MEAN VALUE INEQUALITY
1. Mean Value Inequality
Theorem 1.1. Let B(p, R) be the open ball centered at p of radius R > 0 in Rn and f : B(p, R) →
Rm be a C 1 -function. Suppose that kDf (x)k ≤ M for any x ∈ B(p, R) for some M > 0. Then
kf (x) − f (y)kRm ≤ M kx − ykRn .
Proof. When f (x) = f (y), the above inequality is obviously true. We will prove the inequality in
the case when f (x) 6= f (y).
Let x and y be two points of B(p, R). Then kx−pk < R and ky−pk < R. Define γ(t) = tx+(1−t)y
for t ∈ [0, 1]. Then γ(0) = y and γ(1) = x. By triangle inequality and the property of norms,
kγ(t) − pkRn = ktx + (1 − t)y − pkRn = kt(x − p) + (1 − t)(y − p)kRn
≤ tkx − pkRn + (1 − t)ky − pkRn
< tR + (1 − t)R = R
for any t ∈ [0, 1]. We see that γ(t) ∈ B(p, R) for any t ∈ [0, 1]. Let u be a vector in Rn . Define a
function gu : [0, 1] → R by
gu (t) = hf (γ(t)), uiRm , t ∈ [0, 1].
Then gu (0) = hf (y), uiRm and gu (1) = hf (x), uiRm and
gu0 (t) = hDf (γ(t))(γ 0 (t)), uiRm = hDf (tx + (1 − t)y)(x − y), uiRm .
By mean value theorem, there exists c ∈ [0, 1] so that
gu (1) − gu (0) = gu0 (c).
Since
gu (1) − gu (0) = hf (x), uiRm − hf (y), uiRm = hf (x) − f (y), uiRm ,
and
gu0 (c) = hDf (cx + (1 − c)y)(x − y), uiRm ,
we find that
hf (x) − f (y), ui = hDf (cx + (1 − c)y)(x − y), uiRm .
By the Cauchy-Schwarz inequality and the norm inequality,
|hDf (cx + (1 − c)y)(x − y), uiRm | ≤ kDf (cx + (1 − c)y)(x − y)kRm kukRm
≤ kDf (cx + (1 − c)y)kkx − ykRn kukRm
≤ M kx − ykRn kukRm .
Taking u = (f (x) − f (y))/kf (x) − f (y)kRm , we find that kukRm = 1 and
f (x) − f (y)
hf (x) − f (y), uiRm = f (x) − f (y),
kf (x) − f (y)kRm Rm
hf (x) − f (y), f (x) − f (y)iRm
=
kf (x) − f (y)kRm
kf (x) − f (y)k2Rm
=
= kf (x) − f (y)kRm .
kf (x) − f (y)kRm
When u = (f (x) − f (y))/kf (x) − f (y)kRm , the inequality
|hf (x) − f (y), uiRm | ≤ M kx − ykRn kukRm
1
2
MEAN VALUE INEQUALITY
is equivalent to
kf (x) − f (y)kRm ≤ M kx − ykRn .
In the proof, we only use the fact that the line segment connecting x and y is contained in B(p, R).
Let us introduce the notion of convexity of a set in a real vector space.
Definition 1.1. Let V be a real vector space and C be any nonempty subset of V. We say that C
is convex if for any x, y ∈ C, the line segment
xy = {tx + (1 − t)y ∈ V : t ∈ [0, 1]}
is contained in C.
One can easily see that Theorem 1.1 also holds when we replace the open ball B(p, R) by any
open convex set U.