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For independent and identically distributed random variables $X,Y$ with a centered multivariate normal distribution, a continuously differentiable function $f$ with bounded gradient, and a convex function $\Psi$ for which the expressions are integrable, the displayed inequality holds. First apply Jensen inequality conditionally to $f(X)-f(Y)$. For the Gaussian rotation of independent copies, the chain rule gives $f(X)-f(Y)={\int }_0^{\pi /2}⟨\nabla f(U_{\theta }),V_{\theta }⟩d\theta$. A second application of Jensen inequality to the uniform measure on $[0,\pi /2]$, followed by the Gaussian rotation of independent copies, proves the inequality. The useful choice $\Psi (v)=e^{\lambda |v|}$ yields a Gaussian concentration inequality through an exponential moment of an absolute standard normal variable.

For a cube $Q$ of side length $r$ in ${\mathbb{R}}^n$, $p>n$, and a continuously differentiable function $v$, its average $v_Q=|Q{|}^{-1}{\int }_{Q}v$ satisfiesAveraging the fundamental theorem of calculus along a line segment from $y$ to $z\in Q$ and changing variables gives $|v(y)-v_Q|\le C_n{\int }_Q|Dv(w)||w-y{|}^{1-n}dw$. The Holder inequality applies because $(n-1)p^{\prime }<n$, and the kernel norm is $O(r^{1-n/p})$. Approximation extends the estimate to the continuous representative of a Sobolev space function. On ${\mathbb{R}}^n$ it yields a Hölder continuous function of exponent $1-n/p$ representing every $W^{1,p}$ class.

For $x\ne 0$, all component functions are continuously differentiable and the Jacobian matrix isAt a point $(0,y_0)$, the first component cannot be differentiable: along the first coordinate axis its difference quotient contains $h^{1/3}/h=h^{-2/3}$, which is unbounded as $h\to 0$. Therefore $f$ is continuously differentiable exactly at the points with $x\ne 0$.

A function $f:{\mathbb{R}}^m\to {\mathbb{R}}^r$ is differentiable at $x$ if there is a linear map $L:{\mathbb{R}}^m\to {\mathbb{R}}^r$ such thatThe map $L$, which is unique, is the derivative $f^{\prime }(x)$.

Expanding in a matrix increment $H$ givesTherefore $p$ is differentiable everywhere and

The inverse function theorem states that if $f$ is continuously differentiable near $a$ and $f^{\prime }(a)$ is invertible, then $f$ restricts to a bijection between neighbourhoods of $a$ and $f(a)$; its local inverse is continuously differentiable, with derivativeIn the normalized case $f(0)=0$ and $f^{\prime }(0)=I$, continuity of $f^{\prime }$ gives a closed ball ${\overline{B}}_r(0)$ on whichFor small $y$, define $T_y(x)=x-f(x)+y$. The mean value inequality makes $T_y$ a contraction mapping with constant $q$. If $\Vert y\Vert \le (1-q)r$, thenso $T_y$ maps the ball into itself. The contraction mapping theorem gives a unique fixed point $x_y$, and the fixed-point equation is precisely $f(x_y)=y$. Moreover,so $\Vert x_y-x_z\Vert \le \Vert y-z\Vert /(1-q)$. Thus $y\mapsto x_y$ is a continuous local inverse.

For the polynomial map in the question,The derivative is invertible, so the inverse function theorem supplies some $ϵ>0$ and a continuously differentiable $q:D_{ϵ}(I)\to {\mathcal{M}}_n$ such thatThere can be no inverse on all of ${\mathcal{M}}_n$, becauseso $p$ is not injective.

If $f$ is continuously differentiable, then for every finite partition the fundamental theorem of calculus and the triangle inequality givePart (b) therefore implies $\Vert f\Vert <\infty$.

For a continuously differentiable function $F$ on an interval $I=[t-\delta /2,t+\delta /2]$,To prove it, average $|F(t){|}^2-|F(u){|}^2$ over $u\in I$ using the fundamental theorem of calculus. On either half of $I$, the resulting derivative kernel has absolute value at most $1/2$. Since $|(|F{|}^2)\prime |\le 2|FF\prime |$, the displayed bound follows.