A compact operator with no eigenvalues. A compact operator with no eigenvalues.

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Let X = C([0,1]), v:X® X be the Volterra operator v(f)(x) = ò0x f(t)dt. If S is the closed unit ball of X, then v(S) is equicontinuous and pointwise-bounded, hence by the Arzela-Ascoli theorem, v is compact. If for some l Î C and f ¹ 0 in X, vf = lf, then f(x) = lf¢(x). So l ¹ 0 and ln f(x) = [ x/( l)]+c for some c Î Â. Hence f(x) = f(0)e[ x/( l)] = 0e[ x/( l)] = 0, a contradiction. v has then no eigenvalue.


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On 22 Feb 2001, 00:20.