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For conjugate exponents $p,q>1$, the constraint set is compact, so the continuous product has a maximum. Boundary endpoints have product zero, while $(1,1)$ is feasible with product one. An interior maximum therefore satisfies the Lagrange multiplier equations $y=\lambda x^{p-1}$ and $x=\lambda y^{q-1}$. Multiplying by $x$ and $y$ respectively gives $x^p=y^q$, since the multiplier is nonzero. Substitution in the constraint makes both powers equal one. Thus the maximum is one, attained uniquely at $(1,1)$. The boundary check and compactness are what turn the necessary multiplier equations into a global conclusion.

Hölder's inequality controls a product using separate information about its two factors. If one factor is large on part of a domain, the other factor's complementary Lp norm limits how much their product can accumulate there. The same statement bounds finite sums, integrals and expected values; no independence is required in the probabilistic version.

Let $p,q\in [1,\infty ]$ be conjugate exponents, meaningOn a measure space, if $f\in L^p$ and $g\in L^q$, then $fg\in L^1$ andHere the Lp norms are finite; at exponent $\infty$ the norm uses the essential supremum. For a finite vector, integration with counting measure givesThe first step is the triangle inequality; the second is Hölder's inequality.^[ref]

For finite vectors, the endpoint case $p=1$, $q=\infty$ is $|x^{T}y|\le \Vert x{\Vert }_1\Vert y{\Vert }_{\infty }$: bound every $|y_i|$ by its maximum, then sum the $|x_i|$. Swapping the factors gives the other endpoint. The choice $p=q=2$ gives the Cauchy-Schwarz inequality.

For measurable functions $f,g$ with values in the real numbers or complex numbers on the same measure space $(X,\mathcal{A},\mu )$, and $1\le p\le \infty$, membership in the same Lp space givesThus adding two functions cannot make their combined Lp norm exceed the sum of their separate Lp norms. The measure need not be finite or a probability measure. The hypotheses require finite Lp norms; $p=\infty$ uses the essential supremum of the absolute value, not an ordinary pointwise supremum. For complex numbers, the absolute value is the complex modulus. As usual, Lp spaces identify functions equal almost everywhere. ^[ref]

The Minkowski inequality makes the Lp norm a workable notion of size: it supplies the triangle inequality needed to turn $d_p(f,g)=\Vert f-g{\Vert }_p$ into a metric on Lp space. Applied to $(f-h)+(h-g)$, it says that travelling through an intermediate function $h$ cannot shorten the direct distance measured by this metric. Applied to two error functions, it gives a bound on the total error even when the errors reinforce one another; Minkowski inequality for error aggregation gives explicit examples.

The geometry is already visible in the Euclidean plane. Under counting measure on two coordinates, the closed unit balls for the L1 norm, Euclidean norm, and L-infinity norm are respectively $|x|+|y|\le 1$, $x^2+y^2\le 1$, and $\max (|x|,|y|)\le 1$. Every one is a convex set: if $u,v$ have Lp norm at most one, the Minkowski inequality and absolute homogeneity of a norm give $\Vert tu+(1-t)v{\Vert }_p\le t+(1-t)=1$ for $0\le t\le 1$. The Euclidean norm has no flat boundary segments, whereas the other two norms do. This is the distinction captured by a strictly convex normed space and developed in equality in Minkowski inequality.

A short Hölder's inequality proof outline handles $1<p<\infty$. First $|f+g{|}^p\le 2^{p-1}(|f{|}^p+|g{|}^p)$, so $f+g$ has finite Lp norm before any division. With conjugate exponents $p$ and $q=p/(p-1)$, set $h=|f+g{|}^{p-1}$; then $\Vert h{\Vert }_q=\Vert f+g{\Vert }_p^{p-1}$. The scalar triangle inequality followed by Hölder's inequality givesDivide if $\Vert f+g{\Vert }_p>0$; otherwise the desired bound is immediate. For $p=1$, integrate $|f+g|\le |f|+|g|$. For $p=\infty$, the almost everywhere bounds $|f|\le \Vert f{\Vert }_{\infty }$ and $|g|\le \Vert g{\Vert }_{\infty }$ give the result by the essential supremum. The same scalar triangle inequality works for real numbers and complex numbers.

The conjugate-power constrained product maximum proves Young's inequality for products by rescaling. For nonnegative $x,y$ and conjugate exponents $p,q>1$, let $A=x^p/p+y^q/q$. If $A=0$, both variables vanish. Otherwise the displayed normalized variables satisfy $X^p/p+Y^q/q=1$, hence $XY\le 1$. Since $1/p+1/q=1$, this means $xy/A\le 1$, proving $xy\le x^p/p+y^q/q$. Equality in the constrained maximum requires $X=Y=1$, equivalent to $x^p=y^q$; the zero case gives the same equality criterion. This approach simultaneously gives the sharp constant and every equality case.

For conjugate exponents $p,q\in [1,\infty ]$, $1/p+1/q=1$, Hölder's inequality states that $\int |fg|\le \Vert f{\Vert }_p\Vert g{\Vert }_q$. If $r=\infty$ in the requested product estimate, every $p_i=\infty$ and multiplication of essential bounds proves it directly. Suppose $r<\infty$. Factors with $p_i=\infty$ can again be bounded pointwise by their essential suprema.

Here is an explicit finite-exponent proof of the remaining Generalized Holder inequality. Put $q_i=p_i/r$, so ${\sum }_{i}1/q_i=1$, and normalize $h_i=|f_i{|}^r/\Vert f_i{\Vert }_{p_i}^r$. If one norm is zero the product vanishes almost everywhere. Otherwise $\int h_i^{q_i}=1$. The weighted arithmetic-geometric mean inequality, applied to $h_i^{q_i}$ with weights $1/q_i$, givesIntegrating gives $\int {\prod }_i{h}_i\le 1$. If only one finite exponent remains, its weight is one and the same argument is an equality. Restoring the normalization and taking the $r$th root provesIn particular the product belongs to $L^r$.

For the projection estimate, use the three-dimensional functional Loomis-Whitney inequalityThis permitted three-dimensional result also follows directly by Cauchy-Schwarz inequality: integrate the square root of $g_1{g}_2$ in $x_3$, obtaining the square root of the product of their marginal integrals, and then apply Cauchy-Schwarz inequality in $(x_1,x_2)$ against $\sqrt{g_3}$.

Let $f_j$ be any nonnegative integrable function of the three coordinates excluding $x_j$. For $j=1,2,3$, define $G_j$ by integrating $f_j$ in $x_4$. First Hölder's inequality with three equal exponents yieldsThe fourth factor $f_4(x_1,x_2,x_3)$ is independent of $x_4$. A further Hölder's inequality with exponents $3$ and $3/2$, followed by the three-dimensional estimate, therefore givesThe marginal-integral identifications use Tonelli's theorem. Now take $f_j=1_{K_j}$. The projections are compact and hence measurable with finite measure. Every $x\in K$ has all its projected coordinates in the corresponding $K_j$, so $1_K(x)\le {\prod }_j{1}_{K_j}({\hat{x}}_j{)}^{1/3}$. ConsequentlyThis includes zero-measure projections and is sharp for axis-parallel rectangular boxes.

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 $g=h=0$.

Use the printed bilinear pairing ${\varphi }_u(v)=\int uv$, which is complex linear in both arguments. For the conjugate exponents $p,q$, Holder inequality gives $|{\varphi }_u(v)|\le \Vert u{\Vert }_q\Vert v{\Vert }_p$, proving well-definedness, continuity, and $\Vert {\varphi }_u\Vert \le \Vert u{\Vert }_q$. Linearity of $u\mapsto {\varphi }_u$ follows from linearity of integration.

For $u\ne 0$, takewith the numerator defined as zero where $u=0$. Since $(q-1)p=q$, this has $\Vert v{\Vert }_p=1$ and ${\varphi }_u(v)=\Vert u{\Vert }_q$. Hence $\boxed{\Vert {\varphi }_u\Vert =\Vert u{\Vert }_q}$; the same identity holds for $u=0$. This also proves injectivity.

For surjectivity, let $\Lambda$ be a nonzero bounded linear functional on $L^p$. Choose $f$ with $\Lambda (f)=1$ and let $K=\ker \Lambda$. The closest point theorem for a closed convex subset of Lp supplies the nearest point $h\in K$; write $w=f-h\ne 0$. For every $z\in K$, the whole line $h+tz$ lies in $K$, so the norm derivative at the minimum vanishes for real $t$. Applying this also to $iz$ shows that the complex linear functionalvanishes on $K$. Since $\Lambda (w)=1$, each $v-\Lambda (v)w$ is in $K$, giving $\Psi (v)=\Lambda (v)\Psi (w)$ and $\Psi (w)=\Vert w{\Vert }_p^p$. ThereforeMembership follows from $(p-1)q=p$. The zero functional is represented by $u=0$. This proves the asserted Lp duality isometric isomorphism, including surjectivity, without a separate representation theorem.

Use the bilinear pairing $T_g(f)={\int }_{{\mathbb{R}}^n}fg$, so the map $g\mapsto T_g$ is complex linear. With the sesquilinear convention $\int f\overline{g}$ the representing map would instead be conjugate linear. For conjugate exponents $p,q$, Hölder's inequality gives $|T_g(f)|\le \Vert g{\Vert }_q\Vert f{\Vert }_p$. If $g\ne 0$, take $v=|g{|}^{q-2}\overline{g}$, defined as zero where $g=0$. ThenConsequently $\Vert T_g\Vert =\Vert g{\Vert }_q$. This also proves injectivity of the representation.

For surjectivity, let $\ell$ be a nonzero element of the continuous dual space of $L^p$ and choose $f_0$ with $\ell (f_0)=1$. Its kernel $K$ is a closed linear subspace. Use the allowed best-approximation fact to choose $h\in K$ and put $z=f_0-h$. Thus $\ell (z)=1$ andFor each $f$, the vector $f-\ell (f)z$ belongs to $K$. Therefore $\Phi (f)=\ell (f)\Vert z{\Vert }_p^p$, givingMembership follows from $(p-1)q=p$; the zero functional is represented by zero. This proves the isometric isomorphism $(L^p({\mathbb{R}}^n){)}^{\ast }\cong L^q({\mathbb{R}}^n)$ without using a representation theorem as a substitute for the proof. It is the closed-hyperplane proof of Lp duality.

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 $p$.

The feasible set is closed and contained in the rectangleIt is therefore compact, and the continuous function $xy$ attains a global maximum there by the extreme value theorem. At the two boundary endpoints where a coordinate is zero, the product is zero. Meanwhile $(1,1)$ is feasible because the conjugate exponents satisfy $1/p+1/q=1$, and its product is one. Hence every maximizer has $x,y>0$.

On that positive region the constraint is $C^1$ with nonzero gradient. The Lagrange multiplier equations areThe first two identities imply $xy=\lambda x^p=\lambda y^q$. Since $x,y>0$, the multiplier is nonzero, so $x^p=y^q=r$. The constraint now givesThus the sole interior candidate is $(1,1)$. Existence of a maximum and the boundary comparison prove the unique global resultThis conjugate-power constrained product maximum is valid for all real $p,q>1$ with $1/p+1/q=1$; no integrality of the exponents is needed.

For a nonnegative $f$, first set $g_n(x)=\min (f(x),n)1_{(1/n,n)}(x)$ and $F_n(x)={\int }_0^x{g}_n(t)dt$. These bounded functions vanish near zero and have compact support. Their averages have finite Lp norms: they vanish near zero and decay as $1/x$ beyond $n$. Integration by parts givesThe boundary term $x^{1-p}{F}_n(x{)}^p$ vanishes at zero because $F_n$ is zero there, and at infinity because $F_n$ is eventually constant and $p>1$. The inequality is Hölder's inequality with conjugate exponents $p$ and $p/(p-1)$. Divide when $I_n>0$; the zero case is immediate. Since $g_n\uparrow f$ and $H{g}_n\uparrow Hf$, the monotone convergence theorem proves the full bound without presupposing that $Hf$ has finite Lp norm. Apply it to $|f|$ for the signed or complex case.^[ref]

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 $p$.