Can a function $f$ have $L^p$ norms $\Vert f\Vert_p =p$ for all $1\leq p<\infty$?

Question

I have tried to show that such a function must be in $L^{\infty}$ and thus it is impossible for such a function to exist since, in that case, $$\infty =\lim_{p\rightarrow \infty} \Vert f\Vert_p=\Vert f\Vert_{\infty}.$$ Basic estimates seem to fail and I can't seem to construct a counterexample.

What if we replace $p$ with something that grows at different rates, like $e^p$ or $\log p$?

Answer 1

The magic words are: Riesz-Thorin interpolation Theorem.

Answer 2

There are two cases to consider. Let's first start with $f\in L^p(X)$ where $X$ has finite Lebesgue measure. The $L^p$ spaces are nested. We have that if $f\in L^q(X)$, then $||f||_{L^p}\leq \mu(X)^{\frac{1}{p}-\frac{1}{q}}||f||_{L^q}$

If $||f||_{L^q}=q$ then $\mu(X)^{\frac{1}{p}-\frac{1}{q}}*q$

We want to show that $||f||_{L^p}=p$, if this is true, then $p\leq \mu(X)^{\frac{1}{p}-\frac{1}{q}}*q$. We can rewrite this as $\frac{p}{q}\leq \mu(X)^{\frac{1}{p}-\frac{1}{q}}$. We know $f=0$, which has $L^p$ norm equal $0$ cannot be such a function. For now, lets just take $p=2$ and $q=3$.

We have $\frac{2}{3}\leq \mu(X)^{\frac{1}{6}}$ and so $\frac{2^6}{3^6}\leq \mu(X)$. This means that in order for a function $f$ to have $||f||_{L^2}=2$ and $||f||_{L^3}=3$ we must have the $\frac{2^6}{3^6}\leq \mu(X)$. You can toy around with other $p,q$ values to establish similar bounds.

That said, I am not sure such a function can be constructed. Note that if $||f||_{L^p}=p$, then let's start with $p=1$. As an example, conside $f=exp(2\pi ix)$ this has $L^1([0,1])$ norm of $1$, and even if we insert some scaling factor $aexp(2\pi ix)$, it is impossible to get the desired value for the other $L^p$ norms.

If you are unsure of the algebra, recall you can pull the constant out during integration and so we would need $a^p=p$ and only $a=0,1,$ solve this, and so it is not possible for our function.

I think it is possible to show that no such $f$ can exist, but a clean proof of this evades me at the moment.

The second case is when $\mu(X)=\infty$, and here we don't have nesting of $L^p$ spaces. While I haven't worked it out, I don't think such an $f$ can exist here either.