb{R} \to \mathbb{R}, x \mapsto \frac{a}{b} \chi_{[0, b]}(x)$ behaves well for $||f||_1 = \int_\mathbb{R}\frac{a}{b} \chi_{[0, b]}(x)\text{d} \lambda(x) = a$, but $||f||_2 = \left( \int_\mathbb{R} |\frac{a}{b} \chi_{[0, b]}(x)|^2 \right)^{\frac{1}{2}} = \frac{a}{\sqrt{b}}$ still depends on $a$. I tried a bunch of other functions but couldn't find an example. Is there maybe a non constructive proof for the existence of such a function?


Try out what happens for some scaled characteristic function $f=d\cdot \chi{[0,r]}$. Plug in the desired norms and solve the system for the parameters $d$ and $r$.

  • 1
    $\begingroup$ Okay, thanks! I solved the system and for $d = \frac{b^2}{a}$ and $r = \frac{a^2}{b^2}$ the function $f: \mathbb{R} \to \mathbb{R}, x \mapsto d \chi_{[0,r]}(x)$ does exactly what I was searching for. $\endgroup$
    – Abstract
    Aug 19 '18 at 14:57

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