Mean Value Estimates in R^n

From Derivatives to Estimates on a Segment

The Multivariable Chain Rule lets us differentiate a function after restricting its input to a line. This simple restriction turns a question about changes in $\mathbb{R}^m$ into a one-variable question: how much can the function change as we move from one endpoint of a segment to the other? The resulting mean value estimates are useful even when the derivative is not constant. A bound on the derivative along the segment gives a bound on the total change.

We use the Euclidean norm and the operator norm for linear maps established earlier in this course. If $f:U\to\mathbb{R}^n$ is differentiable at $z$, then $Df(z)$ is a linear map from $\mathbb{R}^m$ to $\mathbb{R}^n$, and $\|Df(z)(h)\|_2\leq\|Df(z)\|_{\mathrm{op}}\|h\|_2$. The estimate below uses a bound on this operator norm along the segment, not merely at its endpoints.

Theorem (Mean Value Estimate Along a Segment): Let $U\subseteq\mathbb{R}^m$ be open, let $f:U\to\mathbb{R}^n$ be differentiable at every point of $U$, and let $x,y\in U$. Suppose the segment $[x,y]=\{x+t(y-x):0\leq t\leq1\}$ is contained in $U$, and suppose that $\|Df(z)\|_{\mathrm{op}}\leq M$ for every $z\in[x,y]$, where $M\geq0$. Then $$ \|f(y)-f(x)\|_2\leq M\|y-x\|_2. $$ If $n=1$ and $x\neq y$, there is also a point $c=x+t_0(y-x)$, with $0<t_0<1$, such that $$ f(y)-f(x)=Df(c)(y-x). $$

Proof. Put $h=y-x$, and define the path $\gamma(t)=x+th$ for $0\leq t\leq1$. The segment hypothesis ensures that every $\gamma(t)$ lies in $U$. By the Multivariable Chain Rule, the function $t\mapsto f(\gamma(t))$ is differentiable for $0<t<1$, with derivative $Df(\gamma(t))(h)$. It is continuous on $[0,1]$, because differentiability of $f$ implies continuity and $\gamma$ is continuous.

First suppose $n=1$ and $x\neq y$. The ordinary one-variable Mean Value Theorem, applied to $t\mapsto f(\gamma(t))$, gives a $t_0\in(0,1)$ such that

f(y)-f(x) = \left.\frac{d}{dt}f(\gamma(t))\right|_{t=t_0} = Df(\gamma(t_0))(h).

Taking absolute values and using the derivative bound gives $|f(y)-f(x)|\leq M\|h\|_2$, which is the asserted estimate when $n=1$. If $x=y$, the estimate holds because both sides are zero.

Now suppose $n$ is any positive integer. If $f(y)=f(x)$, the estimate holds immediately. Otherwise, define $u=(f(y)-f(x))/\|f(y)-f(x)\|_2$, so $\|u\|_2=1$. Apply the one-variable Mean Value Theorem to the scalar function $\psi(t)=u\cdot f(\gamma(t))$. For some $t_0\in(0,1)$,

\|f(y)-f(x)\|_2 = \psi(1)-\psi(0) = u\cdot Df(\gamma(t_0))(h).

By the Cauchy–Schwarz inequality and the operator-norm bound,

\|f(y)-f(x)\|_2 \leq \|u\|_2\|Df(\gamma(t_0))(h)\|_2 \leq M\|h\|_2.

Since $h=y-x$, this proves the estimate. $\square$

The scalar conclusion is an exact formula: one point on the segment gives the full change in the function. For a vector-valued function, the proof instead selects a scalar projection in the direction of the total change. It gives a norm bound, but it does not generally produce one point where the vector difference equals the derivative applied to the segment direction.

Worked Examples: Using the Segment Estimate

Worked Example: An Exact Scalar Mean Value Point

Let $f:\mathbb{R}^2\to\mathbb{R}$ be $f(x,y)=x^2+xy+2y^2$. Take $a=(1,-1)$ and $b=(2,1)$, so $b-a=(1,2)$. The endpoint values are $f(a)=2$ and $f(b)=8$, hence $f(b)-f(a)=6$.

Parameterize the segment by $\gamma(t)=(1+t,-1+2t)$. Substitution gives

f(\gamma(t)) =(1+t)^2+(1+t)(-1+2t)+2(-1+2t)^2 =2-5t+11t^2.

Its derivative with respect to $t$ is $-5+22t$. At $t_0=1/2$, this derivative is $6$, exactly the endpoint difference. The corresponding point is $c=(3/2,0)$. Directly, $Df(x,y)(u,v)=(2x+y)u+(x+4y)v$, so $Df(c)(1,2)=3(1)+\frac32(2)=6$. This verifies the scalar mean value formula at the point found along the segment.

Worked Example: A Vector Estimate Without an Exact Formula

Consider $f:\mathbb{R}\to\mathbb{R}^2$ given by $f(t)=(t^2,t^3)$ on the segment from $0$ to $1$. Its derivative is $Df(t)(h)=(2th,3t^2h)$, whose operator norm is $\sqrt{4t^2+9t^4}\leq\sqrt{13}$ for $0\leq t\leq1$. The Mean Value Estimate therefore gives

\|f(1)-f(0)\|_2=\|(1,1)\|_2=\sqrt{2} \leq \sqrt{13}\,|1-0|.

There is no $t\in(0,1)$ for which $f(1)-f(0)=Df(t)(1)$: that equality would require both $2t=1$ and $3t^2=1$. The first equation forces $t=1/2$, but then $3t^2=3/4\neq1$. This shows why the general vector conclusion is an estimate rather than an exact mean value formula.

Worked Example: A Lipschitz Bound on a Square

Let $f:\mathbb{R}^2\to\mathbb{R}^2$ be $f(x,y)=(x^2+y,xy)$, and consider the square $Q=[-1,1]^2$. Its Jacobian matrix is

J_f(x,y)= \begin{pmatrix} 2x&1\\ y&x \end{pmatrix}.

For $(x,y)\in Q$, the square of the Frobenius norm is $4x^2+1+y^2+x^2=5x^2+y^2+1\leq7$. The Frobenius bound for matrix action therefore gives $\|Df(x,y)\|_{\mathrm{op}}\leq\sqrt{7}$. Since $Q$ is convex, the segment between any two of its points stays in $Q$. Thus, for all $p,q\in Q$,

\|f(p)-f(q)\|_2\leq\sqrt{7}\,\|p-q\|_2.

For example, $p=(1,0)$ and $q=(-1,1)$ give $f(p)=(1,0)$, $f(q)=(2,-1)$, and $\|f(p)-f(q)\|_2=\sqrt{2}$. The input distance is $\sqrt{5}$, and the estimate holds since $\sqrt{2}\leq\sqrt{35}$.

Consequences for Lipschitz Bounds and Linear Approximation

A function is Lipschitz with constant $M$ on a set if the distance between any two output values is at most $M$ times the distance between the corresponding inputs. The segment estimate gives a convenient sufficient condition: on a convex open domain, a uniform bound on the derivative gives a Lipschitz bound. Convexity matters because it keeps every segment between points of the domain inside the region where the derivative bound is known.

Corollary (Derivative Bound Gives a Lipschitz Bound): Let $U\subseteq\mathbb{R}^m$ be open and convex, and let $f:U\to\mathbb{R}^n$ be differentiable at every point. If $\|Df(z)\|_{\mathrm{op}}\leq M$ for every $z\in U$, then $$ \|f(y)-f(x)\|_2\leq M\|y-x\|_2 $$ for all $x,y\in U$.

Proof. For any $x,y\in U$, convexity implies $[x,y]\subseteq U$. The derivative bound holds at every point of that segment, so the Mean Value Estimate Along a Segment applies and gives the stated inequality. $\square$

The same method estimates how far a function differs from a chosen linear map. Fix a linear map $L:\mathbb{R}^m\to\mathbb{R}^n$, and define $g(z)=f(z)-L(z)$. Linearity gives $Dg(z)=Df(z)-L$. Applying the segment estimate to $g$ bounds the difference between the change in $f$ and the change predicted by $L$.

Corollary (Linearization Error Bound on a Segment): Suppose $f:U\to\mathbb{R}^n$ is differentiable on an open set $U$, and $[x,y]\subseteq U$. If $\|Df(z)-L\|_{\mathrm{op}}\leq\varepsilon$ for every $z\in[x,y]$, then $$ \|f(y)-f(x)-L(y-x)\|_2\leq\varepsilon\|y-x\|_2. $$

Proof. Define $g(z)=f(z)-L(z)$. Then $g$ is differentiable and $Dg(z)=Df(z)-L$. The hypothesis gives $\|Dg(z)\|_{\mathrm{op}}\leq\varepsilon$ on the segment. By the Mean Value Estimate,

\|g(y)-g(x)\|_2\leq\varepsilon\|y-x\|_2.

Since $g(y)-g(x)=f(y)-f(x)-L(y-x)$, this is the required bound. $\square$

What the Estimate Requires

The derivative bound must hold along the entire segment, not just at the endpoints. Derivatives can vary between endpoints, and the one-variable Mean Value Theorem samples a point in the interior. If a bound is available only on a smaller region, the segment must stay within that region for the estimate to apply.

Nor does differentiability by itself provide a uniform derivative bound on an arbitrary domain. The theorem explicitly assumes such a bound on the segment. A common way to obtain one is to assume that $Df$ is continuous on a compact convex set: continuity and compactness then ensure that its operator norm is bounded there. This additional continuity assumption is useful for finding a bound, but it is not needed in the segment estimate once the bound is already known.

Finally, do not infer an exact vector-valued mean value formula from the scalar theorem. Projecting onto the direction of $f(y)-f(x)$ proves the norm inequality, but different components can have their one-variable mean value points at different locations. The vector estimate is the robust conclusion needed for Lipschitz bounds and error estimates.

Check Your Understanding

Use the line-segment argument and its hypotheses to answer the following questions.

Why is the path $t\mapsto f(x+t(y-x))$ useful when estimating $f(y)-f(x)$?
In the vector-valued proof, how is the scalar function $\psi$ chosen, and what does its endpoint difference equal?
Why must the derivative bound be known along the segment rather than only at $x$ and $y$?
What role does convexity play in deriving a Lipschitz bound on a domain?
How does applying the segment estimate to $g(z)=f(z)-L(z)$ produce a linearization-error bound?

What You'll Learn