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Posts Tagged ‘uniformly’

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

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.

Categories: Analysis Tags: ,

Traian Lalescu student contest 2011 Problem 3

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

Categories: Analysis, Problem Solving

Convex functions limit Traian Lalescu 2010

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

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

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

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 $E\setminus A_\varepsilon$.
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…