Choose Γ to be a circle of radius greater than the operator norm of "T". As stated above, on such Γ, the resolvent map admits a power series representation  Substituting gives  which is   The δ is the Kronecker delta symbol. For any "f"1 and "f"2 satisfying the appropriate assumptions, the homomorphism property states  We sketch an argument which invokes the first resolvent formula and the assumptions placed on "f". First we choose the Jordan curves such that Γ1 lies in the "inside" of Γ2. The reason for this will become clear below. Start by calculating directly  The last line follows from the fact that ω ∈ Γ2 lies outside of Γ1 and "f"1 is holomorphic on some open neighborhood of σ("T") and therefore the second term vanishes. Therefore, we have:  Let "G" ⊂ C be open with σ("T") ⊂ "G". Suppose a sequence {"fk"} of holomorphic functions on "G" converges uniformly on compact subsets of "G" (this is sometimes called "compact convergence"). Then {"fk"("T")} is convergent in "L"("X"):  Assume for simplicity that Γ consists of only one Jordan curve. We estimate  By combining the uniform convergence assumption and various continuity considerations, we see that the above tends to 0 as "k", "l" → ∞. So {"fk"("T")} is Cauchy, therefore convergent. To summarize, we have shown the holomorphic functional calculus, "f" → "f"("T"), has the following properties:   It can be proved that a calculus satisfying the above properties is unique. We note that, everything discussed so far holds verbatim if the family of bounded operators "L"("X") is replaced by a Banach algebra "A". The functional calculus can be defined in exactly the same way for an element in "A". It is known that the spectral mapping theorem holds for the polynomial functional calculus: for any polynomial "p", "σ"("p"("T")) = "p"("σ"("T")). This can be extended to the holomorphic calculus. To show "f"("σ"("T")) ⊂ "σ"("f"("T")), let μ be any complex number.