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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.

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