[proofplan] We first use homogeneity and the replacement $f \mapsto |f|$ to reduce the inequality to a one-variable estimate with $f(0)=1$ and $f(1)=r \geq 0$. For this normalized function the entropy is an explicit function $\mathcal E_p(r)$, and the desired inequality becomes an upper bound for the quotient $R_p(r)=\mathcal E_p(r)/(r-1)^2$. The main work is a calculus computation showing that $R_p$ increases up to $r=q/p$ and decreases afterwards. Evaluating $R_p$ at this unique maximizer gives exactly $p q/\Lambda(p,q)$, and continuity covers the endpoint cases. [/proofplan]

[step:Reduce the inequality to nonnegative two-point data with one endpoint normalized] Set $q := 1-p$. Define the entropy functional $\operatorname{Ent}_{\nu_p}$ on nonnegative functions $g:\{0,1\}\to[0,\infty)$ by \begin{align*} \operatorname{Ent}_{\nu_p}(g) := p g(1)\log g(1)+q g(0)\log g(0) -\bigl(p g(1)+q g(0)\bigr)\log\bigl(p g(1)+q g(0)\bigr), \end{align*} with $0\log 0:=0$. For every scalar $\lambda \in \mathbb{R}$ and every function $f:\{0,1\}\to\mathbb{R}$, \begin{align*} \operatorname{Ent}_{\nu_p}((\lambda f)^2) = \lambda^2 \operatorname{Ent}_{\nu_p}(f^2), \end{align*} because the two logarithmic contributions containing $\log \lambda^2$ cancel. Also, \begin{align*} \operatorname{Ent}_{\nu_p}(f^2) = \operatorname{Ent}_{\nu_p}(|f|^2), \end{align*} and the [reverse triangle inequality](/theorems/2300) gives \begin{align*} \bigl||f(1)|-|f(0)|\bigr| \leq |f(1)-f(0)|. \end{align*} Thus it is enough to prove the inequality for functions $f:\{0,1\}\to[0,\infty)$. Assume first that $f(0)>0$. Define $r:=f(1)/f(0) \in [0,\infty)$ and define the normalized function $u_r:\{0,1\}\to[0,\infty)$ by $u_r(0)=1$ and $u_r(1)=r$. By homogeneity, the desired inequality for $f$ is equivalent to the desired inequality for $u_r$. If $f(0)=0$, apply the already proved estimate to the functions $f_\varepsilon:\{0,1\}\to[0,\infty)$ defined by $f_\varepsilon(0)=\varepsilon$ and $f_\varepsilon(1)=f(1)$. As $\varepsilon \downarrow 0$, the squared difference $(f(1)-\varepsilon)^2$ converges to $(f(1)-0)^2$. The entropy also converges, because the map $x\mapsto x\log x$ on $[0,\infty)$, with $0\log0:=0$, is continuous at $0$ and continuous on $(0,\infty)$; hence the two-variable expression defining $\operatorname{Ent}_{\nu_p}$ is continuous on $[0,\infty)^2$. Letting $\varepsilon\downarrow0$ gives the estimate for $f$. Therefore it remains to prove, for every $r\geq0$, \begin{align*} \operatorname{Ent}_{\nu_p}(u_r^2) \leq \frac{p q}{\Lambda(p,q)}(r-1)^2. \end{align*} [/step]

[step:Rewrite the normalized inequality as a quotient estimate] For $r\geq0$, define the function $\mathcal E_p:[0,\infty)\to\mathbb{R}$ by \begin{align*} \mathcal E_p(r) := p r^2\log r^2-\bigl(p r^2+q\bigr)\log\bigl(p r^2+q\bigr), \end{align*} again with $0\log0:=0$. Since $u_r(0)=1$ and $u_r(1)=r$, \begin{align*} \operatorname{Ent}_{\nu_p}(u_r^2)=\mathcal E_p(r). \end{align*} Thus the desired estimate is \begin{align*} \mathcal E_p(r)\leq \frac{p q}{\Lambda(p,q)}(r-1)^2. \end{align*} For $r\neq1$, define the entropy quotient as the map $R_p:(0,\infty)\setminus\{1\}\to\mathbb{R}$ by \begin{align*} R_p(r):=\frac{\mathcal E_p(r)}{(r-1)^2}. \end{align*} The function $\mathcal E_p$ is twice differentiable on $(0,\infty)$, and a direct expansion at $r=1$ gives a finite continuous extension of $R_p$ to $r=1$. Hence it is enough to prove \begin{align*} R_p(r)\leq \frac{p q}{\Lambda(p,q)} \end{align*} for all $r>0$, and then pass to $r=0$ by continuity. [/step]

[step:Differentiate the entropy quotient and locate its monotonicity change]For $r>0$, differentiating $\mathcal E_p$ gives \begin{align*} \mathcal E_p'(r) =2pr\log\frac{r^2}{p r^2+q}. \end{align*} Therefore, for $r>0$ with $r\neq1$, \begin{align*} R_p'(r) =\frac{(r-1)\mathcal E_p'(r)-2\mathcal E_p(r)}{(r-1)^3}. \end{align*} Define $H_p:(0,\infty)\to\mathbb{R}$ by \begin{align*} H_p(r):=(q+pr)\log(p r^2+q)-2pr\log r. \end{align*} Substituting the formula for $\mathcal E_p'(r)$ and collecting logarithmic terms yields \begin{align*} (r-1)^3R_p'(r)=2H_p(r). \end{align*} We now determine the sign of $H_p$. The point $r=q/p$ satisfies $H_p(q/p)=0$. Differentiating $H_p$ once more gives \begin{align*} H_p'(r)=p\left(\log\frac{p r^2+q}{r^2}+\frac{2q(r-1)}{p r^2+q}\right), \end{align*} and a second differentiation gives \begin{align*} H_p''(r)=\frac{2pq(r-1)(q-pr^2)}{r(p r^2+q)^2}. \end{align*} Also $H_p(1)=0$ and $H_p'(1)=0$. Assume first that $p<q$. Then $1<\sqrt{q/p}<q/p$. On $(0,1)$ the formula for $H_p''$ gives $H_p''<0$, hence $H_p'$ is decreasing as $r$ increases to $1$; since $H_p'(1)=0$, we have $H_p'(r)>0$ for $0<r<1$, and therefore $H_p(r)<H_p(1)=0$. On $(1,\sqrt{q/p})$ one has $H_p''>0$, so $H_p'>0$ and $H_p>0$. On $(\sqrt{q/p},q/p)$ the function $H_p$ is concave, with positive left endpoint value and right endpoint value $H_p(q/p)=0$, so $H_p>0$ there. Finally, \begin{align*} H_p'\left(\frac{q}{p}\right) =p\left(-\log\frac{q}{p}+2(q-p)\right). \end{align*} To justify the sign, define $\ell:(1,\infty)\to\mathbb{R}$ by \begin{align*} \ell(x):=\log x-\frac{2(x-1)}{x+1}. \end{align*} Then $\lim_{x\downarrow1}\ell(x)=0$ and \begin{align*} \ell'(x)=\frac{(x-1)^2}{x(x+1)^2}>0 \end{align*} for $x>1$. Hence $\log x>2(x-1)/(x+1)$ for $x>1$. Applying this with $x=q/p$ gives $\log(q/p)>2(q-p)$, so $H_p'(q/p)<0$. Since $H_p''<0$ on $(q/p,\infty)$, it follows that $H_p'<0$ and hence $H_p<0$ on $(q/p,\infty)$. If $p=q$, then $q/p=1$ and the displayed formula for $H_p''$ gives $H_p''(r)\leq0$ for all $r>0$. With $H_p(1)=H_p'(1)=0$, concavity gives $H_p(r)<0$ for every $r\neq1$. If $p>q$, use the identity \begin{align*} H_p(r)=r H_q\left(\frac{1}{r}\right), \end{align*} which follows by direct substitution, and apply the already proved case $q<p$ to the parameter $q$. Consequently $H_p(r)$ has the sign of $(r-1)^3(q-pr)$ for every $r>0$, with zeros only at $r=1$ and $r=q/p$. It follows that $R_p'(r)>0$ for $0<r<q/p$ with $r\neq1$, and $R_p'(r)<0$ for $r>q/p$ with $r\neq1$. Hence the continuous extension of $R_p$ is increasing on $(0,q/p)$ and decreasing on $(q/p,\infty)$.[/step]

[guided]The only delicate point is the sign of the derivative of $R_p$, so we isolate it carefully. The normalized entropy is \begin{align*} \mathcal E_p(r) = p r^2\log r^2-\bigl(p r^2+q\bigr)\log\bigl(p r^2+q\bigr). \end{align*} Differentiating the two terms separately gives \begin{align*} \frac{d}{dr}\bigl(p r^2\log r^2\bigr) =2pr\log r^2+2pr, \end{align*} and \begin{align*} \frac{d}{dr}\Bigl(\bigl(p r^2+q\bigr)\log\bigl(p r^2+q\bigr)\Bigr) =2pr\log\bigl(p r^2+q\bigr)+2pr. \end{align*} The two non-logarithmic $2pr$ terms cancel. Hence \begin{align*} \mathcal E_p'(r) =2pr\log\frac{r^2}{p r^2+q}. \end{align*} Now differentiate \begin{align*} R_p(r)=\frac{\mathcal E_p(r)}{(r-1)^2} \end{align*} for $r>0$ and $r\neq1$. The quotient rule gives \begin{align*} R_p'(r) =\frac{(r-1)\mathcal E_p'(r)-2\mathcal E_p(r)}{(r-1)^3}. \end{align*} Substitution and collection of the coefficients of $\log r$ and $\log(p r^2+q)$ give \begin{align*} (r-1)\mathcal E_p'(r)-2\mathcal E_p(r) =2\bigl((q+pr)\log(p r^2+q)-2pr\log r\bigr). \end{align*} Define \begin{align*} H_p:(0,\infty)&\to\mathbb R, \end{align*} by \begin{align*} H_p(r):=(q+pr)\log(p r^2+q)-2pr\log r. \end{align*} Thus the sign of $R_p'(r)$ is determined by comparing the sign of $H_p(r)$ with the sign of $(r-1)^3$. We now prove that comparison rather than assert it. First note that $H_p(1)=0$, $H_p'(1)=0$, and $H_p(q/p)=0$. Direct differentiation gives \begin{align*} H_p'(r)=p\left(\log\frac{p r^2+q}{r^2}+\frac{2q(r-1)}{p r^2+q}\right), \end{align*} and differentiating once more gives \begin{align*} H_p''(r)=\frac{2pq(r-1)(q-pr^2)}{r(p r^2+q)^2}. \end{align*} The denominator is positive for $r>0$, so the sign of $H_p''$ is the sign of $(r-1)(q-pr^2)$. Assume first that $p<q$. Then $1<\sqrt{q/p}<q/p$. On $(0,1)$ we have $H_p''<0$, so $H_p'$ decreases as $r$ increases to $1$. Since $H_p'(1)=0$, this means $H_p'(r)>0$ for $0<r<1$; therefore $H_p$ increases to the value $H_p(1)=0$, and hence $H_p(r)<0$ on $(0,1)$. On $(1,\sqrt{q/p})$ we have $H_p''>0$, so $H_p'>0$ there and $H_p(r)>0$. On $(\sqrt{q/p},q/p)$ the function is concave, while its value at the left endpoint is positive and its value at $q/p$ is $0$; concavity keeps the graph above the chord, so $H_p(r)>0$ throughout this interval. It remains, in the case $p<q$, to handle $r>q/p$. At $r=q/p$ the derivative is \begin{align*} H_p'\left(\frac{q}{p}\right) =p\left(-\log\frac{q}{p}+2(q-p)\right). \end{align*} We prove the elementary logarithmic inequality used to sign this expression. Define $\ell:(1,\infty)\to\mathbb{R}$ by \begin{align*} \ell(x):=\log x-\frac{2(x-1)}{x+1}. \end{align*} Then $\lim_{x\downarrow1}\ell(x)=0$, and differentiation gives \begin{align*} \ell'(x)=\frac{(x-1)^2}{x(x+1)^2}>0 \end{align*} for every $x>1$. Thus $\ell(x)>0$ for every $x>1$, which is the inequality $\log x>2(x-1)/(x+1)$. Applying it with $x=q/p$ gives $\log(q/p)>2(q-p)$. Hence $H_p'(q/p)<0$. Since $H_p''<0$ on $(q/p,\infty)$, the derivative remains negative there, and $H_p(r)<0$ for all $r>q/p$. If $p=q$, then $q/p=1$ and the formula for $H_p''$ gives $H_p''(r)\leq0$ for every $r>0$. With $H_p(1)=H_p'(1)=0$, concavity gives $H_p(r)<0$ for every $r\neq1$. If $p>q$, we reduce to the already proved case by the direct identity \begin{align*} H_p(r)=r H_q\left(\frac{1}{r}\right). \end{align*} Thus, in all cases, $H_p(r)$ has the sign of $(r-1)^3(q-pr)$, with equality only at $r=1$ and $r=q/p$.[/guided]

[step:Evaluate the quotient at its maximum] Since $R_p$ increases up to $q/p$ and decreases after $q/p$, its maximum on $(0,\infty)$ is attained at $r=q/p$. At this point, \begin{align*} p\left(\frac{q}{p}\right)^2+q=\frac{q}{p}. \end{align*} Therefore \begin{align*} \mathcal E_p\left(\frac{q}{p}\right) = p\left(\frac{q}{p}\right)^2\log\left(\frac{q}{p}\right)^2 -\frac{q}{p}\log\frac{q}{p}. \end{align*} Collecting the coefficient of $\log(q/p)$ gives \begin{align*} \mathcal E_p\left(\frac{q}{p}\right) = \frac{q(q-p)}{p}\log\frac{q}{p}. \end{align*} Also, \begin{align*} \left(\frac{q}{p}-1\right)^2 = \frac{(q-p)^2}{p^2}. \end{align*} Thus, when $p\neq q$, \begin{align*} R_p\left(\frac{q}{p}\right) = \frac{p q(\log q-\log p)}{q-p} = \frac{p q}{\Lambda(p,q)}. \end{align*} When $p=q=1/2$, the same value follows by taking the limit $q\to p$, because $\Lambda(p,p)=p$; hence $C_{1/2}=1/2$. [/step]

[step:Return to arbitrary real-valued functions]For every $r>0$, \begin{align*} R_p(r)\leq R_p\left(\frac{q}{p}\right)=\frac{p q}{\Lambda(p,q)}. \end{align*} By the continuous extension of $R_p$ at $r=1$ and by continuity at $r=0$, the same estimate holds for every $r\geq0$. Therefore every normalized nonnegative function $u_r:\{0,1\}\to[0,\infty)$ satisfies \begin{align*} \operatorname{Ent}_{\nu_p}(u_r^2) \leq \frac{p q}{\Lambda(p,q)}(r-1)^2. \end{align*} By the homogeneity reduction, every nonnegative function $f:\{0,1\}\to[0,\infty)$ satisfies \begin{align*} \operatorname{Ent}_{\nu_p}(f^2) \leq \frac{p q}{\Lambda(p,q)}(f(1)-f(0))^2. \end{align*} Finally, for an arbitrary function $f:\{0,1\}\to\mathbb{R}$, applying the nonnegative estimate to $|f|$ gives \begin{align*} \operatorname{Ent}_{\nu_p}(f^2) = \operatorname{Ent}_{\nu_p}(|f|^2) \leq \frac{p q}{\Lambda(p,q)} \bigl(|f(1)|-|f(0)|\bigr)^2. \end{align*} The reverse triangle inequality gives \begin{align*} \bigl(|f(1)|-|f(0)|\bigr)^2 \leq (f(1)-f(0))^2. \end{align*} Combining the last two inequalities proves \begin{align*} \operatorname{Ent}_{\nu_p}(f^2) \leq \frac{p(1-p)}{\Lambda(p,1-p)}(f(1)-f(0))^2, \end{align*} which is the claimed two-point logarithmic Sobolev inequality.[/step]

[guided]The quotient estimate has proved the inequality for every normalized nonnegative function $u_r:\{0,1\}\to[0,\infty)$. By the homogeneity and approximation reduction from the first step, this implies that every nonnegative function $f:\{0,1\}\to[0,\infty)$ satisfies \begin{align*} \operatorname{Ent}_{\nu_p}(f^2) \leq \frac{p q}{\Lambda(p,q)}(f(1)-f(0))^2. \end{align*} Now let $f:\{0,1\}\to\mathbb{R}$ be arbitrary. The entropy term depends only on $f^2$, so \begin{align*} \operatorname{Ent}_{\nu_p}(f^2)=\operatorname{Ent}_{\nu_p}(|f|^2). \end{align*} Applying the nonnegative estimate to the function $|f|:\{0,1\}\to[0,\infty)$ gives \begin{align*} \operatorname{Ent}_{\nu_p}(f^2) \leq \frac{p q}{\Lambda(p,q)}\bigl(|f(1)|-|f(0)|\bigr)^2. \end{align*} The reverse triangle inequality on $\mathbb{R}$ says \begin{align*} \bigl||f(1)|-|f(0)|\bigr|\leq |f(1)-f(0)|, \end{align*} and squaring preserves the inequality because both sides are nonnegative. Therefore \begin{align*} \operatorname{Ent}_{\nu_p}(f^2) \leq \frac{p q}{\Lambda(p,q)}(f(1)-f(0))^2. \end{align*} Since $q=1-p$, this is exactly \begin{align*} \operatorname{Ent}_{\nu_p}(f^2) \leq \frac{p(1-p)}{\Lambda(p,1-p)}(f(1)-f(0))^2. \end{align*}[/guided]