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Take the fixed source and receiver to be $A=(a,h_A)$ and $B=(b,h_B)$ on the same side of the mirror, with $h_A,h_B>0$. In the uniform medium, the Fermat principle minimizes the length through a reflection point $P=(s,0)$:Its stationary equation isThe two fractions are the signed tangential components of the ray directions. Equality of their magnitudes gives $\sin {\theta }_i=\sin {\theta }_r$, where both angles are measured from the normal vector to the mirror and lie in $[0,\pi /2)$. Hence the law of reflection isFor uniqueness, differentiate once more:Thus $L$ is a strictly convex function, its derivative increases from $-2$ to $2$, and it has exactly one stationary point, its global minimum. Explicitly,Substitution gives $(s_{\ast }-a)/h_A=(b-s_{\ast })/h_B$, verifying the reflection angles and the existence of that unique solution. Equivalently it is the point where the line from $A$ to the mirror image $(b,-h_B)$ crosses the mirror.

Differentiating the Gaussian variational bound for an attractive Gaussian well givesThus an interior stationary point obeysThe left side of the first equation increases strictly from zero to infinity, so there is exactly one such point. More explicitly,so $E$ is a strictly convex function on $a>0$ and $a_{\ast }$ is its global minimum. At that minimum the stationarity equation givesThere is a continuous limiting zero at $a=0$, although $y_T=1$ at exactly $a=0$ is not a square-integrable function and hence is not an admissible trial. For a positive zero, division by $\sqrt{a}$ and squaring gives $a(1+a)=4V_0^2$, sois the unique positive zero. The other quadratic root is negative and inadmissible. Near zero,so the graph leaves the origin with slope $-\infty$, falls to its unique negative minimum, rises through $a_z$, and grows as $a/2-V_0+O(a^{-1})$ at infinity.

By the Rayleigh-Ritz variational principle,Thus the attractive well supports a negative-energy bound state for every $V_0>0$.

Put $z=x-\alpha \nabla f(x)$ and $\tau =\alpha \lambda$. The proximal operator for the L1 norm separates into independent scalar minimization problems:The subgradient optimality condition is $0\in p_i-z_i+\tau \partial |p_i|$. By the subgradient of the absolute value, a positive solution must satisfy $p_i=z_i-\tau >0$, and a negative solution must satisfy $p_i=z_i+\tau <0$. At zero, the subgradient optimality condition holds exactly when $|z_i|\le \tau$. The squared term makes the objective function a strictly convex function, so these cases give its unique minimiser.

Thus the proximal operator is the soft-thresholding operatorConsequently the requested proximal gradient mapping, rather than just the proximal operator, isThe corresponding proximal gradient method update is $x_i^{+}=S_{\alpha \lambda }(x_i-\alpha {\partial }_{i}f(x))$. Small coordinates are set exactly to zero. When $\lambda =0$, the soft-thresholding operator is the identity and the proximal gradient mapping reduces to $\nabla f(x)$.

A strictly convex function satisfies that inequality strictly whenever $x\ne y$ and $0<t<1$.

For necessity of the strict supporting-tangent property, convexity gives a nonnegative gap along a segment from $z_0$ to $z$. If the gap at its other endpoint were zero, convexity of the gap and its zero initial value would force it to be zero all along the segment. That would make $F$ affine on this nontrivial segment, contrary to strict convexity. Conversely let $z=(1-t)x+ty$ with distinct endpoints. Apply the strict supporting inequality at $z$ to both endpoints, multiply by $1-t$ and $t$, and add. The linear terms cancel, yielding strict convexity. ThusFor strict Jensen's inequality, put $m=(b-a{)}^{-1}{\int }_a^{b}u$. The supporting gap $F(u)-F(m)-DF(m)\cdot (u-m)$ is nonnegative everywhere and strictly positive where $u\ne m$. That set has positive measure because $u$ is not almost everywhere constant. Its integral is therefore strictly positive. The linear term integrates to zero, giving the requested strict integral Jensen inequality.

The conditions on the cumulative distribution function imply $0<X_n\le 2$ almost surely. With ${\mathcal{H}}_n=\sigma (X_1,\dots ,X_n)$, adjoin $Y_0=1$ and the trivial ${\mathcal{H}}_0$. The random variables $Y_n$ are nonnegative, adapted, and bounded by $2^n$ at each fixed time. Independence and $\mathbb{E}{X}_{n+1}=1$ giveThus this is the product martingale from independent mean-one factors, with $\mathbb{E}{Y}_n=1$. The almost sure supermartingale convergence theorem states that a nonnegative supermartingale converges almost surely to a finite integrable random variable. Applied to this nonnegative martingale, it shows $Y_n\to Y_{\infty }<\infty$ almost surely. Every convergent real sequence is bounded, so ${\sup }_{n\ge 1}{Y}_n<\infty$ almost surely.

The printed maximum can in fact be taken literally here. If $X_1=1$ almost surely, all products equal $1$. Otherwise the strict Jensen equality criterion, applied to the strictly convex function $t\mapsto -\sqrt{t}$ on $(0,\infty )$, gives $q=\mathbb{E}\sqrt{X_1}<1$. Independence yields $\mathbb{E}\sqrt{Y_n}=q^n$. By Markov's inequality, ${\sum }_n\mathbb{P}(Y_n>\epsilon )\le {\epsilon }^{-1/2}{\sum }_n{q}^n<\infty$ for every $\epsilon >0$. The First Borel-Cantelli lemma and the countable choices $\epsilon =1/m$ give $Y_n\to 0$ almost surely. Since $Y_1>0$, a sequence with this limit attains its positive supremum among finitely many initial terms. This is an instance of fractional moment decay of independent mean-one products.

The fixed common temperature, rigid external container and fixed particle numbers on either side make the total Helmholtz free energy the appropriate thermodynamic potential. Let the barrier have cross-sectional area $A$, and let increasing $x$ expand the left compartment. Then $d{V}_1=Adx$ and $d{V}_2=-Adx$. At fixed temperature and particle number, $d{F}_j=-p_{j}d{V}_j$, soThus mechanical equilibrium, $p_1=p_2$, makes $F$ stationary. The physical force on the barrier towards increasing $x$ is $f=A(p_1-p_2)=-F^{\prime }(x)$. If the barrier is displaced slightly from $x_{\ast }$, strict mechanical stability requires a restoring force:ConsequentlyThe Taylor expansion $F(x_{\ast }+\delta x)=F(x_{\ast })+\frac{1}{2}{F}^{\prime \prime }(x_{\ast })\delta x^2+O(\delta x^3)$ proves that thermodynamic equilibrium is a strict local minimum of the Helmholtz free energy, rather than merely a stationary point.

For individually stable gases, the positive isothermal compressibilities ${\kappa }_{T,j}=-(1/V_j)(\partial V_j/\partial p_j{)}_{T,N_j}$ make the sign especially transparent:If these inequalities hold throughout the accessible interval, $F$ is a strictly convex function and an interior stationary point is the unique global minimum of the Helmholtz free energy. Stability at one point alone proves only the local statement. A vanishing restoring derivative describes marginal stability and requires examining higher-order terms.

Apply the mean value theorem, the first-order case of Taylor's theorem, to $f^{\prime }$ between $0$ and $x>1$. It yieldsAlso $f(3/2)\ge -5+9/2+9/4=7/4>0$, so the initial point is to the right of the root $x^{\ast }$. For an increasing strictly convex function, its tangent at a point to the right of a root meets the axis between that root and the point. Hence the exact Newton root-finding iteration remains in $[x^{\ast },3/2]$ and decreases towards $x^{\ast }$.

On every relevant interval, $|f^{\prime \prime }|<3$ and $f^{\prime }>5$, so the stated quadratic convergence bound gives, with $e_n=|x_n-x^{\ast }|$,Iterating the bound givesThus four Newton corrections suffice to reduce the exact-iteration error bound below the floating-point spacing scale. The quadratic convergence doubles the approximate number of correct digits at each late step. Beyond this point further corrections cannot reliably improve a generic floating-point answer. This is a bound on useful accuracy under the usual stable-evaluation model, not a claim that the data specify the exact iteration at which every implementation must stagnate.

For negative $x$, the mean value theorem gives $f^{\prime }(x)=3+x{f}^{\prime \prime }({\xi }_x)$. In particular,Since $f^{\prime }$ is continuous and strictly increasing, it vanishes at exactly one point $x_m\in (-2,-1)$. The positive second derivative makes $f$ a strictly convex function on the entire real line. Its derivative changes from negative to positive at $x_m$, so $x_m$ is the unique global minimum. One may also use $f(x)\ge x^2+3x-5\to \infty$ to verify coercivity.

In the endpoint-bracketing form of golden-section search, begin with $[-2,-1]$, a width-one interval known to contain this unique minimum, and place the two standard interior probes. The attainable position accuracy is different from that for a simple root. Around $x_m$, Taylor's theorem givesHere the minimum value is nonzero and of order one: the lower quadratic bound gives $f(x_m)\ge -29/4$, while $f(x_m)\le f(-1)<-13/2$. Ordinary relative round-off error in these function values therefore hides changes of order $ϵ$, once $h^2$ has that size. The function-value resolution near a nondegenerate minimum is consequently of orderEach golden-section search contraction multiplies the bracket width by $\rho =(\sqrt{5}-1)/2$. Using $2^{-20}$ as the nominal target width givesUnknown evaluation-error constants can change the actual stopping point; $29$ is the standard scale-based count. Requiring $2^{-40}$ position accuracy from these function-value comparisons would disregard the quadratic flatness of the minimum. The same precision effect is documented for one-dimensional numerical minimization.^[ref]

Consider the minimization problemThe objective function is a strictly convex function with its unconstrained optimization minimum at $x=2$. If $b<2$, it decreases throughout the feasible set until the bound $x=b$ is reached. If $b\ge 2$, the feasible set already contains $2$. HenceWith $g(x)=x-b$, the optimization Lagrangian and KKT conditions areSubstituting $x_{\ast }(b)$ gives the unique inequality Lagrange multiplierFor $b<2$, the active constraint has ${\lambda }_{\ast }>0$. For $b>2$, there is strict slack $b-x_{\ast }=b-2$ and ${\lambda }_{\ast }=0$. At $b=2$, the active constraint has zero Lagrange multiplier: the boundary happens to pass through the unconstrained optimum. Feasibility and complementary slackness must both be checked; solving only $2(x-2)+\lambda =0$ would allow infeasible candidates.

An independent certificate follows by completing the square:The Lagrange dual function is a concave function of $\lambda$. Its maximum over $\lambda \ge 0$ occurs at ${\lambda }_{\ast }(b)$ and equals $V(b)$. Thus every case has an explicit lower bound attained by a feasible point; the KKT conditions provide a certificate that can be checked without searching all real numbers.

For probability distributions $P,Q$, relative entropy satisfies $D_{\mathrm{KL}}(P\Vert Q)\ge 0$, with equality exactly when $P=Q$. Thus truthful prediction uniquely minimizes expected log-loss whenever the comparison is well defined, explaining the strictly proper scoring rule property of the logarithmic scoring rule.

For a short proof with probability density functions $f,g$ and $g>0$, set $r=f/g$. Their integrals both equal one, givingThe strictly convex function $r\log r-r+1$ is minimized only at $r=1$, so equality means $f=g$ almost everywhere. Its value at zero is one by $0\log 0=0$. Where $g=0<f$, relative entropy is infinite; where both vanish, the contribution is zero. These conventions extend the argument beyond positive reference probability density functions.

Let $X$ be an integrable random variable taking values in an open interval $I$. Suppose $f:I\to \mathbb{R}$ is a differentiable strictly convex function. Then its expected value has equality in Jensen's inequality exactly when $X$ is constant almost surely.

First $m=\mathbb{E}X$ lies in $I$: at each finite endpoint, taking the expected value of the strictly positive distance to that endpoint gives a strictly positive distance for $m$. A tangent affine function supports $f$ at $m$, soStrict convexity forces $h(t)>0$ for $t\ne m$. Indeed, equality at another point would force the graph to coincide with its supporting line on the connecting line segment, contradicting strict convexity. Thereforeand the right side is zero exactly when $X=m$ almost surely.

The tangent affine function of $X$ is an integrable random variable, so $f(X)$ has an integrable negative part. Thus $\mathbb{E}[f(X)]$ is well defined, possibly $+\infty$, even if absolute integrability was not assumed. Equality to $f(m)$ automatically excludes that infinite case.

The utility ${\sum }_r{n}_r\log x_r$ sums logarithmic utilities over individual flows. Its first-order condition gives proportional fairness. Its negative is a strictly convex function on active flow rates, giving uniqueness of those rates on a convex set of feasible allocations.