Proof of The Gagliardo-Nirenberg Inequality on the Torus in Dimensions at Most Three

Each result cited is universally quantified over the data in its own statement, and is applied here to the data named below. We use silently that the order of $\mathbb{R}$ is reflexive, transitive and antisymmetric, and that equal real numbers satisfy $\le$ in both directions. Throughout, for $i\in[n]$ put

$$\Theta_{i}=|u|+|\partial_{i}u|,\qquad F_{i}=P_{i}\Theta_{i}.$$

In each of the three claims the hypothesis on $n$ gives $n\le3$, so The Slice Average of a Continuous Periodic Function (vii) is available throughout.

(Q1) The functions $\Theta_{i}$ and $F_{i}$. Let $i\in[n]$. Then $\Theta_{i}\in C_{\mathrm{per}}$ and $0\le\Theta_{i}(y)$ for every $y\in\mathbb{R}^{n}$; the map $F_{i}$ lies in $C_{\mathrm{per}}$, satisfies $0\le F_{i}(y)$ for every $y$, and is free of the $i$th coordinate in the sense of The Slice Average of a Continuous Periodic Function; and

$$|u(y)|\le F_{i}(y)\qquad\text{for every }y\in\mathbb{R}^{n}.$$

Indeed, $u$ and $\partial_{i}u$ lie in $C_{\mathrm{per}}$ as recorded in the statement, so $|u|$ and $|\partial_{i}u|$ lie in $C_{\mathrm{per}}$ by The Slice Average of a Continuous Periodic Function (v) and their sum $\Theta_{i}$ does by Elementary Properties of Lattice-Periodic Functions (i); the values $|u(y)|$ and $|\partial_{i}u(y)|$ are nonnegative by claim 1 of Properties of the Absolute Value in an Ordered Field, so $0\le\Theta_{i}(y)$ by claim 2 of Elementary Arithmetic in an Ordered Field. By The Slice Average of a Continuous Periodic Function (i), applied to $\Theta_{i}$, the map $F_{i}$ lies in $C_{\mathrm{per}}$ and takes nonnegative values, and by The Slice Average of a Continuous Periodic Function (ii) it is free of the $i$th coordinate. The final inequality is The One-Dimensional Slice Bound for a Continuously Differentiable Periodic Function (i), applied to $u$ and $i$, whose $\Theta_{i}$ and $P_{i}\Theta_{i}$ are the ones named here.

(Q2) Products and powers. Let $v,w\in C_{\mathrm{per}}$ satisfy $0\le v(y)$ and $0\le w(y)$ for every $y\in\mathbb{R}^{n}$, and let $a$ be a positive real number. Then $vw\in C_{\mathrm{per}}$ with $0\le(vw)(y)$ for every $y$, and $v^{a}\in C_{\mathrm{per}}$ with $0\le v^{a}(y)$ for every $y$. If $j\in[n]$ and $v$ is free of the $j$th coordinate, so is $v^{a}$; if both $v$ and $w$ are free of the $j$th coordinate, so is $vw$.

Indeed, $vw\in C_{\mathrm{per}}$ by Elementary Properties of Lattice-Periodic Functions (i), and $0=0\cdot w(y)\le v(y)w(y)$ by claim 1 of Zero Products and Elementary Identities in a Field and claim 5 of Elementary Arithmetic in an Ordered Field. The assertions about $v^{a}$, and the freeness of $v^{a}$, are The Slice Average of a Continuous Periodic Function (v). Finally $(vw)(x[j{:}s])=v(x[j{:}s])w(x[j{:}s])=v(x)w(x)=(vw)(x)$ when both factors are free of the $j$th coordinate.

(Q3) Weak multiplication of inequalities. Let $\alpha,\beta,\gamma,\delta$ be real numbers with $0\le\alpha$, $\alpha\le\beta$, $0\le\gamma$ and $\gamma\le\delta$. Then $\alpha\gamma\le\beta\delta$.

Indeed, $0\le\beta$ by transitivity, so claim 5 of Elementary Arithmetic in an Ordered Field gives $\alpha\gamma\le\beta\gamma$ and $\beta\gamma\le\beta\delta$; transitivity concludes.

(Q4) Integration over the cell. Let $v\in C_{\mathrm{per}}$ satisfy $0\le v(y)$ for every $y\in\mathbb{R}^{n}$. Then $v|_{Q}$ is measurable with respect to $\mathcal{B}_{Q}$, belongs to $\mathcal{L}^{1}(\mathbb{T}^{n})$, and is integrable with respect to $\lambda_{Q}$.

Indeed, by Elementary Properties of Lattice-Periodic Functions (iii) there is a nonnegative real $M$ with $|v(y)|\le M$ for every $y$, so Continuous Periodic Functions are Power-Integrable and Dense on the Torus (i), applied with $p=1$, gives measurability and membership in $\mathcal{L}^{1}(\mathbb{T}^{n})$. By Power-Integrable Functions and the p-Seminorm (ii) this means $\int_{\mathbb{T}^{n}}|v|_{Q}|^{1}\,dx<\infty$, and $(|v(y)|)^{1}=|v(y)|$ by Properties of Real Powers (ii), so $\int_{\mathbb{T}^{n}}|v|_{Q}|\,dx<\infty$ and $v|_{Q}$ is integrable by the criterion recorded in Measurable Functions and the Lebesgue Integral: Standing Notation (v).

(Q5) Monotonicity over the cell. Let $v,w\in C_{\mathrm{per}}$ take nonnegative values and satisfy $v(y)\le w(y)$ for every $y\in\mathbb{R}^{n}$. Then $\int_{\mathbb{T}^{n}}v|_{Q}\,dx\le\int_{\mathbb{T}^{n}}w|_{Q}\,dx$, and both integrals are nonnegative.

Indeed, both restrictions are integrable by (Q4), so the monotonicity in claim 2 of Linearity and Monotonicity of the Lebesgue Integral, and the Integrability Criterion applies; nonnegativity follows by comparing with the map of constant value $0$ on $Q$, which is integrable with integral $0$ by that same claim applied with both coefficients $0$.

(Q6) A map free of every coordinate. Let $v\in C_{\mathrm{per}}$ take nonnegative values and be free of the $j$th coordinate for every $j\in[n]$. Then $v$ has the constant value $\int_{\mathbb{T}^{n}}v|_{Q}\,dx$.

Indeed, The Slice Average of a Continuous Periodic Function (vi) provides a real number $c$ with $v(y)=c$ for every $y\in\mathbb{R}^{n}$. Then $v|_{Q}$ is $c$ times the indicator of $Q$ formed in the measure space $(Q,\mathcal{B}_{Q},\lambda_{Q})$, whose integral is $\lambda_{Q}(Q)=1$ by The Integral of an Indicator Function is the Measure of the Set and The Flat Torus: Standing Notation (iii); so the linearity in claim 2 of Linearity and Monotonicity of the Lebesgue Integral, and the Integrability Criterion gives $\int_{\mathbb{T}^{n}}v|_{Q}\,dx=c\cdot1=c$.

(Q7) The integral of $F_{i}$. Suppose $n\le3$ and let $i\in[n]$. Then

$$\int_{\mathbb{T}^{n}}F_{i}|_{Q}\,dx=A_{i}.$$

Indeed, The Slice Average of a Continuous Periodic Function (vii), applied with the map $\Theta_{i}$ of (Q1), gives $\int_{\mathbb{T}^{n}}F_{i}|_{Q}\,dx=\int_{\mathbb{T}^{n}}\Theta_{i}|_{Q}\,dx$. The maps $|u|_{Q}|$ and $|(\partial_{i}u)|_{Q}|$, in the notation of Power-Integrable Functions and the p-Seminorm (i) with exponent $1$, are integrable, since $u|_{Q}$ and $(\partial_{i}u)|_{Q}$ lie in $\mathcal{L}^{1}(\mathbb{T}^{n})$ and $(|\tau|)^{1}=|\tau|$ by Properties of Real Powers (ii); and $\Theta_{i}|_{Q}$ is their pointwise sum on $Q$. So the linearity in claim 2 of Linearity and Monotonicity of the Lebesgue Integral, and the Integrability Criterion gives

$$\int_{\mathbb{T}^{n}}\Theta_{i}|_{Q}\,dx=\int_{\mathbb{T}^{n}}|u|_{Q}|\,dx+\int_{\mathbb{T}^{n}}\bigl|(\partial_{i}u)|_{Q}\bigr|\,dx .$$

By Elementary Properties of the p-Seminorm (i), applied with $p=1$, together with $(\lVert f\rVert_{1})^{1}=\lVert f\rVert_{1}$ from Properties of Real Powers (ii), each of these two integrals is the corresponding $1$-seminorm, so the sum is $A_{i}$.

(Q8) The members of the initial segments. Every $j\in[1]$ equals $1$; every $j\in[2]$ equals $1$ or $2$; every $j\in[3]$ equals $1$, $2$ or $3$. Here $2=S(1)$ and $3=S(2)$ for the successor map $S$ of Natural Numbers.

Indeed, $j\in[m]$ means $1\le j$ and $j\le m$. For $m=1$ antisymmetry gives $j=1$. For $m=2$, either $j=2$ or $j\le1$ by claim 5 of Properties of the Order on the Natural Numbers, and in the latter case $j=1$ by antisymmetry. For $m=3$, either $j=3$ or $j\le2$ by that same claim, and the case $m=2$ applies.

Proof of claim 1. Suppose $n=1$. By (Q1) the map $F_{1}$ is free of the first coordinate, and by (Q8) every $j\in[1]$ equals $1$, so $F_{1}$ is free of the $j$th coordinate for every $j\in[n]$. By (Q6) and (Q7) it therefore has the constant value $A_{1}$. Hence, by the last assertion of (Q1), $|u(x)|\le F_{1}(x)=A_{1}$ for every $x\in\mathbb{R}^{n}$.

Proof of claim 2. Suppose $n=2$.

A pointwise bound. Let $y\in\mathbb{R}^{2}$. By Properties of Real Powers (iv) and Properties of Real Powers (ii), $(|u(y)|)^{2}=(|u(y)|)^{1+1}=|u(y)|\cdot|u(y)|$. By (Q1), $0\le|u(y)|\le F_{1}(y)$ and $0\le|u(y)|\le F_{2}(y)$, so (Q3) gives

$$(|u(y)|)^{2}\le F_{1}(y)F_{2}(y).$$

Integration. The map $F_{1}F_{2}$ lies in $C_{\mathrm{per}}$ and takes nonnegative values by (Q2). The map $|u|_{Q}|^{2}$ of Power-Integrable Functions and the p-Seminorm (i) is integrable, since $u|_{Q}\in\mathcal{L}^{2}(\mathbb{T}^{n})$ makes its integral finite and it takes nonnegative values, so that it is integrable by the criterion in Measurable Functions and the Lebesgue Integral: Standing Notation (v) together with $\bigl|(|u(y)|)^{2}\bigr|=(|u(y)|)^{2}$, which holds by claim 1 of Properties of the Absolute Value in an Ordered Field and Properties of Real Powers (i). By the displayed bound, restricted to $Q$, and the monotonicity in claim 2 of Linearity and Monotonicity of the Lebesgue Integral, and the Integrability Criterion together with (Q4) applied to $F_{1}F_{2}$,

$$\int_{\mathbb{T}^{2}}|u|_{Q}|^{2}\,dx\le\int_{\mathbb{T}^{2}}(F_{1}F_{2})|_{Q}\,dx .$$

By Elementary Properties of the p-Seminorm (i), applied with $p=2$, the left-hand side is $\bigl(\lVert u|_{Q}\rVert_{2}\bigr)^{2}$.

Evaluation of the right-hand side. By The Slice Average of a Continuous Periodic Function (vii), applied with $i=2$ and the map $F_{1}F_{2}$,

$$\int_{\mathbb{T}^{2}}(F_{1}F_{2})|_{Q}\,dx=\int_{\mathbb{T}^{2}}\bigl(P_{2}(F_{1}F_{2})\bigr)|_{Q}\,dx .$$

By (Q1) the map $F_{2}$ is free of the second coordinate, so The Slice Average of a Continuous Periodic Function (iii), applied with $i=2$, $v=F_{2}$ and $w=F_{1}$, gives $P_{2}(F_{1}F_{2})=F_{2}\,P_{2}F_{1}$, the maps $F_{1}F_{2}$ and $F_{2}F_{1}$ being equal. The map $P_{2}F_{1}$ lies in $C_{\mathrm{per}}$ and takes nonnegative values by The Slice Average of a Continuous Periodic Function (i), and by The Slice Average of a Continuous Periodic Function (ii) it is free of the second coordinate and, $F_{1}$ being free of the first, also of the first; by (Q8) it is therefore free of the $j$th coordinate for every $j\in[2]$. Hence, by (Q6), it has the constant value

$$\int_{\mathbb{T}^{2}}(P_{2}F_{1})|_{Q}\,dx=\int_{\mathbb{T}^{2}}F_{1}|_{Q}\,dx=A_{1},$$

the first equality by The Slice Average of a Continuous Periodic Function (vii) applied with $i=2$ and the map $F_{1}$, and the second by (Q7). So $P_{2}(F_{1}F_{2})$ is the map $A_{1}F_{2}$, and the linearity in claim 2 of Linearity and Monotonicity of the Lebesgue Integral, and the Integrability Criterion, with (Q4) applied to $F_{2}$, gives

$$\int_{\mathbb{T}^{2}}\bigl(P_{2}(F_{1}F_{2})\bigr)|_{Q}\,dx=A_{1}\int_{\mathbb{T}^{2}}F_{2}|_{Q}\,dx=A_{1}A_{2},$$

the last step by (Q7). Combining the three displays proves claim 2.

Proof of claim 3. Suppose $n=3$ and put

$$G=\bigl(F_{1}F_{2}F_{3}\bigr)^{1/2},$$

where $F_{1}F_{2}F_{3}$ denotes $(F_{1}F_{2})F_{3}$. By (Q2) the maps $F_{1}F_{2}$, $F_{1}F_{2}F_{3}$ and $G$ lie in $C_{\mathrm{per}}$ and take nonnegative values.

A pointwise bound. Let $y\in\mathbb{R}^{3}$ and abbreviate $\mu=|u(y)|$. By Properties of Real Powers (iv) and Properties of Real Powers (ii), $\mu^{3}=\mu^{1+1+1}=\mu\,\mu\,\mu$. By (Q1), $0\le\mu\le F_{1}(y)$, $0\le\mu\le F_{2}(y)$ and $0\le\mu\le F_{3}(y)$, so (Q3) gives first $\mu\mu\le F_{1}(y)F_{2}(y)$ and then, applied again with $\alpha=\mu\mu$ and $\gamma=\mu$,

$$\mu^{3}\le F_{1}(y)F_{2}(y)F_{3}(y),$$

the number $\mu\mu$ being nonnegative by claim 1 of Zero Products and Elementary Identities in a Field and claim 5 of Elementary Arithmetic in an Ordered Field. Applying Properties of Real Powers (v) with exponent $1/2$ and then Properties of Real Powers (iv), which gives $(\mu^{3})^{1/2}=\mu^{3/2}$, we obtain

$$(|u(y)|)^{3/2}\le G(y)\qquad\text{for every }y\in\mathbb{R}^{3}.$$

Integration. The map $|u|_{Q}|^{3/2}$ of Power-Integrable Functions and the p-Seminorm (i) is integrable, since $u|_{Q}\in\mathcal{L}^{3/2}(\mathbb{T}^{n})$ makes its integral finite and its values are nonnegative by Properties of Real Powers (i), so that claim 1 of Properties of the Absolute Value in an Ordered Field and the criterion in Measurable Functions and the Lebesgue Integral: Standing Notation (v) apply. With (Q4) applied to $G$, the monotonicity in claim 2 of Linearity and Monotonicity of the Lebesgue Integral, and the Integrability Criterion and Elementary Properties of the p-Seminorm (i) with $p=3/2$ give

$$\bigl(\lVert u|_{Q}\rVert_{3/2}\bigr)^{3/2}=\int_{\mathbb{T}^{3}}|u|_{Q}|^{3/2}\,dx\le\int_{\mathbb{T}^{3}}G|_{Q}\,dx .$$

It remains to bound the right-hand side by $(A_{1}A_{2}A_{3})^{1/2}$, which is done by removing one coordinate at a time.

Removing the third coordinate. By The Slice Average of a Continuous Periodic Function (vii), applied with $i=3$ and the map $G$,

$$\int_{\mathbb{T}^{3}}G|_{Q}\,dx=\int_{\mathbb{T}^{3}}(P_{3}G)|_{Q}\,dx .$$

Since $F_{1}F_{2}F_{3}$ and $F_{3}(F_{1}F_{2})$ are the same map, Properties of Real Powers (iii) gives $G=F_{3}^{1/2}\,(F_{1}F_{2})^{1/2}$. By (Q1) and (Q2) the map $F_{3}^{1/2}$ lies in $C_{\mathrm{per}}$, takes nonnegative values and is free of the third coordinate, so The Slice Average of a Continuous Periodic Function (iii), applied with $i=3$, gives

$$P_{3}G=F_{3}^{1/2}\,P_{3}\bigl[(F_{1}F_{2})^{1/2}\bigr].$$

By The Slice Average of a Continuous Periodic Function (iv), applied with $i=3$, $v=F_{1}$ and $w=F_{2}$, the value of $P_{3}[(F_{1}F_{2})^{1/2}]$ at any point is at most the product of the values of $(P_{3}F_{1})^{1/2}$ and $(P_{3}F_{2})^{1/2}$ there; multiplying by the nonnegative value of $F_{3}^{1/2}$, using claim 5 of Elementary Arithmetic in an Ordered Field, gives $P_{3}G(y)\le H_{1}(y)$ for every $y\in\mathbb{R}^{3}$, where

$$H_{1}=F_{3}^{1/2}\,(P_{3}F_{1})^{1/2}\,(P_{3}F_{2})^{1/2}.$$

Here $P_{3}F_{1}$ and $P_{3}F_{2}$ lie in $C_{\mathrm{per}}$ and take nonnegative values by The Slice Average of a Continuous Periodic Function (i), so $H_{1}$ does too, by (Q2). Hence (Q5) gives $\int_{\mathbb{T}^{3}}(P_{3}G)|_{Q}\,dx\le\int_{\mathbb{T}^{3}}H_{1}|_{Q}\,dx$.

Removing the second coordinate. By The Slice Average of a Continuous Periodic Function (vii) with $i=2$,

$$\int_{\mathbb{T}^{3}}H_{1}|_{Q}\,dx=\int_{\mathbb{T}^{3}}(P_{2}H_{1})|_{Q}\,dx .$$

The map $F_{2}$ is free of the second coordinate by (Q1), so $P_{3}F_{2}$ is free of the second coordinate by The Slice Average of a Continuous Periodic Function (ii) and $(P_{3}F_{2})^{1/2}$ is by (Q2). Since $H_{1}$ is the product of $(P_{3}F_{2})^{1/2}$ with $F_{3}^{1/2}(P_{3}F_{1})^{1/2}$, which equals $(F_{3}\,P_{3}F_{1})^{1/2}$ by Properties of Real Powers (iii), the clause The Slice Average of a Continuous Periodic Function (iii) with $i=2$ gives

$$P_{2}H_{1}=(P_{3}F_{2})^{1/2}\,P_{2}\bigl[(F_{3}\,P_{3}F_{1})^{1/2}\bigr],$$

and The Slice Average of a Continuous Periodic Function (iv) with $i=2$, $v=F_{3}$ and $w=P_{3}F_{1}$, followed by multiplication by the nonnegative value of $(P_{3}F_{2})^{1/2}$ as above, gives $P_{2}H_{1}(y)\le H_{2}(y)$ for every $y$, where

$$H_{2}=(P_{3}F_{2})^{1/2}\,(P_{2}F_{3})^{1/2}\,(P_{2}P_{3}F_{1})^{1/2}.$$

The map $P_{2}P_{3}F_{1}$ lies in $C_{\mathrm{per}}$ and takes nonnegative values by The Slice Average of a Continuous Periodic Function (i), and it is free of every coordinate: of the first because $F_{1}$ is, by two applications of The Slice Average of a Continuous Periodic Function (ii); of the third because $P_{3}F_{1}$ is, by that same clause applied once more; and of the second by that clause applied to $P_{2}$. By (Q8) this covers every $j\in[3]$, so (Q6) makes it the constant

$$\int_{\mathbb{T}^{3}}(P_{2}P_{3}F_{1})|_{Q}\,dx=\int_{\mathbb{T}^{3}}(P_{3}F_{1})|_{Q}\,dx=\int_{\mathbb{T}^{3}}F_{1}|_{Q}\,dx=A_{1},$$

the first two equalities by The Slice Average of a Continuous Periodic Function (vii) applied with $i=2$ and with $i=3$, and the last by (Q7). Hence $H_{2}=A_{1}^{1/2}K$, where

$$K=(P_{3}F_{2})^{1/2}\,(P_{2}F_{3})^{1/2},$$

a member of $C_{\mathrm{per}}$ with nonnegative values by (Q2), the map $P_{2}F_{3}$ lying in $C_{\mathrm{per}}$ and taking nonnegative values by The Slice Average of a Continuous Periodic Function (i). By (Q5), (Q4) and the linearity in claim 2 of Linearity and Monotonicity of the Lebesgue Integral, and the Integrability Criterion,

$$\int_{\mathbb{T}^{3}}(P_{2}H_{1})|_{Q}\,dx\le\int_{\mathbb{T}^{3}}H_{2}|_{Q}\,dx=A_{1}^{1/2}\int_{\mathbb{T}^{3}}K|_{Q}\,dx .$$

Removing the first coordinate. By Properties of Real Powers (iii), $K=(P_{3}F_{2}\,P_{2}F_{3})^{1/2}$, and by The Slice Average of a Continuous Periodic Function (vii) with $i=1$,

$$\int_{\mathbb{T}^{3}}K|_{Q}\,dx=\int_{\mathbb{T}^{3}}(P_{1}K)|_{Q}\,dx .$$

By The Slice Average of a Continuous Periodic Function (iv) with $i=1$, $v=P_{3}F_{2}$ and $w=P_{2}F_{3}$, the value of $P_{1}K$ at any point is at most the product of the values of $(P_{1}P_{3}F_{2})^{1/2}$ and $(P_{1}P_{2}F_{3})^{1/2}$ there. Exactly as in the previous paragraph, $P_{1}P_{3}F_{2}$ lies in $C_{\mathrm{per}}$, takes nonnegative values and is free of the second coordinate (because $F_{2}$ is), of the third (because $P_{3}F_{2}$ is) and of the first (by the clause applied to $P_{1}$), so by (Q8), (Q6), two applications of The Slice Average of a Continuous Periodic Function (vii) and (Q7) it is the constant $A_{2}$; and likewise $P_{1}P_{2}F_{3}$ is the constant $A_{3}$, being free of the third, second and first coordinates. Hence

$$(P_{1}K)(y)\le A_{2}^{1/2}A_{3}^{1/2}\qquad\text{for every }y\in\mathbb{R}^{3}.$$

Write $\kappa=A_{2}^{1/2}A_{3}^{1/2}$, a nonnegative real by Properties of Real Powers (i) and claim 5 of Elementary Arithmetic in an Ordered Field. The map on $Q$ with constant value $\kappa$ is $\kappa$ times the indicator of $Q$ formed in $(Q,\mathcal{B}_{Q},\lambda_{Q})$, hence integrable with integral $\kappa\,\lambda_{Q}(Q)=\kappa$ by The Integral of an Indicator Function is the Measure of the Set, The Flat Torus: Standing Notation (iii) and claim 2 of Linearity and Monotonicity of the Lebesgue Integral, and the Integrability Criterion; so by (Q4) applied to $P_{1}K$ and the monotonicity in that same claim,

$$\int_{\mathbb{T}^{3}}(P_{1}K)|_{Q}\,dx\le\kappa=A_{2}^{1/2}A_{3}^{1/2}.$$

Conclusion. Chaining the displays of the last four paragraphs,

$$\bigl(\lVert u|_{Q}\rVert_{3/2}\bigr)^{3/2}\le A_{1}^{1/2}A_{2}^{1/2}A_{3}^{1/2},$$

and two applications of Properties of Real Powers (iii) identify the right-hand side with $(A_{1}A_{2}A_{3})^{1/2}$. This proves claim 3.