Suppose an operator $T$ is bounded on $L^2$ and also bounded from $L^{1}$ to $L^{1}$-weak. Then by Marcinkewicz interpolation one gets that $T$ is bounded on every $L^{p}$ for p between 1 and 2. Precise versions of the theorem (see the book of L.Grafakos) give an estimate of the norm of $T$ on $L^{p}$, and of course the norm diverges as $p\to 1$. If not, by a simple argument one could obtain that $T$ is bounded also on $L^{1}$, which is clearly false in general (e.g. singular integrals). By the way, even a weaker assumption on the growth of the norm should allow one to conclude that $T$ is bounded on $L^{1}$, as in extrapolation theorems.


Now suppose you (me) are hard-headed and want to use the general machinery of real interpolation, say the K-method as detailed in the book of Bergh and Löfström. Then $T$ is bounded from $L^{1}$ to $L^{1}$-weak which means the Lorentz space $L^{1,\infty}$, with norm $M\_{0}$, and also from $L^{2}$ to $L^{2}$ with norm $M\_{1}$. Then $T$ is bounded on the corresponding real interpolation spaces with norm $M\_{0}^{1-\theta}M\_{1}^{\theta}$. Real interpolates of Lorentz spaces are Lorentz spaces, see Theorem 5.3.1 in B-L. We conclude that $T$ is bounded on every Lorentz space $L^{p,q}$ with p between 1 and 2, and $1\le q\le \infty$, and in particular on $L^{p,p}=L^{p}$ as expected.


Unfortunately, now I have a uniform bound on the norm of $T$ as $p\to 1$, which would allow me to conclude that $T$ is also bounded on $L^{1}$. Where is the mistake? There must be some inaccuracy in one of the steps, but which one exactly?



A puzzling question on real interpolation
(Note: I don't have my copy of B&L handy, so I'm sort of doing this from memory.)


The problem is that $(L\_{p\_0.q\_0},L\_{p\_1,q\_1})\_{\theta,q} = L\_{p\_\theta,q}$ *under equivalent norm*. There's no saying what the constant of equivalency are. Now, if you look at $T: \{X\_1,X\_2\} \to \{Y\_1,Y\_2\}$ with norm $\{M\_1,M\_2\}$, then you have 
$$ T: (X\_1,X\_2)\_{\theta,q} \to (Y\_1,Y\_2)\_{\theta,q} $$
with norm $M\_1^{(1-\theta)}M\_2^\theta$. Now put in $X\_1 = L\_{1,1}$, $X\_2 = L\_{2,2}$, $Y\_1 = L\_{1,\infty}$ and $Y\_2 = L\_{2,2}$, while the interpolants $(X\_1,X\_2)\_{\theta,q}$ may be equivalent to $L\_{p\_\theta,q}$, and $(Y\_1,Y\_2)\_{\theta,q}$ to $L\_{p\_\theta,q}$, the constants of equivalency may be different. That's where the degeneracy is hiding.


Observe that if $Y\_1$ were $L\_{1,1}$ (or if $X\_1 = L\_{1,\infty}$) instead, then the two interpolation spaces are actually equal, which is what you expect. 


