Find $f$ such that $\|f\|_p=1$ for every $p\geq 1$.

Question

For which measure spaces $(U,\mathscr{F},\mu)$ can I find a function $f\in \bigcap_{p\geq 1} L^p(U)$ such that $\|f\|_p=1$ for every $p$? For example, take $U=\mathbb{R}$ equipped with the Borel $\sigma$-algebra and Lebesgue measure. Are there any functions which satisfy this constraint? My first thought was somehow to use the Gaussian density $\phi$ which is in $\bigcap_{p>0}L_p(\mathbb{R})$, but the pth norm of the Gaussian isn't 1.

Answer 1

I'm going to prove that

Assume that $\|f\|_p=1$ for all $p\geq1$, then we can find a measurable set $E\subset U$ with $m(E)=1$ such that $|f(x)|=1$ for a.e. $x\in E$ and $f(x)=0$ for $x\in U\setminus E$, which is to say, $|f|=\chi_E$ almost everywhere.

Since $\|f\|_p=1$ for all $p\geq1$, we have $$\int_U|f(x)|^p\,dx=1,\qquad \forall p\geq1.$$ Let $E=\{x\in U: |f(x)|>0\}$, and $F(x,p)=|f(x)|^p\chi_E(x)$ for $x\in U$ and $p\in[1,\infty)$, then $$\int_UF(x,p)\,dx=1,\qquad \forall p\geq1.\tag{1}$$ Our idea is to differentiate the above identity with respect to $p$. We have $\frac{\partial F}{\partial p}(x,p)=|f(x)|^p\ln |f(x)|\chi_E(x)$ for all $p>1$ and $x\in U$. We are going to use the following lemma, which is a direct consequence of the dominated convergence theorem:

Lemma. Suppose that $F=F(x,y)$ is a measurable function defined on $U\times (a,b)$. Assume that for fixed $y$, the function $x\mapsto F(x,y)$ is integrable in $U$; and for fixed $x\in U$, the function $y\mapsto F(x,y)$ is differentiable for $y\in (a,b)$. If there exists $G\in L^1(U)$ such that $$\left|\frac{\partial F}{\partial y}(x,y)\right|\leq G(x),\qquad (x,y)\in U\times (a,b),$$ then we have $$\frac d{dy}\int_U F(x,y)\,dx=\int_U\frac{\partial F}{\partial y}(x,y)\,dx,\qquad y\in(a,b).$$

We continue our proof. For $p\in(2,4)$, using $t^p\leq t^2+t^4$ and $|t\ln t|\leq C(1+t^2)$ for all $t>0$, we have $$\left|\frac{\partial F}{\partial p}(x,p)\right|\leq \left(|f(x)|^2+|f(x)|^4\right)\ln|f(x)|\chi_E(x)\leq C\left(|f(x)|+|f(x)|^3\right)\left(1+|f(x)|^2\right)\in L^1(U).$$ Hence we can apply the Lemma to get $$\frac d{dp}\int_U F(x,p)\,dx=\int_U\frac{\partial F}{\partial p}(x,p)\,dx,\qquad p\in(2,4).\tag{2}$$ Recalling $(1)$, the LHS of $(2)$ is zero, hence, $$\int_U|f(x)|^p\ln |f(x)|\chi_E(x)\,dx=0,\qquad p\in(2,4).\tag{3}$$

However, the integrand in $(3)$ is not non-negative, so we can't get that the integrand is zero almost everywhere. So, we differentiate once more with respect to $p$. Using a similar argument as before, we can prove the interchangable of differentiation and integration, so we will deduce $$\int_U|f(x)|^p\left(\ln |f(x)|\right)^2\chi_E(x)\,dx=0,\qquad p\in(2,4).$$ Now, since the integrand is non-negative, taking $p=3$ we have $$|f(x)|^3\left(\ln |f(x)|\right)^2\chi_E(x)=0,\qquad \text{a.e. }x\in U.$$ Since $|f(x)|>0$ for $x\in E$, we have $\ln |f(x)|=0$ for a.e. $x\in E$, and hence $|f(x)|=1$ for a.e. $x\in E$. What's more, we have $$1=\int_U|f(x)|^p\,dx=\int_E|f(x)|^p\,dx=\int_E1\,dx=m(E).$$

Now, the proof is complete.

Remark. This proof indicates that the same conclusion holds if we change $p\in [1,\infty)$ to $p\in(a,b)$ for any interval $(a,b)$. But it fails to answer the following problem: Can we find a non-trivial $f:\mathbb R\to\mathbb C$ such that $\|f\|_{L^{p_n}(\mathbb R)}=1$ for all $n\in\mathbb N$?

Answer 2

The conclusion is true if $\|f\|_{p}=1$ for countably many values $p\ge p_0>0$ (for example $p_0=1$).

Let $P=\{p\ge p_0\,:\,\|f\|_p=1\}.$

Assume the set $P$ is unbounded. Then it contains a sequence $p_n\nearrow \infty.$ We have $$\int\limits_X|f(x)|^{p_n}\,d\mu(x)= 1,\quad n\in \mathbb{N}\quad (*)$$ If for some $\delta>0$ we have $\mu\{ x\in X\,:\, |f(x)|>1+\delta\}>0$ then LHS of $(*)$ tends to infinity. Thus $|f(x)|\le 1$ almost everywhere. In that case LHS of $(*)$ is nonincreasing. If for some $0<\delta <1$ there holds$\mu\{ x\in X\,:\, |f(x)|<1-\delta\}>0$ then LHS in $(*)$ is decreasing, Thus $|f(x)|=1$ almost everywhere.

Assume the set $P$ is bounded. Hence it admits an accumulation point. By the Morera theorem the function $$h(z)= \int\limits_{X}|f(x)|^z\,d\mu(x)$$ is holomorphic in the open right half-plane and $h(p)=1$ for $p\in P.$ Thus $h(z)\equiv 1,$ for $z$ in the open right-half plane. In particular $h(x)=1$ for $x>0,$ i.e. $P$ is unbounded, a contradiction.

Answer 3

Any measure space $(X,\mathscr{B},\mu)$ for which there is a set $A\in\mathscr{B}$ with $\mu(A)=1$ will do: $f=\mathbb{1}_A$. In fact, any function $f$ that satisfies the constrain on the OP is of that form: $$\int |f|^p=\int_{\{0<|f|<1\}}|f|^p + \lambda(|f|=1)+\int_{\{|f|>1\}}|f|^p$$ on $\{|f|>1\}$, $|f|^p\nearrow\infty$ as $p\rightarrow\infty$; thus by monotone convergence, $$\lim_{p\rightarrow\infty}\int_{\{|f|>1\}}|f|^p=\infty\lambda(|f|>1)$$ The constrain $\|f\|_p=1$ for all $p$ implies that $\lambda(|f|>1)=0$.

On $\{0<|f|<1\}$, $|f|^p\searrow0$ as $p\rightarrow\infty$. As $|f|^p\mathbb{1}_{\{0<|f|<1\}}\leq|f|^p$ for $p\geq1$, dominated convergence yields $$\lim_{p\rightarrow\infty}\int_{\{0<|f|<1\}}|f|^p=0$$ Hence $\lambda(|f|=1)=1$. By assumption $\|f\|_p=1$ for all $p$; hence $|f|=1$ $\lambda$-a.s. This means that $|f|=\mathbb{1}_E$ for some measurable set $E$ with $\lambda(E)=1$.