How can I show that a twice continuously differentiable function with a lipschitz continuous hessian with all eigenvalues≥mu is mu strongly convex?
How can I show that a twice continuously differentiable function with a lipschitz continuous hessian with all eigenvalues≥mu is mu strongly convex?
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Question
How can I show that a twice continuously

Transcribed Image Text:Definition 2.2. A function f : R" → R is said to be u-strongly convex if
38
Vx, y E R" :
f (x) + (Vf(x), y – x) + y – x||² < f(y)
(2.4)
![Definition (Lipschitz continuity)
Suppose that
• X and y open sets
• F: X →Y
• || · ||x and || · ||y are norms
Then
• F is Lipschitz continuous at x E X if 3 y(x) such that
|F(z) - F(x) ||ly < Y(x) ||z-제1x
for all z E X.
• F is Lipschitz continuous throughout/in X if 3 y such that
||F(2) – F(x)||» < l|z – x|| x
for all x and z E X.
Theorem (Taylor approximations for real-valued functions)
Let S be an open subset of R", s E R", and suppose that f : S → R is continuously
differentiable throughout S and g = Vf is Lipschitz continuous at x with Lipschitz
constant y (x) for some appropriate vector norm. It follows that if the segment
[x, x + s] E S, then
\F(x + s) – m“ (x + s)| < (x)|||°,
where
m" (x + s) = f(x) +g(x)*s.
If in addition, f is twice continuously differentiable throughout S and H =
Lipschitz continuous at x, with Lipschitz constant y (x), then
= V²f is
F(x + s) – mº (x + s)| < e (x)||s||°,
where
m° (x + s) = f(x) + 8(x)"s + }s'H(x)s.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F39305326-b178-41ef-90ae-6f13f00f1fdf%2F80cb5871-9a52-4f96-a13a-7014fe152e06%2Fmiksjw9_processed.png&w=3840&q=75)
Transcribed Image Text:Definition (Lipschitz continuity)
Suppose that
• X and y open sets
• F: X →Y
• || · ||x and || · ||y are norms
Then
• F is Lipschitz continuous at x E X if 3 y(x) such that
|F(z) - F(x) ||ly < Y(x) ||z-제1x
for all z E X.
• F is Lipschitz continuous throughout/in X if 3 y such that
||F(2) – F(x)||» < l|z – x|| x
for all x and z E X.
Theorem (Taylor approximations for real-valued functions)
Let S be an open subset of R", s E R", and suppose that f : S → R is continuously
differentiable throughout S and g = Vf is Lipschitz continuous at x with Lipschitz
constant y (x) for some appropriate vector norm. It follows that if the segment
[x, x + s] E S, then
\F(x + s) – m“ (x + s)| < (x)|||°,
where
m" (x + s) = f(x) +g(x)*s.
If in addition, f is twice continuously differentiable throughout S and H =
Lipschitz continuous at x, with Lipschitz constant y (x), then
= V²f is
F(x + s) – mº (x + s)| < e (x)||s||°,
where
m° (x + s) = f(x) + 8(x)"s + }s'H(x)s.
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