Prove that $f \in L^1(\Bbb R)$ and that $\|f\|_1 \leq \sqrt{2}\|f\|_2 =\frac{\sqrt{6}}{3} \|x^2f\|_2$

Question

Suppose that $f, x^2f \in L^2(\Bbb R)$. Prove that $f \in L^1(\Bbb R)$ and that $$\|f\|_1 \leq \sqrt{2}\|f\|_2 =\frac{\sqrt{6}}{3} \|x^2f\|_2$$

My Solution so far. Can anyone tell me if I a on right track? In either case any clues how to finish? Thanks

\begin{align*} \|f\|_1 = \int_{\Bbb R} |f| &= \int_{\Bbb R} |f \cdot \frac{x^2}{x^2}|\\ & = \int_{\Bbb R} |x^2 f \cdot \frac{1}{x^2}| && \text{ by Holder}\\ &\leq \|x^2f\|_2 \cdot \Big\|\frac{1}{x^2}\Big\|_2\\ &= \|x^2f\|_2 \cdot \int_{\Bbb R} x^{-4}dx\\ &= -\frac{1}{3x^3} \|x^2 f\|_2 \end{align*}

Answer 1

We note that $$\int_{\mathbb{R}} \lvert f(x)\rvert\,\mathrm{d}x = \int_{[-1, 1]} \lvert f(x)\rvert\,\mathrm{d}x+\int_{[-1, 1]^c = (-\infty, -1)\cup (1, \infty)} \lvert f(x)\rvert\,\mathrm{d}x$$ As $f\in L^2(\mathbb{R})$, $f\in L^2([-1, 1])$, and as $[-1, 1]$ is finite, we have by an application of Hölder that $$\|f\|_{L^1([-1, 1])}\leq m([-1, 1])^{1/2}\|f\|_{L^2([-1, 1])}\leq \sqrt{2}\|f\|_{L^2(\mathbb{R})}$$ Next, we use Hölder again to write $$\|f\|_{L^1([-1, 1]^c)}\leq \|x^2f\|_{L^2([-1, 1]^c)}\left(2\int_1^{\infty} \frac{1}{x^4}\,\mathrm{d}x\right)^{1/2} = \sqrt{\frac{2}{3}}\|x^2f\|_{L^1([-1, 1]^c)}$$ and therefore $$\|f\|_{L^1([-1, 1]^c)}\leq \sqrt{\frac{2}{3}}\|x^2f\|_{L^2(\mathbb{R})}$$ Putting these together, we get $$\|f\|_{L^1(\mathbb{R})}\leq \sqrt{2}\|f\|_{L^2(\mathbb{R})}+\sqrt{\frac{2}{3}}\|x^2f\|_{L^2(\mathbb{R})}$$

Edit: As a side note, we definitely can't have the equality you stated in the question title. Take, for example, $f = \frac{1}{x^4}\mathbf{1}_{(1, \infty)}$. We can see pretty easily that $\|f\|_{L^2(\mathbb{R})}\neq \frac{1}{\sqrt{3}}\|x^2f\|_{L^2(\mathbb{R})}$. My guess is that there's supposed to be a $+$ instead of an $=$.