Looking at $(t,x)\in[0,1]\times\mathbb{R}^2$, I came across the statement (for sufficiently smooth) real-valued $f$ that $$ \|f(t,x)\|_{L^\infty_tL^2_x} \lesssim \|f\|_{L^2_tL^2_x}^{1/2}\|\partial_t f\|_{L^2_tL^2_x}^{1/2} + \|f\|_{L^2_tL^2_x}, $$ which is a "simple calculus identity."

I do not know how to prove this. I tried to let $t\in[0,1]$ and then bound $ \int f(t,x)^2dx$ by the above by writing $f(t,x)$ as $\int_0^t\partial_s f(s,x)ds+f(0,x)$ and similar tricks, but could not get it to work out. I also tried to use the Sobolev embedding of $W^{1,2}(\mathbb{R})$ into $L^\infty(\mathbb{R})$, but could not get that to work (nor is that "simple calculus" in my opinion). Any help would be greatly appreciated, thanks!

(In case it is not standard, the mixed norm notation is $$ \|f(t,x)\|_{L^q_tL^p_x} = \left(\int\|f(t,\cdot)\|^q_{L^p}dt\right)^{1/q}. $$ )

(For reference, the original statement was found in the paper "Sharp Trace Theorems for Null Hypersurfaces on Einstein Metrics with Finite Curvature Flux" by S. Klainerman and I. Rodnianski, Geom. funct. anal. Vol. 16 (2006) 164-229.)


I have an answer! I had tried to rewrite $f$ using the fundamental theorem of calculus when I should have been rewriting $f^2$. Explicitly, for any $t_0$ in $I=[0,1]$ $$ f(t_0,x)^2 = f(t,x)^2 + \int_t^{t_0}2f(s,x)f_t(s,x)ds $$ for $t\in I$ as well. Then we can integrate in $t$, noting that $[0,1]$ has measure 1 and that the left hand side does not depend on $t$ such that $$ f(t_0,x)^2 = \int_I f(t,x)^2dx + \int_0^1\int_t^{t_0}2f(s,x)f_t(s,x)dsdt. $$ Then using Fubini's theorem we can switch the $t$ and $s$ integral, and, empluying absolute values and the like we see that $$ |f(t_0,x)|^2 \lesssim \int_I |f(t,x)|^2dx + \int_0^1\int_0^s|f(s,x)||f_t(s,x)|dtds. $$ Performing the first integration on the second term on the right-hand side gives an $s$ which is bounded by 1 such that $$ |f(t_0,x)|^2 \lesssim \int_I|f(t,x)|^2dx + \int_I|f(s,x)||f_t(s,x)|ds. $$ Integrating next in $x$ (and employing Fubini again) $$ \|f\|_{L^\infty_tL^2_x}^2 \lesssim \|f\|^2_{L^2_tL^2_x} + \int_I\int_{\mathbb{R}^2} |f(s,x)||f_t(s,x)|dxds. $$ Using Holder's inequality (Schwarz) on the second term on the right-hand side gives the desired result except with every term squared. The subadditivity of the square-root finishes the proof.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.