Let {fn}1 be a sequence of continous functions on an interval [a, b]. If {fn}1 converges uniformly to a function f on [a, b], then f is continous on [a, b]. (a) (i) Discuss the convergence of the...


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Let {fn}1 be a sequence of continous functions on an interval [a, b]. If {fn}1 converges<br>uniformly to a function f on [a, b], then f is continous on [a, b].<br>(a) (i) Discuss the convergence of the sequence of functions fn, when fn(x)<br>1+ xn<br>where r E [0, 1].<br>(ii) Show that the sequence {fn}, where fn(x) =<br>1<br>is not uniformly convergent<br>1+ xn<br>on r E (0, 1].<br>(b) Let fn(x) =<br>is not uniform.<br>14 „2n X E [0, 1]. Use the above theorem to show that the convergence<br>(c) Is the converse of the above theorem necessary true? Justify your answer using the<br>sequence fn(x) = nxe-n, x E [0, 1].<br>

Extracted text: Let {fn}1 be a sequence of continous functions on an interval [a, b]. If {fn}1 converges uniformly to a function f on [a, b], then f is continous on [a, b]. (a) (i) Discuss the convergence of the sequence of functions fn, when fn(x) 1+ xn where r E [0, 1]. (ii) Show that the sequence {fn}, where fn(x) = 1 is not uniformly convergent 1+ xn on r E (0, 1]. (b) Let fn(x) = is not uniform. 14 „2n X E [0, 1]. Use the above theorem to show that the convergence (c) Is the converse of the above theorem necessary true? Justify your answer using the sequence fn(x) = nxe-n, x E [0, 1].

Jun 04, 2022
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