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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.

Evaluating the function at the endpoint gives the displacementThereforeThe force-induced displacement $h$ is a linear map of $F$.

Suppose first that $X$ is connected and $f:X\to \mathbb{Z}$ is continuous. Its image is connected by the continuous image of a connected space, but $\mathbb{Z}$ is discrete, so its only connected nonempty subsets are singletons. Hence $f$ is constant.

Conversely, if $X=U\cup V$ is a disconnection into nonempty disjoint open sets, the function equal to zero on $U$ and one on $V$ is continuous and nonconstant. This proves the integer-valued function criterion for connectedness.

Now let $f:X\to \mathbb{Z}$ be continuous under the hypotheses on the family $\mathcal{A}$. Each restriction $f{|}_A$ is constant because $A$ is connected. If $A,B\in \mathcal{A}$, a point of $A\cap B$ shows that their two constants agree. Since the sets cover $X$, $f$ is constant on $X$, and the criterion proves that $X$ is connected. This is the pairwise-intersecting connected cover argument.

Finally, fix $y_0\in Y$. For each $x\in X$, the setis connected: its two connected pieces meet at $(x,y_0)$. The sets $A_x$ cover $X\times Y$ and any two share $X\times \{y_0\}$. The preceding result proves that $X\times Y$ is connected, giving the product of connected spaces result.

The function is odd, so only sine coefficients occur. For $n\ge 1$,Integration by parts givesandThe terms of order $1/n$ cancel, leavingThus

The Parseval identity givesDirect integration yieldsThereforeas recorded in the Fourier series of x cubed minus pi squared x.

UseStationarity givesPutting $t=\lambda /2$, this becomesand the constraint requiresThe left side is strictly increasing, and $t=1$ solves the equation. Hence

The objective is convex and the constraint is affine. Its tangent-plane inequality at $(1,1,1)$ gives, for every feasible $(x,y,z)$,so the Lagrange point is globally optimal. Moreover, at the dual value $\lambda =2$, the infimum of the Lagrangian function in constrained optimization is attained at the same point and equals three. The primal and dual values coincide, so strong duality holds.

For the value function, the multiplier convention above gives the derivative of a constrained value functionAt $b=6$, therefore,

For $a\in U$, the function $f:U\to {\mathbb{R}}^n$ is differentiable at $a$ if there is a linear map $A:{\mathbb{R}}^m\to {\mathbb{R}}^n$ such thatThe linear map is unique and is the Fréchet derivative $D{f}_a=A$. The function is continuously differentiable at $a$ if it is differentiable on a neighbourhood of $a$ and the map $x\mapsto D{f}_x$, with values in the space of linear maps equipped with the operator norm, is continuous at $a$.

If $T:{\mathbb{R}}^m\to {\mathbb{R}}^n$ is linear, thenConsequently $D{T}_a=T$ at every $a$. The derivative is a constant function of $a$, hence is continuous, so every linear map is continuously differentiable everywhere.

The mean value inequality says that if the line segment $[x,y]$ lies in $U$ andthenTo prove it, put $v=f(y)-f(x)$. The claim is immediate if $v=0$. Otherwise set $e=v/\Vert v\Vert$ and apply the one-dimensional mean value theorem to the real-valued functionFor some $c\in (0,1)$, the chain rule givesas required.

Now suppose that $U$ is open and connected and that $D{f}_a=0$ for every $a\in U$. Every point has an open ball contained in $U$. The mean value inequality with $M=0$ shows that $f$ is constant on each such ball, so $f$ is locally constant. Fix $a_0\in U$. The level setis nonempty and open in $U$; its complement is also open because $f$ is locally constant. Since $U$ is connected, $E=U$. This proves the zero derivative on a connected open set result: $f$ is constant.

The inverse function theorem states that if $f:U\to {\mathbb{R}}^m$ is continuously differentiable, $a\in U$, and $D{f}_a$ is an invertible linear map, then there are open neighbourhoods $V$ of $a$ and $W$ of $f(a)$ such that $f{|}_V:V\to W$ is a bijection whose inverse is continuously differentiable.

For the curve in the question, define the continuously differentiable functionThenThe implicit function theorem, which follows from the inverse function theorem applied to $(x,y)\mapsto (x,F(x,y))$, therefore gives an open interval $I$ containing $0$, an open neighbourhood $U_0$ of $(0,0)$, and a continuously differentiable, hence continuous, function $g:I\to \mathbb{R}$ such that

Rouché's theorem states that if $g$ and $h$ are holomorphic functions on a neighbourhood of the closure of a bounded domain andon its positively oriented boundary, then $g$ and $g+h$ have the same number of zeros in the domain, counted with multiplicity.

The Open mapping theorem states that a nonconstant holomorphic function on a domain maps every open subset to an open subset. To prove it, fix $z_0$ and put $w_0=f(z_0)$. By the identity theorem, the zeros of $f-w_0$ are isolated. Choose $r>0$ so that the closed disc $\overline{D(z_0,r)}$ lies in the domain and $z_0$ is the only zero of $f-w_0$ in that disc. ThenWhenever $|w-w_0|<\delta$, the constant $w_0-w$ has modulus smaller than $f-w_0$ on the circle. The Rouche theorem therefore says thathas the same positive number of zeros in the disc as $f-w_0$. Thus every $w\in D(w_0,\delta )$ lies in the image of $f$, proving that the image is open.

If $|f|$ had a local maximum at $z_0$, take a small open disc $D$ on which $|f(z)|\le |f(z_0)|$. The restriction of $f$ to $D$ cannot be constant, since the identity theorem would then make $f$ constant on the whole domain. The complex open mapping theorem says that $f(D)$ is an open neighbourhood of $f(z_0)$, and such a neighbourhood contains points of modulus greater than $|f(z_0)|$, a contradiction. This proves the maximum modulus principle from the complex open mapping theorem.

Now let $\Omega$ be bounded. Its closure is compact, so the continuous function $|f|$ attains a maximum there. If $f$ is nonconstant, the maximum modulus principle excludes an interior maximum; if $f$ is constant, the claimed bound is immediate. Hencewhich is the maximum modulus principle on a bounded domain.

Finally let $\Omega =\{z:\mathrm{Re}z>1\}$ and suppose $|f|\le K$ throughout $\overline{\Omega }$. Fix $z_0\in \Omega$. The principal complex logarithm is holomorphic on the right half-plane, so for each positive integer $n$ the functionis holomorphic on $\Omega$ and continuous on its closure. Given $\epsilon >0$, choose $R>|z_0|$ so large thatApply the bounded-domain result to $g_n$ on $\Omega \cap D(0,R)$. On the vertical part of the boundary, $|z|\ge 1$ and hence $|g_n(z)|\le M$. On the circular part,It follows thatLetting $n\to \infty$ and then $\epsilon \to 0$ gives $|f(z_0)|\le M$. Since $z_0$ was arbitrary, this proves the bounded half-plane maximum principle.

The boundedness assumption is necessary. The functionis holomorphic on $\Omega$ and continuous on its closure, and $|f(1+iy)|=e$ on the boundary, but $|f(x)|=e^x$ is unbounded for real $x>1$. Thus the boundary estimate does not control an unbounded holomorphic function on this unbounded domain.

A function $f:D\to \mathbb{R}$ on a convex set $D$ is convex whenfor all $x,y\in D$ and $0\le t\le 1$.

If $f$ is once differentiable, this is equivalent to monotonicity of its gradient:Equivalently, every tangent hyperplane supports the graph:If $f$ is twice differentiable, convexity is equivalent to the Hessian matrix being positive semidefinite throughout $D$.

Forthe Hessian isA real symmetric two-by-two matrix is positive semidefinite exactly when its two diagonal entries and its determinant are nonnegative. ThusThe largest convexity domain is thereforeFor $A\ne 0$, its boundary is the hyperbola $y=A^2/(36x)$ in the first quadrant and the domain lies above it. For $A=0$, it is the closed first quadrant. This is the convexity domain of x cubed plus y cubed plus Axy.

The spatial wavefunction $\chi (x)$ is the position amplitude of a stationary quantum state, so $|\chi (x){|}^2$ is its probability density after normalization. The real function $U(x)$ is the potential energy, $E$ is the energy eigenvalue, $m$ is the particle mass, and $\hslash$ is the reduced Planck constant.

The function $f$ is differentiable at $a$ when there is a real number $L$ such thatEquivalently, the difference quotient has a finite limit, and

No. At $a=0$, letThe function $g$ is differentiable at $f(0)=0$, andso the composite has derivative one at zero. Butis unbounded, so $f$ is not differentiable. This is a nondifferentiable inner function with a differentiable nondegenerate composite.

For $k\ge 1$, let $E_k$ be the set of numbers whose first digit $5$ occurs in position $k$. The first $k-1$ digits each have nine allowed values, while the $k$th is fixed, so $E_k$ is a union of decimal intervals of total lengthThe set $E_{\infty }$ of numbers having no digit $5$ can, after $N$ places, be covered by intervals of total length $(9/10{)}^N$; hence it has length zero.

For each fixed truncation level, the function is constant on finitely many decimal cylinders apart from the remaining set, whose covering length can be made arbitrarily small. The Riemann integrability criterion therefore shows that every truncation is Riemann integrable.

For integer $N$,The final term tends to zero, and monotonicity handles noninteger truncation levels. Therefore the first occurrence of a decimal digit calculation gives

No. Enumerate the rationals in $[0,1]$ asand defineThe function is nonnegative and unbounded, while it is identically zero on $[0,1]\setminus A$. For any $r>0$, however, $f_r$ is positive on the dense set of rationals and zero on the dense set of irrationals. Every interval therefore has positive oscillation, and the upper and lower Darboux integrals differ. Thus even the truncations need not be Riemann integrable.

The zero function is a solution. On an interval where $y\ne 0$, putThenThe differential equation becomesHence $z=Cx$, and integration givesThus all nonzero solutions aretogether with $\boxed{y\equiv 0}$.

Set $x=y=0$ in the functional equation:The denominator is defined. If $f(0)\ne 0$, cancellation would give $f(0{)}^2=-1$, impossible for a real-valued function. Hence

Independence givesThis is the moment generating function of a Poisson distribution with mean $n$. Hence