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Posts Tagged ‘uniformly’

Approximation by polynomials of a function defined on the entire real line

April 27, 2012 Leave a comment

Suppose {f: \Bbb{R}\rightarrow \Bbb{R}} is a function which can be uniformly approximated by polynomials on {\Bbb{R}}. Then {f} is also a polynomial.

Note the striking difference between this result and the Weierstrass approximation theorem.

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Categories: Analysis Tags: ,

Traian Lalescu student contest 2011 Problem 3

May 14, 2011 Leave a comment

For a continuous function f : [0,\infty) \to [0,\infty) such that \displaystyle \int_0^\infty f(x)dx <\infty prove that if f is uniformly continuous, then it is bounded. Prove also that the converse of the previous statement is not true.

Traian Lalescu student contest 2011

Convex functions limit Traian Lalescu 2010

May 17, 2010 2 comments

Suppose that the sequence of convex functions (f_n) converges pointwise to 0 on [0,1]. Prove that it converges uniformly to 0 on [0,1].
Traian Lalescu student contest 2010, Iasi, Romania

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Uniform limit of polynomials

January 11, 2010 1 comment

Suppose that the sequence of polynomials (p_n) converges uniformly on [0,1] to a function f which is not a polynomial function. Prove that the sequence a_n=\deg p_n is unbounded.
J. Marsden, Elementary Classical Analysis
It is possible that the pointwise convergence siffice for the purpose of the problem.

This means that the space \mathcal{P}_n=\{ p \ \text{polynomial}\ : \deg p \leq n\} is closed if we take the uniform convergence norm \| f-g \| =sup_{x \in [0,1]} |f(x)-g(x)|. Read more…

Measurable function again

January 7, 2010 2 comments

Suppose f:[a,b]\to \mathbb{R} is a measurable function. Prove that there exists a sequence of polynomials (P_n) such that P_n \to f almost uniformly, which means that for any \varepsilon >0, we can find a set A_\varepsilon with measure smaller than \varepsilon such that P_n \to f uniformly on [a,b]\setminus A_\varepsilon.
Hint: Use Lusin’s theorem and Weierstrass’ approximation theorem, for continuous functions.

Egorov’s Theorem

January 6, 2010 Leave a comment

Suppose (f_k) is a sequence of measurable functions defined on a measurable set E with m(E)<\infty , we can find a closed set A_\varepsilon \subset E such that m(E\setminus A_\varepsilon)\leq \varepsilon and f_k \to f uniformly on E\setminus A_\varepsilon.

This theorem states that in spaces with finite measure, almost everywhere convergence is equivalent to almost uniformly convergence.

Barbalat’s Lemma

October 1, 2009 5 comments

Prove that if we have f: [0,\infty) \to [0,\infty) a function which is uniformly continuous on [0,\infty) with \int_0^\infty f(t)\text{d}t <\infty then \lim\limits_{t\to \infty} f(t)=0. Read more…

Categories: Analysis Tags: ,
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