Past exam of the mathematics course of the University of Cambridge / 2005 / iii / Paper 5 / 4

The interpolated exponent is strictly between $p_0$ and $p_1$. Apply Hölder's inequality to $|f{|}^p=|f{|}^{p(1-\theta )}|f{|}^{p\theta }$ with conjugate exponents $p_0/[p(1-\theta )]$ and $p_1/(p\theta )$. Their reciprocals sum to one, soThis gives $L^{p_0}\cap L^{p_1}\subseteq L^p$.

For the other inclusion take any $a>0$ and use the threshold decomposition between Lp spacesBecause $p_0-p<0<p_1-p$,Thus $g\in L^{p_0}$, $h\in L^{p_1}$ and $f=g+h$, proving $L^p\subseteq L^{p_0}+L^{p_1}$. Raising the two norm bounds to powers $1-\theta$ and $\theta$ gives a power of $a$ equal toand a power of $\Vert f{\Vert }_p$ equal to one. HenceFor later use, choose $a=\Vert f{\Vert }_p$ when $f\ne 0$. Each endpoint norm is then at most $\Vert f{\Vert }_p$, so $\Vert f{\Vert }_{L^{p_0}+L^{p_1}}\le 2\Vert f{\Vert }_p$ for the sum of Lp spaces norm. If $f=0$, take $g=h=0$.

Put $C=M_0^{1-\theta }{M}_1^{\theta }$. Since $p<\infty$, choose simple functions $f_n$ of finite-measure support with $f_n\to f$ in $L^p$. These belong to both endpoint domain spaces. The given simple-function estimate applies also to their differences, soBy completeness of Lp spaces, including $L^{\infty }$, there is $u\in L^q$ with $T{f}_n\to u$ in $L^q$. Taking limits of the simple-function bound yields $\Vert u{\Vert }_q\le C\Vert f{\Vert }_p$.

It remains to identify $u$ with the already defined $Tf$, rather than just define an unrelated extension. Write $X=L^{p_0}+L^{p_1}$ and $Y=L^{q_0}+L^{q_1}$ with their sum norms. The endpoint estimates implyIndeed, split $v=v_0+v_1$, bound the two images in their endpoint norms, and take the infimum over all splits. Part (i) gives $\Vert f_n-f{\Vert }_X\le 2\Vert f_n-f{\Vert }_p$, hence $T{f}_n\to Tf$ in $Y$.

The inclusion $L^q\subseteq Y$ is also continuous. For distinct finite endpoints this is the same threshold argument as part (i), after ordering the endpoints. If the larger endpoint is infinity, splitting at $a=\Vert v{\Vert }_q$ bounds both the lower endpoint norm of the large-value part and the supremum norm of the small-value part by $\Vert v{\Vert }_q$. If the endpoints coincide, the inclusion is immediate. Thus $T{f}_n\to u$ also in $Y$. Uniqueness of limits in the sum norm gives $u=Tf$ almost everywhere. Consequently the Riesz-Thorin theorem estimate extends as

Consider the complex-linear two-coordinate map $H(x,y)=(x+y,x-y)$. The triangle inequality shows that its operator norm from ${\ell }^1$ to ${\ell }^{\infty }$ is one; equality is attained at $(1,0)$. The complex parallelogram identityshows that its norm from ${\ell }^2$ to ${\ell }^2$ is $\sqrt{2}$. Interpolate these endpoint bounds by the Riesz-Thorin theorem with parameter $\eta =2/p$. ThenDividing the output norm by $2^{1/p}$ proves the scalar estimate:Apply it pointwise to $f,g$ and integrate its $p$th power. With $r=p/p^{\prime }=p-1>1$, this givesBy Minkowski inequality, the expression on the right is at mostTherefore the requested function inequality holds, namely the dual-exponent Clarkson inequality.