Does $f^2\in\mathcal{L}^1(\mathbb{R})$ imply $f\in\mathcal{L}^1(\mathbb{R})$?

Question

Suppose $f^2\in\mathcal{L}^1(\mathbb{R})$.

i.e. $\int |f^2|~d\lambda=\int f^2~d\lambda<\infty$ where $\lambda$ is the Lebesgue measure.

Is it true that $f\in\mathcal{L}^1(\mathbb{R})$?

Answer 1

No, $L^2(\Bbb R)\nsubseteq L^1(\Bbb R)$. It would be the case if you were considering a space of finite measure, though.

Answer 2

Clearly $f^2$ is in $\mathcal{L}^1(\mathbb{R})$ if and only if $f$ is in $\mathcal{L}^2(\mathbb{R})$, so your question is equivalent to asking whether $\mathcal{L}^2(\mathbb{R})$ is a subset of $\mathcal{L}^1(\mathbb{R})$.

This is false, with a particular example being $$f(x) = \left\{\array{0&x\in[-1,1]\\\frac{1}{|x|}&\text{otherwise}}\right.$$

Now, $f^2(x)$ is integrable (with integral $2$), but $f$ is not.