Proof of The Function $s\log s$: Continuity, Young's Inequality and Lower Bounds, with the Elementary Bounds for the Exponential and the Logarithm

Each result cited below is universally quantified over the data in its own statement. Throughout, we use the identities $\log(\exp(u))=u$ for $u\in\mathbb{R}$ and $\exp(\log t)=t$ for positive $t$, and the rule $\log(st)=\log s+\log t$ for positive $s,t$, all recorded in The Natural Logarithm. By claim 1 of Basic Properties of the Exponential Function, $\exp(0)=1$, hence $\log 1=\log(\exp(0))=0$.

Claim 1. Let $u\in\mathbb{R}$. If $0\le u$, then $1+u\le\exp(u)$ by claim 4 of Basic Properties of the Exponential Function. Suppose now $u<0$, and let $g:[u,0]\to\mathbb{R}$ be the restriction of $\exp$ to the closed interval $[u,0]$. The set $\mathbb{R}$ is an interval, and each $x\in\mathbb{R}$ is an interior point of it, since $x-1<x<x+1$. By claim 3 of Basic Properties of the Exponential Function, $\exp$ is differentiable at every $x\in\mathbb{R}$ with derivative $\exp(x)$; hence, by Differentiability at an Interior Point Implies Continuity There, $\exp$ is continuous at every $x$ relative to $\mathbb{R}$. Consequently $g$ is continuous on $[u,0]$, because the condition defining continuity relative to $[u,0]$ only concerns points $y$ of $[u,0]\subseteq\mathbb{R}$. Likewise, for every $c$ with $u<c<0$ the point $c$ is an interior point of $[u,0]$, and $g$ is differentiable at $c$ with $g'(c)=\exp(c)$: a $\delta$ that serves for $\exp$ at $c$ in the definition of the derivative also serves for $g$, because the condition for $g$ only concerns those $h$ with $c+h\in[u,0]$. By Mean Value Theorem on a Closed Real Interval there is $c$ with $u<c<0$ and

Since $\exp$ is strictly increasing by claim 4 of Basic Properties of the Exponential Function and $c<0$, we have $\exp(c)<\exp(0)=1$. As $0\le-u$, multiplying $\exp(c)\le1$ by $-u$ (claim 5 of Elementary Arithmetic in an Ordered Field) gives $1-\exp(u)=\exp(c)(-u)\le-u$, that is, $1+u\le\exp(u)$.

Claim 2. Let $t$ be positive. Claim 1 with $u=\log t$ gives $1+\log t\le\exp(\log t)=t$, so $\log t\le t-1$. By claim 7 of Elementary Order Arithmetic in an Ordered Field, $t^{-1}$ exists and is positive, and $0=\log 1=\log(t\,t^{-1})=\log t+\log(t^{-1})$, so $\log(t^{-1})=-\log t$. The upper bound just proved, applied to $t^{-1}$, gives $-\log t\le t^{-1}-1$, that is, $1-t^{-1}\le\log t$.

Claim 3. Step 1 (continuity of the logarithm). Let $t$ be positive and let $0<\varepsilon$. Put $a=\exp(\log t-\varepsilon)$ and $b=\exp(\log t+\varepsilon)$; since $\exp$ is strictly increasing (claim 4 of Basic Properties of the Exponential Function) and $\exp(\log t)=t$, we have $0<a<t<b$, where $0<a$ is claim 2 there. By claim 9 of Elementary Order Arithmetic in an Ordered Field let $\delta$ be the smaller of $t-a$ and $b-t$; then $0<\delta$. Let $y$ be positive with $|y-t|<\delta$. By claim 9 of Properties of the Absolute Value in an Ordered Field, $t-\delta<y<t+\delta$, and as $a\le t-\delta$ and $t+\delta\le b$, claim 2 of Elementary Order Arithmetic in an Ordered Field gives $a<y<b$. Since $\log$ is strictly increasing on $(0,\infty)$ by claim 2 of Real Powers Through the Exponential, and Elementary Asymptotic Tools: Monotonicity, Null Sequences of Negative Powers, Exponential Domination, Integer Rounding, and Square-Root and Exponential Inequalities, and $\log a=\log t-\varepsilon$, $\log b=\log t+\varepsilon$, we get $\log t-\varepsilon<\log y<\log t+\varepsilon$, so $|\log y-\log t|<\varepsilon$ by claim 9 of Properties of the Absolute Value in an Ordered Field. Thus $\log$ is continuous at $t$ relative to $(0,\infty)$, as a map from the subset $(0,\infty)$ of the real line $(\mathbb{R},d_{\mathbb{R}})$ of The Absolute Value Metric on the Real Line into $(\mathbb{R},d_{\mathbb{R}})$.

Step 2 (continuity at a positive point). Let $s$ be positive and put $A=(0,\infty)$. The map $y\mapsto y$ on $A$ is continuous at $s$ relative to $A$ (given $0<\varepsilon$, take $\delta=\varepsilon$), and so is $\log$ by Step 1; hence their product $y\mapsto y\log y$, which agrees with $\phi$ on $A$, is continuous at $s$ relative to $A$ by claim 3 of Continuity of Sums and Products of Real-Valued Functions on a Metric Space. Let $0<\varepsilon$, let $\delta_{1}$ be positive with $|\phi(y)-\phi(s)|<\varepsilon$ for every $y\in A$ with $|y-s|<\delta_{1}$, and let $\delta$ be the smaller of $\delta_{1}$ and $s$ (claim 9 of Elementary Order Arithmetic in an Ordered Field). If $y\in[0,\infty)$ and $|y-s|<\delta$, then $s-y\le|s-y|=|y-s|<s$, by claim 3 of Properties of the Absolute Value in an Ordered Field applied to $s-y$, claim 2 there (as $s-y=-(y-s)$), and claim 2 of Elementary Order Arithmetic in an Ordered Field (as $|y-s|<\delta\le s$); so $0<y$ by claim 1 of Elementary Order Arithmetic in an Ordered Field, and $y\in A$, and $|y-s|<\delta_{1}$; hence $|\phi(y)-\phi(s)|<\varepsilon$. So $\phi$ is continuous at $s$ relative to $[0,\infty)$.

Step 3 (continuity at $0$). First let $0<y<1$; we show $|\phi(y)|\le2r$, where $r$ is the nonnegative real number with $r^{2}=y$ given by Existence and Uniqueness of the Nonnegative Square Root. Since $r^{2}=y\ne0$, $r\ne0$, so $r$ is positive; and $r<1$ by claim 1 of Monotonicity of Squaring on the Nonnegative Elements of an Ordered Field, since $r^{2}=y<1=1^{2}$. Now $\log y=\log(r\,r)=2\log r$, and $1-r^{-1}\le\log r$ by Claim 2, so multiplying by $2$, which is positive by claim 8 of Elementary Order Arithmetic in an Ordered Field, gives $2-2r^{-1}\le\log y$ (claim 5 of Elementary Arithmetic in an Ordered Field); multiplying by $y=r^{2}$, which is nonnegative (claim 5 of Elementary Arithmetic in an Ordered Field), gives $2r^{2}-2r\le y\log y$, and $2r^{2}-2r\ge-2r$ because $0<2r^{2}$ (claim 5 of Elementary Order Arithmetic in an Ordered Field, as $0<2$ and $0<r^{2}$). On the other hand $\log y\le y-1<0$ by Claim 2, so multiplying $\log y\le0$ by $y$ (claim 5 of Elementary Arithmetic in an Ordered Field) gives $y\log y\le0\le 2r$, where $0<2r$ by claim 5 of Elementary Order Arithmetic in an Ordered Field. Hence $-2r\le\phi(y)\le2r$, and $|\phi(y)|\le2r$ by claim 6 of Properties of the Absolute Value in an Ordered Field.

Now let $0<\varepsilon$, and by claim 9 of Elementary Order Arithmetic in an Ordered Field let $\delta$ be the smaller of $1$ and $\varepsilon^{2}/4$; then $0<\delta$. Let $y\in[0,\infty)$ with $|y-0|<\delta$. If $y=0$, then $|\phi(y)-\phi(0)|=0<\varepsilon$. Otherwise $0<y<1$, and with $r$ as above, $r^{2}=y<\varepsilon^{2}/4=(\varepsilon/2)^{2}$, so $r<\varepsilon/2$ by claim 1 of Monotonicity of Squaring on the Nonnegative Elements of an Ordered Field; hence $|\phi(y)-\phi(0)|=|\phi(y)|\le2r<\varepsilon$. So $\phi$ is continuous at $0$ relative to $[0,\infty)$, and with Step 2, $\phi$ is continuous on $[0,\infty)$.

Step 4 (measurability). For $x\in\mathbb{R}$ let $x^{+}=x$ if $0\le x$ and $x^{+}=0$ otherwise, so that $x^{+}\in[0,\infty)$. For all $x,y\in\mathbb{R}$ we have $|y^{+}-x^{+}|\le|y-x|$: if $x,y$ are both nonnegative the two sides are equal; if both are negative the left side is $0$; if $0\le y$ and $x<0$ the left side is $y\le y-x=|y-x|$; and the remaining case is symmetric. Let $\psi:\mathbb{R}\to\mathbb{R}$ be given by $\psi(x)=\phi(x^{+})$. Then $\psi$ is continuous on $\mathbb{R}$: given $x\in\mathbb{R}$ and $0<\varepsilon$, let $\delta$ be positive with $|\phi(z)-\phi(x^{+})|<\varepsilon$ for all $z\in[0,\infty)$ with $|z-x^{+}|<\delta$ (Steps 2 and 3); if $|y-x|<\delta$ then $|y^{+}-x^{+}|<\delta$ by claim 2 of Elementary Order Arithmetic in an Ordered Field, so $|\psi(y)-\psi(x)|<\varepsilon$.

By claim 2 of Borel Measurability and Bounded Integration on a Metric Space, the Borel $\sigma$-algebra of the metric space $(\mathbb{R},d_{\mathbb{R}})$ is $\mathcal{B}(\mathbb{R})$; so by claim 3 of that lemma, applied with both metric spaces equal to $(\mathbb{R},d_{\mathbb{R}})$, $\psi$ is measurable with respect to $\mathcal{B}(\mathbb{R})$ and $\mathcal{B}(\mathbb{R})$. Let $(X,\mathcal{F},\mu)$ be a measure space and let $f:X\to\mathbb{R}$ be measurable with $0\le f(x)$ for every $x\in X$. By Measure Spaces and the Lebesgue Integral: Standing Notation §measurable, $f$ is measurable with respect to $\mathcal{F}$ and $\mathcal{B}(\mathbb{R})$. By claim 4 of Borel Measurability and Bounded Integration on a Metric Space, applied with the metric space $(\mathbb{R},d_{\mathbb{R}})$, the measurable space $(X,\mathcal{F})$ in the role of $(\Omega,\mathcal{F})$, the measurable space $(\mathbb{R},\mathcal{B}(\mathbb{R}))$ in the role of $(Z,\mathcal{G})$, and the maps $f$ and $\psi$, the composite $\psi\circ f$ is measurable with respect to $\mathcal{F}$ and $\mathcal{B}(\mathbb{R})$, that is, measurable in the sense of Measure Spaces and the Lebesgue Integral: Standing Notation §measurable. Since $f(x)^{+}=f(x)$ for every $x\in X$, $\psi\circ f=\phi\circ f$, which is therefore measurable.

Claim 4. Let $a\in\mathbb{R}$ and $s\in[0,\infty)$. If $s=0$, then $a\,s=0<\exp(a-1)=\phi(0)+\exp(a-1)$ by claim 2 of Basic Properties of the Exponential Function. Let $s$ be positive. By claim 1 of Basic Properties of the Exponential Function,

By Claim 1, $a-\log s=1+(a-1-\log s)\le\exp(a-1-\log s)$, and multiplying by $s$ (claim 5 of Elementary Arithmetic in an Ordered Field) gives $a\,s-s\log s\le\exp(a-1)$, that is, $a\,s\le\phi(s)+\exp(a-1)$.

Claim 5. Let $s\in[0,\infty)$. If $s=0$, then $s-1=-1<0=\phi(s)$. If $s$ is positive, then $1-s^{-1}\le\log s$ by Claim 2, and multiplying by $s$ (claim 5 of Elementary Arithmetic in an Ordered Field) gives $s-1\le s\log s=\phi(s)$. Finally, Claim 4 with $a=0$ gives $0\le\phi(s)+\exp(-1)$, that is, $-\exp(-1)\le\phi(s)$.