Existence of a cts. function $f \geq 0$ satisfying $\int f(x)^n \, dx =1$ for all $n \geq 1$ Does there exist a continuous function $f:\mathbb{R}\to(0,\infty)$ that satisfies
$$\int_{\mathbb{R}} f^n d L^1 = 1 $$
for every natural $n$? ($L^1$ is Lebesgue measure on $\mathbb{R}$)
 A: No.  Suppose such $f$ does exist to reach a contradiction.  Then $\sup\limits_{x\in \mathbb R}f(x)=\lim\limits_{n\to\infty}\left(\int_{\mathbb R} f^n dL\right)^{1/n}=1$, so $f(x)\leq 1$ for all $x$.  Hence $f-f^2$ is a nonnegative function, and by continuity $f-f^2$ is not always zero.  This implies (again using continuity) that $\int_{\mathbb R} f-f^2 dL>0$.
A: Suppose there were a function $f$ with the desired properties. Let us first prove that $$f \leq 1.$$
Indeed, if $\{f>1\}$ had strictly positive Lebesgue measure, then we could find $\epsilon>0$ such that $L^1(\{f \geq 1+\epsilon\})>0$, and so
$$\int_{\mathbb{R}^n} f(x)^n \, dx \geq \int_{\{f \geq 1+\epsilon\}} f(x)^n \, dx \geq (1+\epsilon)^n L^1(\{f \geq 1+ \epsilon\}).$$ Since $\epsilon>0$, the right-hand side converges to $\infty$ as $n \to \infty$, and this contradicts our assumption that $\int f(x)^n \, dx \leq 1$ for all $n \geq 1$. Hence, $f \leq 1$ Lebesgue almost everywhere. By continuity of $f$, this gives $f \leq 1$ (everywhere).
Since $f \leq 1$, the function $f-f^2$ is non-negative. On the other hand, we have
$$\int (f(x)-f(x)^2) \, dx = \int f(x) \, dx - \int f(x)^2 \, dx = 1-1=0.$$
This implies $f-f^2=0$ almost everywhere. Again we can use continuity to get $f-f^2=0$ (everywhere). Hence,
$$f(x) (1-f(x))=0, \qquad x \in \mathbb{R}.$$
Since $0 \leq f \leq 1$, we conclude that $f(x) \in \{0,1\}$ for all $x$ which is impossible because $f$ is continuous.
Remark:  The continuity of $f$ is only needed to discuss away exceptional sets and at the very end of the proof. Thus, the proof actually gives the following more general result.

Let $f \geq 0$ be a measurable function (not necessarily continuous) such that $$\int f(x)^n \, dx = 1, \qquad n \in \mathbb{N}.$$ Then there exists a set $A$ of Lebesgue measure $1$ such that $f(x)=1_A(x)$.

Proof: Following the above proof, we get $$f(x) (1-f(x)) =0 \quad \text{and} \quad 0 \leq f(x) \leq 1$$ Lebesgue almost everywhere. Thus $f(x) \in \{0,1\}$ almost everywhere. If we set $A=\{f=1\}$, the assertion follows.
