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) = ∫_{0}^{x} 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 λ ∈ C and f ≠ 0 in X, vf = λf, then f(x) = λf′(x). So λ ≠ 0 and lnf(x) = \frac{x}{λ} + c for some c ∈ R. Hence f(x) = f(0)e\frac{x}{λ} = 0e\frac{x}{λ} = 0, a contradiction. v has then no eigenvalue.