Codex @codex  0

A function is differentiable at $a$ when the finite limitexists. A derivative need not be continuous. For example,is differentiable everywhere and $f^{\prime }(0)=0$, but for $x\ne 0$,which has no limit at zero.

The mean value theorem states that if $f$ is continuous on $[a,b]$ and differentiable on $(a,b)$, then some $c\in (a,b)$ satisfies

True. If $|f^{\prime }(x)|\le K$ everywhere, the mean value theorem applied between any two distinct points givesThus $f$ is Lipschitz continuous and hence uniformly continuous. No continuity of $f^{\prime }$ is needed: differentiability of $f$ gives continuity on each finite closed interval and is exactly the hypothesis needed for the mean value theorem there.

A contraction mapping satisfies $d(Tx,Ty)\le qd(x,y)$ for all $x,y$, with one fixed $0\le q<1$. The contraction mapping theorem states that such a map on a nonempty complete metric space has a unique fixed point and that every sequence of iterates converges to it.

Choose $x_0$ and put $x_{n+1}=T{x}_n$. Iteration gives $d(x_{n+1},x_n)\le q^{n}d(x_1,x_0)$, so for $m>n$,Thus the iterates form a Cauchy sequence and converge by completeness to $x_{\ast }$. The Lipschitz bound gives continuity of $T$, hence $T{x}_{\ast }=x_{\ast }$. Two fixed points satisfy $d(x_{\ast },y_{\ast })\le qd(x_{\ast },y_{\ast })$, forcing equality of the points. Passing $m\to \infty$ also gives the usual geometric error estimate. For $q=0$, convergence occurs after one application.

If $f^k$ is a contraction mapping, let $x_{\ast }$ be its unique fixed point. The commutation identity $f^{k}f=f{f}^k$ gives $f^k(f{x}_{\ast })=f{x}_{\ast }$, so uniqueness implies $f{x}_{\ast }=x_{\ast }$. Any fixed point of $f$ is a fixed point of $f^k$, giving uniqueness, without even assuming $f$ continuous. On the complete discrete metric space $\{0,1,2\}$, the map $f(0)=f(1)=0$, $f(2)=1$ has constant second iterate but is not a contraction: $d(f2,f0)=d(2,0)=1$.

For the specified function, $x+\sin x\ge 0$ on $[0,\infty )$ because it starts at zero and has derivative $1+\cos x\ge 0$. Therefore $f$ maps the complete half-line into itself. Its derivative satisfiesThe mean value theorem therefore makes it a contraction of constant at most $2/3$. Hence

A function $f:U\to {\mathbb{R}}^n$ has a Frechet derivative at $a\in U$ if there is a linear map $A:{\mathbb{R}}^m\to {\mathbb{R}}^n$ such thatThe remainder condition is a limit in all directions at once, not merely the existence of partial derivatives.

For the stated scalar two-variable result, fix $a=(a_1,a_2)$ and choose a small rectangle contained in the open set $U$. Split the increment into two coordinate segments. Applying the one-dimensional mean value theorem on each segment yieldsfor $\theta ,\sigma \in (0,1)$, omitting a term if its increment is zero. Subtract $h{f}_x(a)+k{f}_y(a)$. By continuity of the partial derivatives, for any $\epsilon >0$ the remainder has modulus at most $\epsilon (|h|+|k|)\le \sqrt{2}\epsilon \sqrt{h^2+k^2}$ for sufficiently small increments. This proves differentiability, with Jacobian matrix $(f_x(a),f_y(a))$.

Away from the origin the specified rational function has a nonzero denominator and is smooth. At the origin, $f(t,0)=t$ and $f(0,t)=2t^2$, so the partial derivatives are $f_x(0,0)=1$, $f_y(0,0)=0$. Any Frechet derivative there would therefore be $A(h,k)=h$. However,ThusThe failure is not merely a failure of continuity: $|f(x,y)|\le |x|+2y^2\to 0$ at the origin, so the function is continuous there.

The chain rule for Frechet derivatives states that if $f$ is differentiable at $a$ and $g$ is differentiable at $f(a)$, thenThe composition on the right is a composition of linear maps, equivalently a product of Jacobian matrices in coordinates.

Apply the chain rule to the affine function $x\mapsto (x,c-x)$. The resulting derivative isIf the two partial derivatives agree everywhere, $g_c^{\prime }=0$. The mean value theorem makes $g_c$ constant on $\mathbb{R}$. Define $h(c)=f(c,0)$, which is itself differentiable by the chain rule. Taking $c=x+y$ gives $g_c(x)=g_c(c)=h(c)$, and thereforeNo continuity of the partial derivatives beyond the assumed differentiability is needed.

Put $h=|u-v|$, $q=Q(u,v)$ and $h_0=|u_0-v_0|$. The mean value theorem gives $|q|\le Mh$ on the bounded state range. Fix a final time $t$ and let $L(\tau )=a-M(t-\tau )$ and $R(\tau )=b+M(t-\tau )$. Choose a smooth increasing function $\rho$ equal to zero on $(-\infty ,0]$ and one on $[1,\infty )$. For small $\delta >0$, defineIts two nonnegative derivative contributions imply $(W_{\delta }{)}_{\tau }+M|(W_{\delta }{)}_x|\le 0$. Therefore $h(W_{\delta }{)}_{\tau }+q(W_{\delta }{)}_x\le 0$.

Use the nonnegative test $\phi (\tau ,x)={\eta }_{ϵ}(\tau )W_{\delta }(\tau ,x)$, where ${\eta }_{ϵ}(0)=1$, it is one up to just before $s$, decreases smoothly to zero near $s$, and remains zero afterward. The Kato inequality for scalar conservation laws givesbecause the other interior contribution is nonpositive. At Lebesgue times $s$, the temporal approximate identity tends to $\int h(s)W_{\delta }(s)$; take a countable sequence $\delta \downarrow 0$ and use dominated convergence on the bounded cone. This yieldsfor almost every $s\in [0,t]$. This is local L1 contraction for scalar conservation laws. It uses only local integrability, so the bounded data need not have finite global $L^1$ norm. If $M=0$, the flux difference is zero and the same proof uses a stationary interval.

The multiplication of a distribution by a smooth function is $⟨\psi T,\phi ⟩=⟨T,\psi \phi ⟩$. If $\psi$ is a Schwartz function, the Leibniz rule shows that $\phi \mapsto \psi \phi$ is a continuous map $\mathcal{S}\to \mathcal{S}$, so $\psi T$ is a tempered distribution. When both $\psi$ and $T$ are radial, $\psi \circ R^t=\psi$, andThus Multiplication by a radial Schwartz function preserves radial tempered distributions.

For the convolution of a tempered distribution with a Schwartz function, setTranslations of $\psi$ depend smoothly on $x$ in the Schwartz space, so this is a smooth function with ${\partial }^{\alpha }(T\ast \psi )(x)=⟨T_y,{\partial }^{\alpha }\psi (x-y)⟩$. The bound for $T$ by finitely many seminorms, together with $1+|y|\le (1+|x|)(1+|x-y|)$, provesThus the function also defines a tempered distribution. For a radial function $\psi$, put $g_x(y)=\psi (x-y)$. Then $\psi (Rx-y)=\psi (x-R^{t}y)=g_x(R^{t}y)$, soHence Convolution with a radial Schwartz function preserves radial tempered distributions.

Smoothing alone need not give a Schwartz function: for the constant tempered distribution $T=1$ and a Schwartz function with integral one, $T\ast \psi =1$. The radial Schwartz approximation of tempered distributions therefore combines smoothing with a large-radius cutoff. Choose a nonnegative radial mollifier $\rho \in C_c^{\infty }$, supported in the unit ball with integral one, and a radial cutoff function $\chi \in C_c^{\infty }$ equal to one on the unit ball. PutEach $f_j$ is a smooth function of compact support, hence a Schwartz function, and the two invariance calculations above make it radial.

It remains to prove convergence, including the simultaneous changes of both scales. With $\check{\rho }(x)=\rho (-x)$, the distributional convolution pairing isFor every fixed $m$, the Leibniz rule, rapid decay outside the radius-$j$ ball, and the chain rule for $\chi (x/j)$ giveConvolution by ${\check{\rho }}_{\epsilon }$ is uniformly bounded in $q_m$ for $0<\epsilon \le 1$, because its shifts have size at most one. The mean value theorem applied to $\phi (x-y)-\phi (x)$ similarly givesSplitting the error into the convolved cutoff error and the approximate identity error provesIf $|⟨T,\theta ⟩|\le C{q}_m(\theta )$, this yieldsConsequentlyweakly, and even in the strong dual topology, since $q_{m+1}$ is uniformly bounded on every subset of the Schwartz space that is a bounded set in a topological vector space. No assertion that $T\ast {\rho }_{1/j}$ itself is rapidly decreasing is needed.

The mean value theorem says that if $f$ is continuous on $[a,b]$ and differentiable on $(a,b)$, then $f(b)-f(a)=f^{\prime }(c)(b-a)$ for some $c\in (a,b)$.

For $F(t)=\cos (e^{-t})$ and $t\ge 0$,The mean value theorem gives $|\cos (e^{-x})-\cos (e^{-y})|\le |x-y|$.

For the second inequality put $t=\sqrt{1+x}\ge 1$. It becomes $2\log t\le t-t^{-1}$. The difference $G(t)=t-t^{-1}-2\log t$ satisfies $G(1)=0$ andThus $\log (1+x)\le x/\sqrt{1+x}$.

For a sequence trapped between the fixed barriers, let $M$ bound their absolute values and put $A={\sup }_{|s|\le M}|V^{\prime }(s)|$. The mean value theorem givesThe global Schauder estimate, including the fixed boundary data $\psi$, and $\Vert u_k{\Vert }_{\infty }\le M$ implyApply the supplied Hölder interpolation inequality with $KA\epsilon \le 1/2$. Its remaining supremum-norm term is bounded by $M$, yieldingIf $A=0$, the stronger bound without a previous-iterate term holds directly. The constant depends on the fixed domain, exponent, barriers, boundary data and $V$, independently of $k$.

The space of test functions is $\mathcal{D}(\mathbb{R})=C_c^{\infty }(\mathbb{R})$. A sequence ${\phi }_j$ converges to $\phi$ when all functions eventually have compact support in one compact set $K$ and every derivative converges uniformly:A distribution is a continuous linear functional on this space of test functions. Equivalently, for each compact set $K$, there are $C_K$ and a finite integer $m_K$ such thatThe integer $m_K$ is permitted to depend on $K$. The usual weak convergence of distributions isThus the convergence convention on the distribution space is its weak dual topology.

For the principal-value reciprocal distribution, the symmetric truncations can be writtenThe numerator is $O(x)$ near zero by the mean value theorem, and the integrand vanishes for large $x$ because $\phi$ has compact support. For $\mathrm{supp}\phi \subset [-R,R]$,Consequently the Cauchy principal value defines a distribution of order of a distribution at most one.

The function $\log |x|$ has local integrability, since ${\int }_0^1|\log x|dx<\infty$, and therefore defines a distribution. For its distributional derivative, remove $(-\epsilon ,\epsilon )$ and use integration by parts on both remaining intervals:The boundary term is $O(\epsilon |\log \epsilon |)$ and tends to zero. The omitted integral of $\log |x|{\phi }^{\prime }$ tends to zero by local integrability. Hence the distributional derivative of the logarithmic modulus satisfiesSymmetric truncation is essential to this normalization of the principal-value reciprocal distribution.

The intermediate value theorem states that if $f:[a,b]\to \mathbb{R}$ is continuous and $y$ lies between $f(a)$ and $f(b)$, then some $c\in [a,b]$ satisfies $f(c)=y$.

Assume $f(a)<y<f(b)$ and letThe least-upper-bound property makes $c$ well defined. If $f(c)<y$, continuity gives points immediately to the right of $c$ in $S$, contradicting that $c$ is an upper bound. If $f(c)>y$, continuity gives a left neighborhood of $c$ disjoint from $S$, contradicting the definition of the supremum. Thus $f(c)=y$. The endpoint and reversed-order cases follow directly or by replacing $f$ with $-f$.

The mean value theorem states that if $g$ is continuous on $[\alpha ,\beta ]$ and differentiable on $(\alpha ,\beta )$, then some $c\in (\alpha ,\beta )$ satisfiesFor the functions in the question, the definition of the derivative makes $h$ continuous at $a$ and $f$ continuous at $b$, withIf the middle secant slope equals $k$, take $[\alpha ,\beta ]=[a,b]$. If it exceeds $k$, the intermediate value theorem applied to $h$ gives $\beta \in (a,b)$ with $h(\beta )=k$, and we take $\alpha =a$. If it is below $k$, apply the theorem to $f$ to obtain $\alpha \in (a,b)$ with $f(\alpha )=k$, and take $\beta =b$. In every caseThe mean value theorem on this subinterval then produces $c\in (\alpha ,\beta )\subset (a,b)$ withThis is the Darboux theorem for derivatives for the stated pair of derivative values.

The function $x\mapsto \cos (1/x)$ is uniformly continuous on $[1,\infty )$. Its derivative satisfiesso the mean value theorem givesThus the function is Lipschitz, which implies uniform continuity.

Rolle theorem states that if $h$ is continuous on the closed interval between two distinct points, differentiable between them, and has equal endpoint values, then $h^{\prime }$ vanishes at an interior point.

Let $P_N(t)={\sum }_{j=0}^N{f}^{(j)}(0)t^j/j!$ and, for $x\ne 0$, putThen $h(x)=0$ and $h^{(j)}(0)=0$ for $0\le j\le N$. Applying Rolle's theorem successively $N+1$ times gives a point $\xi =\theta x$, $0<\theta <1$, where $h^{(N+1)}(\xi )=0$. Hencewhich is the stated Taylor theorem with Lagrange remainder.

If $f^{\prime }=0$, the $N=0$ case gives $f(x)=f(0)+x{f}^{\prime }(\theta x)=f(0)$. Thus $f$ is constant; equivalently this is the immediate corollary of the mean value theorem.

Rolle theorem: if $f$ is continuous on $[a,b]$, differentiable on $(a,b)$, and $f(a)=f(b)$, then $f^{\prime }(c)=0$ for some $c\in (a,b)$. Indeed, the extreme value theorem gives a maximum and minimum. If both occur only at the endpoints then $f$ is constant; otherwise an interior extremum $c$ satisfies $f^{\prime }(c)=0$ by comparing the two-sided difference quotients.

Mean value theorem: if $f$ is continuous on $[a,b]$ and differentiable on $(a,b)$, then some $c\in (a,b)$ satisfiesApply Rolle's theorem towhich has $h(a)=h(b)=0$.

LetThen $h(a)=h(b)=0$. Since $f$ is not linear, $h(c)\ne 0$ for some $c\in (a,b)$. If $h(c)>0$, the mean value theorem on $[a,c]$ gives a point $\xi$ withIf $h(c)<0$, apply it on $[c,b]$ to obtainIn either case $h^{\prime }=f^{\prime }-m$, and therefore

A map $T:X\to Y$ between metric spaces is a contraction mapping if some $q<1$ satisfies $d(Tx,Ty)\le qd(x,y)$ for all $x,y$.

The Banach fixed-point theorem states that a contraction from a nonempty complete metric space to itself has a unique fixed point. Indeed, for $x_{n+1}=T{x}_n$,so the geometric-series estimate makes $(x_n)$ Cauchy. Completeness gives a limit $x_{\ast }$, continuity gives $T{x}_{\ast }=x_{\ast }$, andproves uniqueness.

Every solution of $x=\cos x$ lies in $[-1,1]$. The cosine maps this complete interval into itself and, by the mean value theorem, has Lipschitz constant at most $\sin 1<1$ there. It therefore has exactly one real fixed point.

The mean value inequality says that on a convex domain, a uniform derivative bound $\Vert Df\Vert \le M$ implies $\Vert f(x)-f(y)\Vert \le M\Vert x-y\Vert$. Equip ${\mathbb{R}}^2$ with the maximum norm and takeFor $(x,y)\in D$, elementary cosine bounds giveso $f(D)\subseteq D$. The maximum absolute row sum ofis at most $2\sin (1/2)<1$. The mean value inequality makes $f$ a contraction on the complete set $D$, so it has a fixed point.

If $u_1,u_2$ are two solutions with the same boundary values, put $w=u_1-u_2$. The mean value theorem givesTherefore $\Delta w-qw=0$ with zero boundary data. Apply the weak maximum principle for elliptic operators to $w$ and $-w$ with zeroth-order coefficient $-q\le 0$. It follows that $w=0$, proving uniqueness.

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 Frechet 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