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Differentiability is the existence of a derivative at a point. For maps between finite-dimensional normed vector spaces, the existence of a Frechet derivative requires a single linear map approximating the increment with error $o(\Vert h\Vert )$. Merely having partial derivatives need not provide this approximation; continuous partial derivatives on a neighbourhood do suffice.

For $p>n$, the Hölder continuous function representative $u^{\ast }$ of $u\in W^{1,p}({\mathbb{R}}^n)$ has a classical Frechet derivative at almost every point, equal to its weak derivative. At a point $x$ where the $p$-mean oscillation of $Du$ tends to zero, apply the Morrey inequality on a cube to $u^{\ast }(y)-u^{\ast }(x)-Du(x)\cdot (y-x)$. On a cube of side comparable to $|y-x|$, its oscillation is at mostThe required points have full measure by the Lebesgue differentiation theorem.

This convex function penalizes positive densities using negative entropy, with $0\log 0=0$ and infinite value for negative inputs. Its scalar derivative is $\log x$ on $x>0$. On a bounded domain, restricting to uniformly positive bounded functions gives a simple setting for a Frechet derivative. Its Bregman distance is the generalized Kullback–Leibler divergence. With fixed total mass, minimizing this negative-entropy expression maximizes the corresponding entropy.

The Frechet derivative at $p=(x_0,y_0)$ is a linear map $L:{\mathbb{R}}^2\to \mathbb{R}$ for whichThis is differentiability; if it holds, $L(h_1,h_2)=D_{1}f(p)h_1+D_{2}f(p)h_2$.

To prove the sufficient condition, take a rectangle about $p$ contained in the open domain. Split the increment along its coordinate directions:The one-variable mean value theorem, applicable because the restrictions are differentiable, represents these two terms asThe zero-increment cases are interpreted directly. Subtract $D_{1}f(p)h+D_{2}f(p)k$. By continuity of both partial derivatives at $p$, the remainder is at most $\epsilon (|h|+|k|)\le \sqrt{2}\epsilon \sqrt{h^2+k^2}$ once the increment is small. This is the required little-oh remainder and proves differentiability.

The converse is false. Set $f(x,y)=x^2\sin (1/x)$ for $x\ne 0$, and $f(0,y)=0$. At each point with $x=0$, its Frechet derivative is zero, since the function increment has magnitude at most $h^2=o(\sqrt{h^2+k^2})$. It is smooth away from this line. Butis not continuous there. Thus a function can be differentiable everywhere while its partial derivatives are discontinuous.

For $h$, every point with $x\ne 0$ is a smooth point. At the origin, $|h(x,y)|\le |xy|\le (x^2+y^2)/2$, so $h$ is differentiable with zero derivative. At $(0,y_0)$ with $y_0\ne 0$, the putative first partial derivative would require the limit $y_0\sin (1/x)$, which does not exist. Consequently the exact set of differentiability is

Put $r=\Vert X\Vert <1$. By submultiplicativity of the operator norm, $\Vert X^k\Vert \le r^k$, so the Neumann series converges absolutely in the complete finite-dimensional matrix space. Its partial sums satisfyTaking limits gives the inverse and a quantitative remainder bound:To establish two actual Frechet derivatives, use the resolvent identityFor fixed $X\in U$ and sufficiently small $H$, factor $I-X-H=(I-X)(I-f(X)H)$ and apply the Neumann series to the second factor. This proves that $X+H\in U$, that $f(X+H)$ is locally bounded, and that $f(X+H)-f(X)=O(\Vert H\Vert )$. Subtracting $f(X)Hf(X)$ in the identity leaves $O(\Vert H{\Vert }^2)$. Thus $Df(X)[H]=f(X)Hf(X)$ is the Frechet derivative.

Near zero, $f(X)=I+X+O(\Vert X{\Vert }^2)$, uniformly in the operator norm. Hence, as a linear operator in $H$,This proves differentiability of $Df$ at zero, givingThe order of multiplication matters: the second Frechet derivative is the symmetric bilinear map $HK+KH$, not $2HK$.

Differentiability here is understood in the sense of the Frechet derivative: there is a linear map $L:{\mathbb{R}}^n\to {\mathbb{R}}^m$ such thatThe map $L=Df(x)$ is the Frechet derivative. In finite dimensions a linear map is bounded, so $\Vert Lh\Vert \le C\Vert h\Vert$ for some $C$. The defining estimate therefore impliesThis is precisely continuity at $x$: differentiability implies continuity.

First suppose each $g_j$ is differentiable at $x_j$. Write $g_j(x_j+t)=g_j(x_j)+v_{j}t+r_j(t)$, where $v_j\in {\mathbb{R}}^m$ and $r_j(t)=o(|t|)$. Define the linear map $Lh={\sum }_j{v}_j{h}_j$. Given $\epsilon >0$, choose a common sufficiently small neighbourhood so that $\Vert r_j(h_j)\Vert \le \epsilon |h_j|$ for all $j$, with $r_j(0)=0$. ThenThis proves differentiability, withConversely, suppose $f$ is differentiable with Frechet derivative $L$. Along the $j$th coordinate direction $e_j$, the other summands cancel exactly:Thus $g_j$ is differentiable at $x_j$, with $g_j^{\prime }(x_j)=L{e}_j$. The equivalence works because each summand varies in only one coordinate; it is stronger than merely having partial derivatives.

For the maximum-entropy regularization functional, differentiation of the scalar integrand gives $(s\log s-s{)}^{\prime }=\log s$. Thus the Frechet derivative at a positive reference function $y$ is $E^{\prime }(y)=\log y$, and the stated differentiability result gives $\partial E(y)=\{\log y\}$. Consequently,This is the generalized Kullback–Leibler divergence. When both functions integrate to the same mass, the last two terms cancel after integration; for probability density functions it becomes the usual Kullback-Leibler divergence. For two positive functions, adding the reverse distance givesA sufficient setting for the differentiability argument is $L^{\infty }(\Omega )$ with $y$ bounded away from zero. For nonnegative $x$, use $0\log (0/y)=0$ wherever appropriate. Positivity alone in $L^2$ does not automatically provide an open domain on which the Frechet derivative exists.