Given $f\in L^1(\mathbb{R})$ with $||f||_1 < \infty$, is it true that $\int_{\mathbb{R}} ||f||_1 - f(x) \, dx = 0$?

Question

According to my intuition so far, the answer should be yes, hinging very important on the assumption that $||f||_1 < \infty$.

To speak very roughly, if the $L^1$ norm of $f$ is finite, it seems to make sense to me that the integral over the reals of the $L^1$ norm of $f$ minus $f$ should come out to be zero, as its fluctuations about its norm should "cancel out".

When I try to formalize this conjecture, however,

$$\int_\mathbb{R} ||f||_1 - f(x) \, dx$$

is not an integral I can see how to evaluate.

Is my conjecture correct? How might one go about proving or disproving it?

Answer 1

Not true at all. As a matter of fact, if $\|f\|_1\not= 0$, the integral diverges!

For example, consider $f(x)=\chi_{(0,1)}$. Then $\|f(x)\|_1=1$ and

$$\int_{-\infty}^{\infty} 1-\chi_{(0,1)}dx=2\int_1^{\infty}1dx$$