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Write $D=\mathbb{E}{\int }_0^{\infty }{e}^{-bt/R}{Z}_t^{1-1/R}dt$ and use the product measure $d\mathbb{P}dt$. For $0<R<1$, the conjugate exponents are $p=1/(1-R)$ and $q=1/R$. Factor the utility integrand asThe Holder inequality on this product measure givesDivide by $1-R$ to obtain the Hölder bound for discounted CRRA consumption:For a finite right side this is ordinary Hölder; on the infinite measure space one can first truncate time and then pass by monotone convergence theorem. If $D=\infty$ and $X_0>0$, the bound is valid but uninformative. If $X_0=0$, the budget forces $c=0$ almost everywhere, so the utility integral is zero; handle this separately instead of writing an undefined $0\cdot \infty$.

When $D$ is finite and the bound is attained, equality requires full budget use and $c_t=(X_0/D)e^{-bt/R}{Z}_t^{-1/R}$. This is the equality pattern of Hölder, not an assertion that every market can finance that process.

Choose a nonzero $f\in L^p({\mathbb{R}}^n)\cap L^1({\mathbb{R}}^n)$ whose transform belongs to $L^q$, for example a Gaussian by the Fourier transform of a Gaussian. For $\lambda >0$, set $f_{\lambda }(x)=f(\lambda x)$. The scaling property of the Fourier transform givesThe assumed estimate, applied for every $\lambda >0$, becomesSince both norms are nonzero, this can hold as $\lambda$ tends both to zero and to infinity only if the powers agree. Thus the scaling necessity for an Lp to Lq Fourier bound yieldsHence $q$ is uniquely the conjugate exponent of $p$.

If a nontrivial uniform estimateholds for all integrable $L^p$ functions on ${\mathbb{R}}^n$, dilation by every $\lambda >0$ forcesThus $q$ must be the conjugate exponent of $p$.