Let $U$ be a normal operator. Prove, that there exists a sequence $\{ U_n \}$ which converges to $U$ uniformly.

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I was told to use the "uniform limit theorem" here, but this theorem is just true for functions, and not operators?

What I am actually doing is to prove that $f(U)$ is normal if $f$ is a function from the disk algebra.

So what I did here was to use the spectral theorem:

$$f(U) = \int_{\sigma(U)} f(z) dE(z)$$

where $dE(z)$ is the spectral measure for $U$.

And if $f(U)$ is continuous i.e. there exists such a $\{ f_n \}$ and $\{ U_n \}$ that $f_n (U_n) \rightarrow f(U)$ uniformly, then $f(U)$ will also be a normal operator.

I already found my sequence $\{ f_n \}$ for disk algebra functions, but I am having trouble finding a sequence $\{ U_n \}$

Any ideas?