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Suppose I have a real-valued function $f \in L^{\infty}([0,2\pi)^2)$. Suppose also that, if $\| f \|_{p}$ denotes the $L^{p}([0,2\pi)^2)$ norm of $f$, then the following bounds are known:

$$\displaystyle C_{1}r^{1/2} \leqslant \|f\|_{1} \leqslant \|f\|_{2} \leqslant C_{2}r^{1/2},$$

where $r \gg 1$. I would like to know if the following is possible:

$$\displaystyle \|f\|_{4} \leqslant C r^{1/4}.$$

This seems counter-intuitive as I expected the norms to have the same or greater bound. Moreover, by the generalised Holder's inequality, this seems impossible, since then we would have

$$\displaystyle C_{1}^{2}r \leqslant \|f\|_{2}^{2} = \|f \cdot f \cdot 1 \cdot 1\|_{1} \leqslant \|f\|_{4}\|f\|_{4}\|1\|_{4}\|1\|_{4} \leqslant Cr^{1/2},$$

but the left hand side is bounded below by $r$, which indicates that my claimed upper bound for $\|f\|_4$ is impossible, and the above implies that $r^{1/2}$ is a lower bound for $\|f\|_4$ -- right?

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There is no upper bound on $L^4$ norm in terms of the $L^2$ norm. Generally, on a finite measure space $L^p$ controls $L^q$ for $p>q$, but not for $p

As a concrete example, consider $f = \epsilon^{-1/2}\chi_E$ where $E$ has measure $\epsilon$. Then $\|f\|_2=1$ but $\|f\|_4 = \epsilon^{-1/4}$ which can be arbitrarily large.

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    I see. How about lower bounds, then? Supposing I know that the $L^1$ and $L^2$ norms are bounded below (and above) by $Cr^{1/2}$ -- is the upper bound of $Cr^{1/4}$ for the $L^4$ norm still possible? – 2017-01-28
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    Nothing you know about $L^1$ and $L^2$ norms will give you an upper bound about $L^4$ norm. The higher-exponent norms see things that the lower exponents do not. A lower bound on $L^4$ norm is in your post: Hölder's inequality. – 2017-01-28
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    But if the lower bound in question (due to Hölder's inequality) is $r^{1/2}$, how can it have $r^{1/4}$ as an upper bound? Or have I misunderstood what you are saying? – 2017-01-28
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    There is no upper bound on $L^4$ norm in terms of $L^2$ norm. No upper bound on $L^4$ norm can be given on the basis of $L^2$ norm. Knowing $L^2$ norm is not enough to obtain an upper bound on $L^4$ norm. (Am I getting through?) – 2017-01-28
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    Yes, I understand that. So is it correct to say that a lower bound on the $L^2$ norm places absolutely no restrictions on what the upper bound of the $L^4$ norm could be? – 2017-01-28
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    @user363087: Just look at $a\cdot \varepsilon^{-1/2}\chi_{(0,\varepsilon)}$ for $a,\varepsilon>0$, $\varepsilon\leq 2\pi$. The $L^2$ norm will be exactly $a$, but the $L^4$ norm can be as big as you want it to be. – 2017-01-31
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    I agree -- but in your example, due to your restriction on the size of $\epsilon$, that also means that $\|f\|_4 \geqslant (2\pi)^{-1/4}$. This automatically means that we cannot have $(2\pi)^{-1/2}$ as an upper bound for $\|f\|_4$, for instance. But according to what I wrote in the question, we have $$\displaystyle C_{1}r \leqslant \|f\|_{2}^{2} = \|f \cdot f \cdot 1 \cdot 1\|_{1} \leqslant \|f\|_{4}\|f\|_{4}\|1\|_{4}\|1\|_{4} \leqslant C_{3}r^{1/2},$$ and for large $r$, this can't be true -- so my suggested upper bound for $\|f\|_4$ is too small, right? – 2017-01-31