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Assume the holonomic constraints have linearly independent gradients, so their common level set is locally a smooth constraint manifold parametrized by $q_1,\dots ,q_n$. The multiplier form of the Euler-Lagrange equations isAt fixed time ${\sum }_A{\partial }_A{f}_{\alpha }{\partial }_i{x}_A=0$. The chain rule for the reduced Lagrangian givesConversely, if these reduced equations hold, the Cartesian Euler-Lagrange residual annihilates every tangent direction. Linear independence of the constraint gradients implies that it is a linear combination of those gradients, giving the required Lagrange multipliers. This proves the local equivalence, including time-dependent constraints; singular constraints would need separate analysis.

For the bead, $\dot{y}=f^{\prime }(x)\dot{x}$, soThe autonomous conservation of energy gives $E=\frac{1}{2}(1+f^{\prime 2}){\dot{x}}^2$. If $E>0$, the motion has a fixed sign $s=\mathrm{sgn}\dot{x}$ andThe integral is the signed arc length divided by the constant speed, and is locally invertible. The special solution $E=0$ is a stationary bead, for which writing time as a function of its unchanging position is not meaningful.

This can fail in the same example as in part (i): $g\circ f$ is injective on the singleton $X$, but $g(a)=g(b)$.

For the final count, $g\circ f={\mathrm{id}}_X$ forces $f$ to be injective. There aresuch functions $f$. Once $f$ is chosen, $g$ is forced on the $m$ elements of $f(X)$ and has $m$ independent choices at each of the other $n-m$ elements of $Y$. Hence the number of pairs is

The intermediate value theorem says that a continuous $f:[a,b]\to \mathbb{R}$ takes every value between $f(a)$ and $f(b)$. For $f(a)<y<f(b)$, the closed sets where $f\le y$ and $f\ge y$ cannot separate the connected interval; equivalently, taking the supremum of $\{x:f(x)\le y\}$ and using continuity gives a point with value $y$.

Set $\varphi (a)=0$ and $\varphi (x)=\sin (1/(x-a))$ for $x>a$. Every interval $[a,b]$ contains a zero $c<b$; continuity on $[c,b]$ makes $\varphi$ take every value between $0=\varphi (a)$ and $\varphi (b)$, yet $\varphi$ is discontinuous at $a$.

A monotone function can be discontinuous only by a jump. If it had a jump at $c$, any number strictly between the left and right limits would lie between $f(a)$ and $f(b)$ but would not be attained, contrary to the hypothesis. Thus it is continuous.

For the last assertion, pass to subsequences with $x_n,y_n$ both within $1/n$ of $a$ and $g(x_n)$ near $l$, $g(y_n)$ near $L$. For $\lambda \in (l,L)$, the intermediate value theorem on the interval joining $x_n$ and $y_n$ gives $z_n$ with $g(z_n)=\lambda$; then $z_n\to a$. The endpoint cases use the original sequences.

The ordinary mean value theorem gives ${\Delta }_{h}f(a)=h{f}^{\prime }(b_1)$ for some $b_1\in (a,a+h)$. More generally, applying it to the difference of two translated functions shows inductively thatfor some $b_k\in (a,a+kh)$. Taking $k=n$ proves the result.

The continuous approximation of a Riemann-integrable function follows here by choosing partitions $P_n$ with upper-minus-lower sum below $1/n$. On each subinterval choose the midpoint of the infimum and supremum to form a step function $s_n$. ThenReplace each of the finitely many jumps of $s_n$ by a linear transition on intervals of sufficiently small total length, obtaining a continuous ${\varphi }_n$ with $\int |s_n-{\varphi }_n|<1/n$. Hence $\int |g-{\varphi }_n|<2/n$, and uniformly for every subinterval $[\alpha ,\beta ]$,

No. TakeBoth functions are nonconstant on every interval and nondifferentiable at zero, but $g(f(x))=2x$ is differentiable.

No. Take $f(x)=x^2$, $a=0$, and $g(y)=|y|$. The function $g$ is not differentiable at $f(0)=0$, but $g(f(x))=x^2$ is differentiable. Neither function is constant on an interval.

Differentiating the moment-generating function at zero gives

The probability generating function isTermwise differentiation givesThe coefficients are nonnegative, and $\mathbb{P}(X\ge 2)>0$, so both derivatives are positive and nondecreasing on $(0,1]$.

For a compactly supported test function $f$,Thus this is the distributional derivative of the Heaviside step function: $H^{\prime }=\delta$.

For every test function $\phi$ and nonzero $p$, substitution $u=pt$ with the orientation handled by $|p|$ givesHence the Dirac delta scaling identity is $\delta (pt)=\delta (t)/|p|$.

If $A_n$ are countable, choose enumerations and map each element of ${\bigcup }_n{A}_n$ to the first pair $(n,k)$ at which it occurs. Since ${\mathbb{N}}^2$ is countable, the union is countable.

A periodic function of period $k$ is determined by its $k$ values on a complete residue system. Thus all periodic functions form a countable union over $k$ of the countable sets ${\mathbb{N}}^k$, and are countable.

The set of bijections $\mathbb{N}\to \mathbb{N}$ is uncountable. Indeed, each binary sequence determines a bijection that swaps $2j-1$ and $2j$ exactly when its $j$th bit is one. This is an injection from the uncountable set of binary sequences.

A Cauchy sequence $(x_n)$ in a metric space $(M,d)$ satisfies: for every $\epsilon >0$ there is $N$ such that $d(x_m,x_n)<\epsilon$ whenever $m,n\ge N$. A complete metric space is one in which every Cauchy sequence converges to a point of the space.

Every Cauchy sequence is bounded. Choose $N$ such that $d(x_n,x_N)<1$ for $n\ge N$, and putThen every term lies in the ball $B(x_N,R)$.

Now suppose $M$ is complete and $(F_n)$ is a decreasing sequence of nonempty closed sets with $\mathrm{diam}{F}_n\to 0$. Choose $x_n\in F_n$. Given $\epsilon >0$, choose $N$ with $\mathrm{diam}{F}_N<\epsilon$. For $m,n\ge N$, both points lie in $F_N$, so $d(x_m,x_n)<\epsilon$. Completeness gives $x_n\to x\in M$. For each fixed $N$, the tail lies in the closed set $F_N$, hence $x\in F_N$. Therefore

Conversely, assume the nested-set property and let $(x_n)$ be Cauchy. DefineThese sets are nonempty, closed, and decreasing. The Cauchy property implies $\mathrm{diam}{F}_n\to 0$; taking a closure does not change the diameter. Choose $x\in {\bigcap }_n{F}_n$. Since $x_n,x\in F_n$,Thus every Cauchy sequence converges and $M$ is complete. This proves the Cantor intersection theorem characterization.

The contraction mapping theorem states that a contraction of a nonempty complete metric space has a unique fixed point.

For each $\lambda \in \Lambda$, the map $T_{\lambda }(x)=T(\lambda ,x)$ is a contraction with the common constant $k$, so it has a unique fixed point $x^{\ast }(\lambda )$. This defines the required unique function. Fix ${\lambda }_0$. The fixed-point identities and the triangle inequality giveConsequentlyThe numerator tends to zero as $\lambda \to {\lambda }_0$ by the assumed continuity for the fixed point $x^{\ast }({\lambda }_0)$. Hence $x^{\ast }$ is continuous, an instance of continuous dependence of the fixed point of a uniform contraction.

The residue theorem states that if a meromorphic function has finitely many poles inside a positively oriented simple closed contour and none on it, then its contour integral is $2\pi i$ times the sum of the enclosed residues.

Apply it toon a keyhole contour around the positive real axis. The outer and inner circles vanish as their radii tend to infinity and zero because $\alpha <1$ and $\alpha >-1$, respectively. On the upper bank the numerator tends to $x^{\alpha }$, while on the lower bank it tends to $e^{2\pi i\alpha }{x}^{\alpha }$ and the direction is reversed. Therefore the limiting contour integral is

The only enclosed pole is the double pole at $z=-1$. Since $L(-1)=i\pi$,The residue theorem now givesFor $\alpha \ne 0$, division andyieldAt $\alpha =0$, the integral is one, agreeing with the continuous limit. This is the positive-axis keyhole beta integral.

For a differentiable wave function,Thus $[\hat{x},\hat{p}]=i\hslash I$. Substitution into part (a) yields

The function is not Lipschitz on $\mathbb{R}$. For $\epsilon >0$,which is unbounded as $\epsilon \downarrow 0$.

The functionis continuous, satisfies the equation away from $\tilde{t}=y_0$, and has unequal one-sided derivatives there.

It is unique in the stated class. While a solution is positive its derivative is $-1$, so continuity forces it to follow $y_0-t$ until it first reaches zero at $t=y_0$. It cannot subsequently make a positive excursion: on any connected interval $(a,b)$ where $y>0$ and $y(a)=0$, integration of $y^{\prime }=-1$ gives $y(t)=-(t-a)<0$. A negative excursion similarly contradicts $y^{\prime }=1$. Hence the solution remains zero. This is the sign-decay differential equation.

A sequence $f_n:S\to \mathbb{R}$ converges uniformly to $f$ if, for every $\epsilon >0$, there is $N$ such thatfor every $x\in S$ and every $n\ge N$. A map $h:M\to N$ is uniformly continuous if, for every $\epsilon >0$, there is $\delta >0$ such thatfor all $x,y\in M$.

If $f\in C_0({\mathbb{R}}^d)$, choose a ball outside which $|f|<1$. On the closed ball, continuity gives boundedness, so $f$ is bounded everywhere.

Now let $(f_n)$ be Cauchy in the uniform metric. For each $x$, $(f_n(x))$ is Cauchy in $\mathbb{R}$; let its limit be $f(x)$. Passing to the pointwise limit in the uniform Cauchy estimate shows that $f_n\to f$ uniformly. Hence $f$ is continuous. Given $\epsilon >0$, choose $n$ with $\Vert f-f_n{\Vert }_{\infty }<\epsilon /2$, and then choose $K$ so that $|f_n(x)|<\epsilon /2$ for $\Vert x\Vert >K$. It follows that $f$ also vanishes at infinity. Thus $C_0({\mathbb{R}}^d)$ is complete, as in completeness of continuous functions vanishing at infinity.

Every $f\in C_0({\mathbb{R}}^d)$ is uniformly continuous. Given $\epsilon >0$, choose $R$ so that $|f(x)|<\epsilon /2$ outside the ball of radius $R$. On the compact ball of radius $R+1$, $f$ is uniformly continuous; choose the corresponding $\delta \le 1$. If two points at distance below $\delta$ are not both in that ball, then both lie outside the ball of radius $R$, and their function values differ by less than $\epsilon$.

For the final sequence, continuity of $\epsilon$ at zero gives, for each fixed $x$,Thus pointwise convergence is compulsory. Uniform convergence need not hold: with $\epsilon (t)=t^2$,which is unbounded as a function of $x$ for every fixed $n$.

Under the additional bound $\epsilon (t)\le M|t|$, however,for $x\ne 0$, and the difference is zero at $x=0$. Hence convergence is uniform, by the uniform square-root perturbation under linear growth estimate. The pointwise answer remains yes.