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I'm reactivating this, but it's an extended comment at best. (It started out as what I thought was a complete answer, until I realized that I hadn't read the question carefully enough.) The interesting part of the question remains open.

There are no such functions if $b-a>1$: Indeed, $f(x)=\|f\|_x\ge (b-a)^{1/x}\min f$, which leads to a contradiction if we take a point $x$ for which the min is assumed. Similarly, if $b-a=1$ and $f$ is not constant, then we obtain strict inequality here, so this isn't possible either. The case $b=\infty$ is also easily ruled out because then $f(x)\ge \left(\int_a^{a+1} f^x \right)^{1/x} \ge c>0$, so $f\notin L^p$. Similarly, $f\in L^p$ for $p\in [a,b]$ and $b-a<1$ is ruled out in the same way, by estimating $f$ at its maximum.

This leaves us with the following slightly more focused version of the question: Is there an $f\in \bigcap_{a\le p<b}L^p$ with $f(x)=\|f\|_x$ for all $a\le x<b$. Here necessarily $b-a<1$ and $f\notin L^b$.

I'm reactivating this, but it's an extended comment at best. (It started out as what I thought was a complete answer, until I realized that I hadn't read the question carefully enough.) The interesting part of the question remains open.

There are no such functions if $b-a>1$: Indeed, $f(x)=\|f\|_x\ge (b-a)^{1/x}\min f$, which leads to a contradiction if we take a point $x$ for which the min is assumed. Similarly, if $b-a=1$ and $f$ is not constant, then we obtain strict inequality here, so this isn't possible either. The case $b=\infty$ is also easily ruled out because then $f(x)\ge \min_{a+1\le t\le a+2}f(t)>0$, so $f\notin L^p$. Similarly, $f\in L^p$ for $p\in [a,b]$ and $b-a<1$ is ruled out in the same way, by estimating $f$ at its maximum.

This leaves us with the following slightly more focused version of the question: Is there an $f\in \bigcap_{a\le p<b}L^p$ with $f(x)=\|f\|_x$ for all $a\le x<b$? Here necessarily $b-a<1$ and $f\notin L^b$.

These ($f=c>0$ on $[a,b]$ with $b-a=1$) are the only examples. Write $m=\min f$, $M=\max f$. Then a solution to $f(p)=\|f\|_p$ would have to satisfy $$ m(b-a)^{1/x}\le f(x)\le M(b-a)^{1/x}, \quad\quad\quad (1) $$ and since $x$ can be a point where the min/max is assumed, this rules out $b-a\not=1$. If $b-a=1$ and $f$ is not constant, then we obtain strict inequalities in (1) and the same argument still works.

I'm reactivating this, but it's an extended comment at best. (It started out as what I thought was a complete answer, until I realized that I hadn't read the question carefully enough.) The interesting part of the question remains open.

There are no such functions if $b-a>1$: Indeed, $f(x)=\|f\|_x\ge (b-a)^{1/x}\min f$, which leads to a contradiction if we take a point $x$ for which the min is assumed. Similarly, if $b-a=1$ and $f$ is not constant, then we obtain strict inequality here, so this isn't possible either. The case $b=\infty$ is also easily ruled out because then $f(x)\ge \left(\int_a^{a+1} f^x \right)^{1/x} \ge c>0$, so $f\notin L^p$. Similarly, $f\in L^p$ for $p\in [a,b]$ and $b-a<1$ is ruled out in the same way, by estimating $f$ at its maximum.

This leaves us with the following slightly more focused version of the question: Is there an $f\in \bigcap_{a\le p<b}L^p$ with $f(x)=\|f\|_x$ for all $a\le x<b$. Here necessarily $b-a<1$ and $f\notin L^b$.

These ($f=c>0$ on $(a,b)$ with $b-a=1$) are the only examples. Write $m=\min f$, $M=\max f$. Then a solution to $f(p)=\|f\|_p$ would have to satisfy $$ m(b-a)^{1/x}\le f(x)\le M(b-a)^{1/x}, \quad\quad\quad (1) $$ and since $x$ can be a point where the min/max is assumed, this rules out $b-a\not=1$. If $b-a=1$ and $f$ is not constant, then we obtain strict inequalities in (1) and the same argument still works.

These ($f=c>0$ on $[a,b]$ with $b-a=1$) are the only examples. Write $m=\min f$, $M=\max f$. Then a solution to $f(p)=\|f\|_p$ would have to satisfy $$ m(b-a)^{1/x}\le f(x)\le M(b-a)^{1/x}, \quad\quad\quad (1) $$ and since $x$ can be a point where the min/max is assumed, this rules out $b-a\not=1$. If $b-a=1$ and $f$ is not constant, then we obtain strict inequalities in (1) and the same argument still works.