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For a bounded real function on $[a,b]$, take a partition $a=t_0<\cdots <t_m=b$. Its lower and upper Darboux sums areThe function is Riemann integrable when ${\sup }_{P}L(f,P)={\inf }_{P}U(f,P)$; the common value is its Riemann integral. Equivalently, for every $\epsilon >0$ there is a partition with $U(f,P)-L(f,P)<\epsilon$. A continuous function is bounded and Riemann integrable on each compact interval, so all the following integrals exist.

Set $h(x)=x^{-1}{\int }_0^{x}f(t)dt$ for $x>0$. Strict decrease gives $f(x)<f(t)<f(0)$ for $0<t<x$. Integrating, with strict inequalities because the difference is positive on interior subintervals, givesThe intermediate value theorem produces $g(x)\in (0,x)$ with $f(g(x))=h(x)$, and strict monotonicity makes this point unique. Thus there is exactly one such $g(x)$ in $(0,x)$.

For the given exponential function, $h(x)=(1-e^{-x})/x$, soThe preceding strict bounds already prove $0<g(x)<x$; no limiting value at zero needs to be included in its domain.

For differentiability, a continuous strictly monotone function has a continuous inverse on its range. If it is differentiable at an interior point $t$ with $f^{\prime }(t)\ne 0$, the inverse is differentiable at $y=f(t)$, with derivative $1/f^{\prime }(t)$. Indeed, setting $u=f^{-1}(y^{\prime })$, inverse continuity gives $u\to t$ as $y^{\prime }\to y$, andThis one-dimensional inverse result needs no assumption that $f^{\prime }$ is continuous. It is enough here that $f$ is differentiable and $f^{\prime }<0$ at every positive argument.

The fundamental theorem of calculus and the quotient rule giveNow $g=f^{-1}\circ h$ and $g(x)>0$, so the inverse derivative formula applies at every relevant point. The monotone mean-value point of an integral average satisfiesBoth numerator and denominator are negative. For the exponential example, this also agrees with $g^{\prime }(x)=1/x-1/(e^x-1)>0$, since $e^x>1+x$ for $x>0$.

The substitution is nonsingular in the interior $x>0$, $0<y<1$, $0<z<x$. Its inverse isThus $y<1$ is equivalent to $\beta <\alpha$, while $z<x$ is equivalent to $\gamma <xy=\beta$. Positivity gives the full transformed domainThe forward Jacobian determinant isTherefore $xdxdydz=\frac{1}{2}d\alpha d\beta d\gamma$. The boundary faces where the formula fails have measure zero. The change of variables formula givesThe integrand is nonnegative, so Tonelli's theorem permits reordering the integration even before convergence has been established. Integrating first in $\alpha$, then in $\beta$, givesThe assumptions $A,B,C>0$ make all three integrals of exponential functions converge. Equivalently, the ordered-cone substitution by products and ratios can be followed by the increment variables $\alpha -\beta$, $\beta -\gamma$, $\gamma$, all independently positive and with Jacobian determinant one; the exponent then separates into three linear terms with coefficients $A$, $A+B$ and $A+B+C$.

For finite sets $A_1,\dots ,A_n$, the inclusion-exclusion principle statesTo prove it, fix an element in exactly $r$ of the sets. If $r=0$, it contributes to neither side. If $r\ge 1$, its contribution to the right side isusing the binomial theorem. This is exactly its contribution to the union. Summing over the finite ambient union proves the formula. Equivalently, the proof is an identity of elementwise membership indicators, so no unproved rule for overlapping sets is being assumed.

Now let the ambient set be all $n!$ permutations, and let $E_i$ be those fixing element $i$. A set intersection of $k$ specified events $E_i$ has $(n-k)!$ elements: the specified elements are fixed and the others may be permuted freely. Apply inclusion-exclusion to the complement of the union of these events. There are $\left(\genfrac{}{}{0ex}{}{n}{k}\right)$ possible choices of the fixed positions, hence the derangement count isThe $k=0$ term counts the whole ambient set before subtraction; it is not an omitted empty intersection. The formula also gives $f(0)=1$ and $f(1)=0$.

Dividing by the factorial and using the convergent power series for the exponential function at $-1$ givesThe alternating-series estimate supplies the explicit error bound $|f(n)/n!-e^{-1}|\le 1/(n+1)!$. Thus the limiting proportion is established, not merely identified from the first few values.

Expose the vertices in order, revealing at step $i$ every edge from vertex $i$ to an earlier vertex. With the resulting filtration ${\mathcal{F}}_i$, putThis is the vertex exposure for chromatic number martingale. Its increments $D_i=M_i-M_{i-1}$ have conditional mean zero and satisfy $|D_i|\le 1$, by the permitted Lipschitz condition.

Here is the required exponential-moment proof of the Azuma-Hoeffding inequality. More generally, suppose $|D_i|\le c_i$ for deterministic $c_i$. Convexity of the exponential function gives, for $-c_i\le d\le c_i$,Taking conditional expectation and using the zero conditional mean yieldsFor the last inequality, integrate $(\log \cosh u{)}^{\prime }=\tanh u\le u$ for $u\ge 0$, and use evenness. A zero $c_i$ simply means a zero increment. Iterated conditional expectation now givesFor $s,t>0$, the strict Markov bound on the event $M_n-M_0>s$ givesThe strict inequality follows directly by integrating $e^{t(M_n-M_0)}>e^{ts}$ on that event; it is also immediate if the event is empty. Optimize at $t=s/{\sum }_i{c}_i^2$ and apply the same argument to the negative increments. For $c_i=1$ and $s=\lambda \sqrt{n}$, the union bound proves, for every $\lambda >0$,At $\lambda =0$ the displayed strict bound is trivial.

For the second conclusion write $a_n=n/\log n$ and choose a fixed $C>\sqrt{2\log 10}$. If $\mu \ge a_n+C\sqrt{n}$, then $\{\chi (G)<a_n\}\subseteq \{\chi (G)-\mu <-C\sqrt{n}\}$, whose probability is less than $e^{-C^2/2}<1/10$. Thus the given positive probability forces $\mu <a_n+C\sqrt{n}$. For sufficiently large $n$,Here the non-strict one-sided bound follows by the usual non-strict Markov inequality. Consequently the asserted upper bound holds with high probability, uniformly over any sequence $p=p(n)$ satisfying the stated assumption.

The exponent printed in the original PDF is too strong under the stated increment bound. Take $M_0=0$, let $M_1$ equal $1$ or $-1$ with equal probabilities, and set $M_k=M_1$ for $k\ge 1$. This is a martingale with $c_1=1$ and $c_k=0$ for $k>1$. For $n=1$ and $x=1$,Thus the printed inequality is false. The intended two-sided Azuma-Hoeffding inequality isHere the $c_k$ are deterministic increment bounds. We prove this corrected assertion from conditional exponential moments.

Let $D_k=M_k-M_{k-1}$. The martingale property gives $\mathbb{E}[D_k\mid {\mathcal{F}}_{k-1}]=0$. For $c_k>0$, convexity of the exponential function gives the secant bound on $[-c_k,c_k]$,Taking conditional expectation removes the terms containing $D_k$, soThe last inequality follows term by term from $(2j)!\ge 2^{j}j!$ in the power series for $\cosh z$ and $e^{z^2/2}$. If $c_k=0$ the increment is zero and the same bound holds directly. This is the bounded-increment case of the Hoeffding lemma.

Repeated conditioning, without any independence assumption, now givesWrite $C={\sum }_{k=1}^n{c}_k^2$. If $C>0$, the Markov inequality gives for $\theta >0$,Minimizing at $\theta =x/C$ gives $e^{-x^2/(2C)}$. Apply the same argument to the martingale $-M$ for the lower tail, and use the union bound to obtain the boxed result. If $C=0$, all increments vanish almost surely, so the deviation probability is zero for every $x>0$. This handles the degenerate case without dividing by zero and proves the correct bound in full.

Here $X_1\le 1$, so for every $\tau >0$,The Jensen inequality for the strictly convex exponential function givesIn fact the inequality is strict: $\mathbb{P}(X_1=1)>0$ and $\mathbb{E}{X}_1=0$ exclude a constant random variable. In particular, $\boxed{1\le M(\tau )<\infty }$ as required.

For $Z_n=e^{\tau S_n}/M(\tau {)}^n$, independence and identical distribution giveThus $Z_n$ is positive, integrable with mean one, and adapted to the natural filtration. Conditioning on the next increment givesConsequently $\boxed{(Z_n)\text{ is a martingale}.}$ This is the exponential martingale of a random walk.

The exponents are real, as required for this family of real continuous functions. We prove the exponential Chebyshev system property by induction on the number of functions. A single nonzero multiple of an exponential function has no zero, so the base case holds.

Suppose the result holds for $n$ distinct real exponents, and consider a nontrivial combination $F(x)={\sum }_{j=0}^n{c}_j{e}^{{\lambda }_{j}x}$. If it had $n+1$ distinct zeros, so wouldsince the prefactor is everywhere positive. Applying Rolle's theorem between consecutive zeros gives at least $n$ distinct zeros ofThe $n$ exponents in this derivative are distinct. By induction, a nontrivial combination of them has at most $n-1$ zeros. Therefore every coefficient $c_j({\lambda }_j-{\lambda }_0)$ must be zero. Distinctness gives $c_j=0$ for $j\ge 1$. Then $g=c_0$ has a zero only if $c_0=0$, contradicting the original nontriviality. Consequently $F$ has at most $n$ zeros, and

Let $p$ be the best uniform approximation to the exponential function from the degree-at-most-$n-1$ polynomial space. Suppose that $E_{n-1}(e^x)=E_n(e^x)$. Then the same $p$ is best in the larger space. Its error $g(x)=e^x-p(x)$ is nonzero, since the exponential is not a polynomial. The Chebyshev alternation theorem supplies $n+2$ ordered points where $g$ has alternating signs. By the intermediate value theorem, $g$ has at least one zero in each of the $n+1$ intervening intervals, hence at least $n+1$ distinct zeros.

Applying Rolle's theorem successively $n$ times forces a zero of $g^{(n)}$. But $\mathrm{deg}p\le n-1$, soeverywhere. This is a contradiction. Together with the automatic non-increase of best errors, it provesThe argument is an instance of strict polynomial error decrease from nonvanishing derivatives.

Use left Grassmann derivatives and define the supersymmetric covariant derivatives $D_{\pm }={\partial }_{{\theta }^{\pm }}-i{\theta }^{\pm }{\partial }_{\pm }$. Multiplication in the indicated order gives exactly the printed differential operator:Thus the equation is $i{D}_{-}{D}_{+}\Phi =W(\Phi )$ in two-dimensional N=(1,1) superspace.

Under a Lorentz boost, take $x^{\pm }\mapsto e^{\pm \omega }{x}^{\pm }$ and ${\theta }^{\pm }\mapsto e^{\pm \omega /2}{\theta }^{\pm }$. The derivatives ${\partial }_{\pm }$ have boost weights $\mp 1$, and $D_{\pm }$ have weights $\mp 1/2$. Consequently $D_{-}{D}_{+}$ has weight zero, and is a Lorentz scalar. Both $\Phi$ and the ordinary function $W(\Phi )$ are scalar superfields, so the equation is Lorentz invariant.

For supersymmetry, $\{{\partial }_{\theta },\theta \}=1$ and anticommutation of distinct odd coordinates giveMoving either supercharge through the two odd derivatives introduces two minus signs, hence $[Q_t,D_{-}{D}_{+}]=0$. If $\delta \Phi ={ϵ}^{+}{Q}_{+}\Phi +{ϵ}^{-}{Q}_{-}\Phi$, with odd constant parameters, the chain rule for an even scalar superfield givesTherefore a solution is mapped to a solution, proving supersymmetry invariance.

To compute the supersymmetric Liouville equation, order the odd coordinates as ${\theta }^{-}{\theta }^{+}$ and anticommute the fermionic component fields with both coordinates. In particular, ${\partial }_{{\theta }^{+}}({\theta }^{-}{\theta }^{+})=-{\theta }^{-}$. Direct application of the operator givesLet $N=\Phi -\varphi$. Only the two products of its degree-one terms survive in $N^2$; for example, $(i{\theta }^{-}{\psi }_{+})(i{\theta }^{+}{\psi }_{-})={\theta }^{-}{\theta }^{+}{\psi }_{+}{\psi }_{-}$. Hence $N^2=2{\theta }^{-}{\theta }^{+}{\psi }_{+}{\psi }_{-}$ and $N^3=0$. The exponential function expands exactly asEquating the constant terms fixes the auxiliary field, $\boxed{F=e^{\varphi }}$. The linear and quadratic terms then give the requested component field equations of motion:The signs here follow from the printed operator, the printed component expansion, and left Grassmann derivatives; no convention change has been made in eliminating the auxiliary field.

The exponential function series gives Euler's number $e={\sum }_{k=0}^{\infty }1/k!$. Suppose $e=a/b$ with integers $a,b$ and $b>0$. Choose $n\ge \max (b,2)$. Since $b$ divides $n!$, the numberis an integer. But its positive tail satisfiesThe inequality is strict because the factors after the first exceed $n+1$ when $j\ge 2$. No integer lies strictly between zero and one. This proves the irrationality of e:

For $x\ge 0$, the function $h(x)=e^x-1-x$ has $h(0)=0$ and $h^{\prime }(x)=e^x-1\ge 0$, because the exponential function is increasing and $e^0=1$. Therefore $\boxed{e^x\ge 1+x}$.

Write $S_n={\sum }_{j=1}^n{a}_j$ and $P_n={\prod }_{j=1}^n(1+a_j)$. Expanding the product produces $1+S_n$ and additional nonnegative terms, giving $P_n\ge 1+S_n\ge S_n$. Applying the exponential inequality to every factor givesBoth sequences are increasing, since the terms $a_j$ are positive. If $S_n$ converges to a finite value $S$, then $P_n\le e^S$, so the convergence theorem for monotone bounded sequences makes $P_n$ converge. Conversely, if $P_n$ has a finite limit, then $S_n\le P_n$ is bounded and increasing, hence converges by the same theorem. Thus the positive series and the product both converge to finite limits, or both diverge to $+\infty$. This proves the positive sum-product convergence criterion. The word limit here means a finite real limit; allowing extended infinite limits would make the conclusion vacuous for these monotone sequences.

The Malgrange–Ehrenpreis theorem states that every nonzero constant-coefficient linear differential operator on Euclidean space has a distributional fundamental solution. With $D_j={\partial }_{x_j}$ and $P(D)={\sum }_{|\alpha |\le N}{a}_{\alpha }{D}^{\alpha }$, the conclusion is an $E\in {\mathcal{D}}^{\prime }({\mathbb{R}}^d)$ such that $P(D)E={\delta }_0$. If $N=0$, simply take $E=a_0^{-1}{\delta }_0$; hence assume $N\ge 1$.

We construct a Hörmander staircase in frequency space. The highest-degree homogeneous part $P_N$ is not identically zero on real vectors, since a polynomial vanishing on all real vectors has all coefficients zero. After an orthogonal transformation of coordinates, we can arrange that $P_N(e_1)\ne 0$. Thus, for each real ${\xi }^{\prime }\in {\mathbb{R}}^{d-1}$, the polynomialhas degree $N$ in $z$ and the same nonzero leading coefficient $b=i^N{P}_N(e_1)$, independent of ${\xi }^{\prime }$. This constant leading coefficient of a polynomial is what makes a uniform staircase possible.

For a fixed ${\xi }^{\prime }$, factor $p(z,{\xi }^{\prime })=b{\prod }_{\nu =1}^N(z-z_{\nu })$, counting repeated roots of a polynomial. Consider the $N+1$ heights $h_j=3j$, $j=0,\dots ,N$. A given root has imaginary part within distance less than one of at most one of these heights. By the pigeonhole principle, some height avoids every root, and thenThis is finite-height polynomial root avoidance. Root labels need not be chosen continuously or even measurably. Instead define the closed setsContinuity in $s$ extends the inequality from rational $q$ to every real $s$. These Borel sets partition ${\mathbb{R}}^{d-1}$. The Hörmander staircase assigns the horizontal contour $s+i{h}_j$ over $A_j$; its height is bounded by $3N$, and the denominator has the uniform lower bound $|b|$. For $d=1$, the transverse space is a single point and only one horizontal contour is needed.

Use the Fourier transform convention $\hat{\phi }(\zeta )=\int e^{-ix\cdot \zeta }\phi (x)dx$, with inverse factor $(2\pi {)}^{-d}$. Define the candidate fundamental solution of a linear differential operator byFor a test function supported in a fixed compact set $K$, Fourier decay in a bounded complex strip gives, for any integer $M$,Indeed, this is the real Fourier transform of $e^{-h_j{x}_1}\phi (x)$ with reversed frequency, and repeated integration by parts with $(1-\Delta {)}^M$ proves the estimate. Taking $2M>d$ and using the denominator bound proves absolute convergence and a continuity estimate on ${\mathcal{D}}_K$. Thus $E$ is a distribution; no unsupported interpretation of a divergent inverse Fourier transform is being used.

For its distributional derivatives, the formal transpose of a differential operator is $P(-D)$. Sinceapplying $P(D)$ cancels the denominator. For each fixed ${\xi }^{\prime }$, the numerator is an entire function of $z$. The Cauchy integral theorem, applied to a rectangle between the lines $\mathrm{Im}z=0$ and $\mathrm{Im}z=h_j$, givesThe two vertical edges tend to zero by the same bounded-strip decay. The resulting contour deformation is performed separately for each transverse frequency, so discontinuities of the staircase height introduce no additional boundary terms. Absolute convergence permits integration over the partition $A_j$. The Fourier inversion theorem then yieldsThis proves the Malgrange–Ehrenpreis theorem. Undoing the orthogonal change of coordinates gives the fundamental solution of a linear differential operator for the original operator; the Dirac delta distribution is unchanged by that change of coordinates.

For a one-dimensional operator $L=P(D)$ of degree $N\ge 1$, with leading coefficient $a_N$, we may choose the single staircase contour below every pole. Take $\gamma >\max \{\mathrm{Re}\lambda :P(\lambda )=0\}$ and integrate on $\mathrm{Im}\zeta =-\gamma$. The corresponding formula is the Bromwich contour version of the construction above, after putting $\lambda =i\zeta$. Its poles all lie above the frequency contour. For $x<0$, close that contour downwards; the exponential function $e^{ix\zeta }$ decays and there are no enclosed poles. For $x>0$, close upwards, where the exponential function again decays, and apply the residue theorem. The bound $1/P(i\zeta )=O(|\zeta {|}^{-N})$ justifies the large arcs, including $N=1$ by the Jordan lemma away from their endpoints. Thus away from $x=0$ the retarded fundamental solution is $H(x)u(x)$, whereThe sum is over distinct roots of a polynomial, with the residue including the whole multiplicity. For simple characteristic roots of a constant-coefficient differential equation, it reduces to $u(x)={\sum }_{\lambda }{e}^{\lambda x}/P^{\prime }(\lambda )$. Repeated roots give exponential polynomial solutions of a constant-coefficient differential equation through differentiation of $e^{zx}$.

To establish the equality also at the origin, rather than leave a possible point-supported term undecided, verify the distributional jump formula for a Heaviside product. The residue expression is an entire function of $x$, andIts initial derivatives areThese identities follow by integrating over a large circle: for $j\le N-2$ the integrand is $O(|z{|}^{-2})$, whereas for $j=N-1$ its coefficient of $z^{-1}$ is $a_N^{-1}$. The distributional jump formula for a Heaviside product therefore gives $L(Hu)={\delta }_0$. Moreover, since $e^{-\gamma x}Hu$ is a tempered distribution, its Fourier transform satisfiesThere are no real zeros of this polynomial, so its reciprocal is the transform. This is exactly the shifted-contour construction, proving equality there as a distribution as well. HenceThe sign of $a_N$ matters: for the operator in the preceding solution, $a_N=-1$ and the last initial derivative is $-1$.

Put $Z=\lfloor N^{2/5}\rfloor$ and $S={\sum }_{N<n\le N+M}{n}^{-it}$. For any positive integer shift $h\le Z^2$, translating the interval changes its sum by at most $2h$. Averaging the shifts $h=xy$ gives the bilinear shift averaging for a logarithmic phase identitySince $xy/n\le N^{-1/5}$, the alternating Taylor expansion of the logarithm has remainder at most $(xy/n{)}^{r+1}/(r+1)$. ThusFor $r=\lfloor 5.01\log t/\log N\rfloor$, we have $r+1>5.01\log t/\log N$, hence the error is at most $t^{-1/500}$. The exponential function on an imaginary argument changes by at most the change in that argument. ThereforeUse $Z^2\asymp N^{4/5}$ and $N<n\le N+M\le 2N$. Taking absolute values provesThe boundary error comes from integer shifts, so this argument also covers intervals shorter than a shift. Here $x,y$ range over positive integers; no zero term is needed.

Consider the flat functionAway from zero every derivative has the form $P_n(1/x)e^{-1/x^2}$ for a polynomial $P_n$: differentiating preserves this form. For every $m\ge 0$,because an exponential function decays faster than any power. Inductively, extend each displayed derivative by zero at zero. It is continuous there, and its difference quotient at zero also tends to zero by the same estimate with one extra power of $|x{|}^{-1}$. Thus each extension is the derivative of the preceding extension. This proves $f\in C^{\infty }(\mathbb{R})$ and $f^{(n)}(0)=0$ for all $n$.

Its Taylor series at zero is identically zero, whereas $f(x)>0$ for every $x\ne 0$. It is smooth everywhere but not real analytic at zero.

For $x$ sufficiently close to zero, $1+x>0$. Apply the mean value theorem to the logarithm between $1$ and $1+x$: there is a point ${\xi }_x$ between these endpoints withAs $x\to 0$ from either side, ${\xi }_x\to 1$, so this quotient tends to $1$. The exponential function is continuous, and thereforeThe hypotheses of the mean value theorem hold because $\log t$ is continuous and differentiable for $t>0$, with derivative $1/t$.

Use the Fourier transform convention $\hat{\phi }(\xi )=\int e^{-ix\cdot \xi }\phi (x)dx$ and $D_j=-i{\partial }_j$, so $\widehat{P(D)u}=P(\xi )\hat{u}$. All distribution pairings use complex bilinearity, and the formal transpose is $P(-D)$, without complex conjugation of the coefficients. If $D_j$ instead denotes ${\partial }_j$, the Fourier multiplier is $P(i\xi )$; the argument below is unchanged after this substitution.

For a nonempty compact convex set $K\subset {\mathbb{R}}^n$, let $H_K(\eta )={\sup }_{x\in K}x\cdot \eta$ be its support function. The Paley–Wiener–Schwartz theorem says that the Fourier transform is a bijection between distributions supported in $K$ and entire functions $F$ on ${\mathbb{C}}^n$ for whichfor some $C>0$ and nonnegative integer $N$. In particular, support in ${\overline{B}}_R$ is equivalent to the bound with $e^{R|\mathrm{Im}\zeta |}$. Convexity is essential in the precise support formulation: the support function of a set equals that of its convex hull.

Suppose first that $u$ is a compactly supported distribution with support in $K$. Define $F(\zeta )=⟨u,e^{-ix\cdot \zeta }⟩$, inserting any cutoff function equal to one near $K$ when viewing $u$ on test functions. A fixed cutoff and the finite order of a distribution estimate permit differentiation of the pairing in each complex variable, giving ${\partial }_{\zeta }^{\alpha }F=⟨u,(-ix{)}^{\alpha }{e}^{-ix\cdot \zeta }⟩$. The exponential function's power series converges with every required derivative on a fixed compact set; hence $F$ is an entire function. On real frequencies it agrees with the Fourier transform of a compactly supported distribution.

The exact exponential indicator needs a shrinking cutoff function, rather than one fixed outside $K$. Choose ${\chi }_{\epsilon }=1$ near $K$, with support in $K+B_{2\epsilon }$ and $\Vert {\partial }^{\alpha }{\chi }_{\epsilon }{\Vert }_{\infty }\le C_{\alpha }{\epsilon }^{-|\alpha |}$ for $0<\epsilon \le 1$. Such a cutoff is obtained by convolving the indicator of $K+B_{\epsilon }$ with a mollifier supported in $B_{\epsilon /2}$. On one fixed ball containing all these supports, the finite order of a distribution bound isThe Leibniz rule applied to $\theta ={\chi }_{\epsilon }{e}^{-ix\cdot \zeta }$ therefore givesTaking $\epsilon =(1+|\zeta |{)}^{-1}$ proves the required bound, with $N=2m$ and a fixed additional factor at most $e^2$.