Insights Blog
-- Browse All Articles --
Physics Articles
Physics Tutorials
Physics Guides
Physics FAQ
Math Articles
Math Tutorials
Math Guides
Math FAQ
Education Articles
Education Guides
Bio/Chem Articles
Technology Guides
Computer Science Tutorials
Forums
Intro Physics Homework Help
Advanced Physics Homework Help
Precalculus Homework Help
Calculus Homework Help
Bio/Chem Homework Help
Engineering Homework Help
Trending
Featured Threads
Log in
Register
What's new
Search
Search
Search titles only
By:
Intro Physics Homework Help
Advanced Physics Homework Help
Precalculus Homework Help
Calculus Homework Help
Bio/Chem Homework Help
Engineering Homework Help
Menu
Log in
Register
Navigation
More options
Contact us
Close Menu
JavaScript is disabled. For a better experience, please enable JavaScript in your browser before proceeding.
You are using an out of date browser. It may not display this or other websites correctly.
You should upgrade or use an
alternative browser
.
Forums
Homework Help
Calculus and Beyond Homework Help
Uniform convergence of a sequence of functions
Reply to thread
Message
[QUOTE="AllRelative, post: 6073388, member: 540873"] [h2]Homework Statement[/h2] This is a translation so sorry in advance if there are funky words in here[/B] f: ℝ→ℝ a function 2 time differentiable on ℝ. The second derivative f'' is bounded on ℝ. Show that the sequence on functions $$ n[f(x + 1/n) - f(x)] $$ converges uniformly on f'(x) on ℝ. [h2]Homework Equations[/h2] In my attempt I used the definition of differentiation: $$\lim_{x\to\infty} \frac{f(x+h) - f(x)}{h} $$ and I used a criteria for uniform convergence of sequences of functions: f[SUB]n[/SUB] fonverges to f uniformely on A if and on if for all ε > 0, ∃ N ∈ ℕ, for which $$\lim_{x\to\infty} sup|fn - f(x)| \leqslant \varepsilon$$ for all n≥ N, for all x ∈ A [h2]The Attempt at a Solution[/h2] [/B] I arrived to an answer but I fear I got sidetracked somewhere because I did not use the bounded second derivative. I rewrote $$ n[f(x + 1/n) - f(x)] = \frac{[f(x + 1/n) - f(x)]}{1/n} $$ Now this looks awefully like the derivative of f[SUB]n[/SUB] for all x which is: $$\lim_{n\to\infty} \frac{[f(x + 1/n) - f(x)]}{1/n}$$ And now I applied the definition of the uniform convergence which is: $$\lim_{n\to\infty} sup| \frac{[f(x + 1/n) - f(x)]}{1/n} - f'(x)| \leqslant \varepsilon $$ And therefore, I proved the uniform convergence to f'(x) on ℝ. (I am missing a few for all ε belonging to...and stuff, I just wanted to write it quickly) Thank you [/QUOTE]
Insert quotes…
Post reply
Forums
Homework Help
Calculus and Beyond Homework Help
Uniform convergence of a sequence of functions
Back
Top