Functions which have same even Lp norms have same shape

Question

Is it true that the all the functions which have the same $L^{2p}$ norms must be the same, up to isometries? More precisely I would like to show that given $f_1,f_2 : \mathbb R \to \mathbb R$ smooth as needed, $1$-periodic, and s.t. $$ \int_0^1 f_1^{2p} = \int_0^1 f_2^{2p}, \quad \forall p \in \mathbb N $$ then there is $t \in \mathbb R$ s.t. $f_1(x) = f_2(x+t)$ or $f_1(x) = f_2(t-x)$, $\forall x \in [0,1]$.

Answer 1

As commenters pointed out, this is not true. To clarify the picture, introduce the distribution function $$\lambda_f(t) = \mu(\{x : |f(x)|\ge t\}),\qquad t>0$$ One can show that $$ \|f\|_p^p = \int_0^\infty t^{p-1} \lambda_f(t)\,dt $$ Hence, the equality $\lambda_f = \lambda_g$ implies $\|f\|_p=\|g\|_p$ for all $p\in [1,\infty)$ (actually, for $p=\infty$ too).

To visualize what can be done with $f$ while preserving its distribution function, think of sliding the level sets of $f$ to the left or right, ensuring they remain properly stacked. For example, if the graph of $f$ is a triangle, it can be deformed to any other triangle of the same base and height.