From First-Order Approximation to Nearby Values
The Jacobian Matrix tutorial described the derivative as the linear map that predicts the first-order change in a function. The key consequence for this tutorial is that the prediction becomes accurate relative to the size of the input displacement as that displacement shrinks. In particular, the output cannot jump abruptly at a point where this approximation holds.
Let \(U\) be open in \(\mathbb{R}^m\), let \(f:U\to\mathbb{R}^n\), and let \(a\in U\). Differentiability at \(a\) means that there is a linear map \(L:\mathbb{R}^m\to\mathbb{R}^n\) for which the error after subtracting the linear approximation is small compared with the input displacement. In the notation used earlier in this course, \(L=Df(a)\).
The theorem is pointwise: differentiability at \(a\) gives continuity at \(a\). If \(f\) is differentiable at every point of \(U\), applying the theorem separately at each point shows that \(f\) is continuous throughout \(U\). Neither conclusion requires the derivative to be continuous as a function of the point.
Continuity at one point is already enough for useful local control. For example, it prevents the function values from becoming arbitrarily large arbitrarily close to that point. It also ensures that a nonzero value cannot suddenly become zero at points arbitrarily close by.
Worked Examples of Pointwise Continuity
Worked Example: Checking a First-Order Approximation
Define \(F:\mathbb{R}^2\to\mathbb{R}^2\) by \(F(x,y)=(x^2+xy,\ y^2-x)\), and consider \(a=(1,0)\). Direct substitution gives \(F(1,0)=(1,-1)\). For an input displacement \(h=(u,v)\), expansion gives $$ F(1+u,v)=(1+2u+u^2+v+uv,\ v^2-1-u). $$ The linear part of the change is \(L(u,v)=(2u+v,-u)\), represented by the matrix $$ \begin{pmatrix}2&1\\-1&0\end{pmatrix}. $$
The remainder after subtracting this linear part is \(r(u,v)=(u^2+uv,v^2)\). If \(\rho=\sqrt{u^2+v^2}\), then \(|u|\leq\rho\) and \(|v|\leq\rho\), so \(|u^2+uv|\leq 2\rho^2\) and \(|v^2|\leq\rho^2\). Therefore \(\|r(u,v)\|_2\leq\sqrt{5}\rho^2\), and, for \(\rho>0\), \(\|r(u,v)\|_2/\rho\leq\sqrt{5}\rho\to0\) as \((u,v)\to(0,0)\). This verifies differentiability at \((1,0)\). The recalled theorem then gives continuity there: inputs approaching \((1,0)\) have outputs approaching \((1,-1)\).
Worked Example: A Differentiable Function on an Interval
Let \(g:\mathbb{R}\to\mathbb{R}\) be \(g(t)=t^3-2t\), and take \(a=2\). Its derivative is \(g'(t)=3t^2-2\), so \(g'(2)=10\); in particular, \(g\) is differentiable at \(2\). Also \(g(2)=8-4=4\). By differentiability implying continuity, for every \(\varepsilon>0\) there is a \(\delta>0\) such that \(|t-2|<\delta\) implies \(|g(t)-4|<\varepsilon\). For example, taking \(\varepsilon=1\) guarantees \(3<g(t)<5\) whenever \(t\) is sufficiently close to \(2\). This conclusion concerns all nearby inputs, not only inputs on one selected side of \(2\).
These examples illustrate the pointwise conclusion: once differentiability at a point is known, continuity controls the function values for every sufficiently small displacement. The following consequences make that control explicit.
Local Boundedness Near a Differentiability Point
Proof. By the theorem that differentiability implies continuity, \(f\) is continuous at \(a\). Apply continuity with output tolerance \(1\). There is \(\delta>0\) such that, for \(x\in U\), if \(\|x-a\|_2<\delta\), then \(\|f(x)-f(a)\|_2<1\). Since \(U\) is open and contains \(a\), there is \(s>0\) such that \(B_s^{(2)}(a)\subseteq U\). Set \(r=\min\{\delta,s\}\), which is positive. For every \(x\in B_r^{(2)}(a)\), the triangle inequality gives $$ \|f(x)\|_2 \leq \|f(a)\|_2+\|f(x)-f(a)\|_2 <\|f(a)\|_2+1. $$ Thus \(M=\|f(a)\|_2+1\) is a positive bound on the ball, as required. \(\square\)
The radius in this result may depend on both the point and the function. The conclusion is local: it does not say that \(f\) is bounded on all of \(U\), which may be unbounded or may contain points where the function grows without limit. It says that near a fixed differentiability point, the values stay within a finite distance of \(f(a)\).
Worked Example: An Explicit Local Bound
Let \(H:\mathbb{R}^2\to\mathbb{R}\) be \(H(x,y)=x^2+3y\), and let \(a=(0,0)\). The function is differentiable and \(H(0,0)=0\). We can obtain an explicit bound on a neighborhood. If \(\|(x,y)\|_2<1\), then \(|x|<1\) and \(|y|<1\). Consequently, $$ |H(x,y)|=|x^2+3y| \leq |x|^2+3|y|<1+3=4. $$ Thus \(H\) is bounded on the open unit ball about \((0,0)\). The theorem guarantees such a neighborhood from differentiability; the direct estimates here give a particular radius and bound.
Nonzero Values Persist Nearby
Proof. Differentiability implies continuity at \(a\). Set \(\varepsilon=\|f(a)\|_2/2\), which is positive because \(f(a)\neq0\). Continuity gives \(\delta>0\) such that, for \(x\in U\), \(\|x-a\|_2<\delta\) implies \(\|f(x)-f(a)\|_2<\|f(a)\|_2/2\). Openness of \(U\) gives \(s>0\) with \(B_s^{(2)}(a)\subseteq U\). Let \(r=\min\{\delta,s\}\). For \(x\in B_r^{(2)}(a)\), the reverse triangle inequality yields $$ \|f(x)\|_2 \geq \|f(a)\|_2-\|f(x)-f(a)\|_2 >\frac{\|f(a)\|_2}{2}>0. $$ Therefore \(f(x)\neq0\) throughout this ball. \(\square\)
For a real-valued function, the same continuity argument shows that a strictly positive value remains positive in a sufficiently small neighborhood. If \(f(a)>0\), choose \(\varepsilon=f(a)/2\); then \(|f(x)-f(a)|<f(a)/2\) implies \(f(x)>f(a)/2>0\). If \(f(a)<0\), choosing \(\varepsilon=|f(a)|/2\) ensures \(f(x)<f(a)/2<0\). Thus differentiability at a point guarantees local persistence of the sign whenever the value there is nonzero.
Worked Example: Preserving a Nonzero Vector Value
Define \(P:\mathbb{R}^2\to\mathbb{R}^2\) by \(P(x,y)=(3+x,\,4+y)\), and consider \(a=(0,0)\). This affine map is differentiable, and \(P(a)=(3,4)\), whose Euclidean norm is \(5\). If \(\|(x,y)\|_2<1\), then \(\|P(x,y)-P(0,0)\|_2=\|(x,y)\|_2<1\). The reverse triangle inequality gives \(\|P(x,y)\|_2>5-1=4\). In particular, \(P(x,y)\neq(0,0)\) throughout the unit ball. This explicit estimate illustrates the local separation from zero theorem.
Why the Converse Fails
Differentiability is sufficient for continuity, but continuity alone does not guarantee differentiability. A continuous function can fail to have a linear first-order approximation at a point. In one dimension, that approximation would have the form \(f(a)+ch\), with a single slope \(c\) that works for small displacements from both sides.
Worked Example: Continuous but Not Differentiable at the Origin
Consider \(q:\mathbb{R}\to\mathbb{R}\) defined by \(q(t)=|t|\). Since \(||t|-|s||\leq|t-s|\) for all real \(s,t\), \(q\) is continuous everywhere, including at \(0\). But the difference quotient at \(0\) is $$ \frac{q(h)-q(0)}{h} = \begin{cases} 1,&h>0,\\ -1,&h<0. \end{cases} $$ The right-hand and left-hand limits differ, so the derivative at \(0\) does not exist. Continuity controls how close the function values are to \(q(0)\); differentiability requires the stronger condition that one linear map correctly describes the first-order change.
There is also a converse pitfall involving partial derivatives: their existence at a point does not by itself establish differentiability there. The Jacobian Matrices tutorial demonstrated this with a function whose coordinate partial derivatives exist but whose change along a diagonal is not negligible compared with the displacement. Thus neither continuity nor the mere existence of partial derivatives should be substituted for the full differentiability condition.
Using the Implication Carefully
The practical direction of the theorem is one-way. When differentiability at \(a\) has been established, continuity at \(a\) follows, and the local boundedness and nonvanishing results above are available. When only continuity is known, the example \(q(t)=|t|\) shows that differentiability cannot be inferred. When only partial derivatives are known, the multivariable remainder condition still needs to be checked.
It is also important to keep the location of the conclusion clear. Differentiability at one point guarantees continuity at that point; it does not by itself assert continuity at every other point of the domain. If differentiability is known at every point, then continuity holds at every point by applying the theorem pointwise. No continuity of the derivative is needed for this conclusion.
Check Your Understanding
Use the theorem and its consequences to answer each question.
- What is the pointwise conclusion of differentiability at \(a\), and what extra assumption gives continuity throughout \(U\)?
- How does continuity with tolerance \(1\) produce a bound for \(f\) near a differentiability point?
- Why does choosing an output tolerance smaller than \(\|f(a)\|_2\) ensure that nearby values of \(f\) are nonzero?
- For a scalar-valued function with \(f(a)<0\), what tolerance can be used to show that \(f\) remains negative near \(a\)?
- Why does continuity of \(q(t)=|t|\) at \(0\) not imply differentiability there?