Let {fn} be a sequence of continuous functions which converges uniformly to a function f on a set E. Show that for xn ∈ E such that xn → x ∈ E, it follows that fn(xn) → f(x). Is the converse of this true?
Added by James L.
Step 1
We are given that the sequence of functions {fn} converges uniformly to a function f on a set E. This means that for any ε > 0, there exists an N such that for all n ≥ N and for all x ∈ E, we have |fn(x) - f(x)| < ε. Show more…
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A sequence of functions {fn} defined on a set E is said to be equicontinuous on E if for every ε > 0, there exists a δ > 0 such that |fn(x) - fn(y)| < ε whenever |x - y| < δ, x ∈ E, y ∈ E and n ∈ N. Prove that, if an equicontinuous sequence of functions {fn} converges pointwise to f on a set E, then f is uniformly continuous on E.
Adi S.
(a) Suppose that $\left\{f_{n}\right\}$ is a sequence of continuous functions on $[a, b]$ that converges uniformly to $f .$ Prove that if $x_{n}$ approaches $x,$ then $f_{n}\left(x_{n}\right)$ approaches $f(x)$ (b) Is this statement true without assuming that the $f_{n}$ are continuous? (c) Prove the converse of part (a): If $f$ is continuous on $[a, b]$ and $\left\{f_{n}\right\}$ is a sequence with the property that $f_{n}\left(x_{n}\right)$ approaches $f(x)$ whenever $x_{n}$ approaches $x,$ then $f_{n}$ converges uniformly to $f$ on $[a, b]$ Hint: If not, there is an $\varepsilon>0$ and a sequence $x_{n}$ with $| f_{n}\left(x_{n}\right)-$ $f\left(x_{n}\right) |>\varepsilon .$ Then use the Bolzano-Weierstrass theorem.
Madhur L.
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