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Analysis Theorems

5 Series

  • Lemma 5.1 (Raabe’s Test). For the series \(\sum _0^\infty c_n\), \(c_i \in \RR \) if \(R_n = n(1-\frac {c_{n+1}}{c_n})\) is larger than \(R>1\) for sufficiently large \(n\), the series is absolutely convergent.

  • Proof. WLOG \(c_n \geq 0\). Then note that \(R_n-1 = \frac {1}{c_n}(c_n(n-1)-c_{n+1}n)\) so \(0 \leq c_n = \frac {c_n(n-1)-c_{n+1}n}{R_n-1}\). By hypothesis \(R_n-1\) stays away from \(0\) for large \(n\) so it suffices to show \(c_n(n-1)-c_{n+1}n\), which is positive for large \(n\), converges. But \(\sum _1^mc_n(n-1)-c_{n+1}n = -c_{m+1}m\) is monotonically increasing for large \(m\) and bounded above by \(0\) so converges to a limit. □

  • Proposition 5.2. Every power series \(\sum _0^\infty c_nz^n\), \(c_n \in \CC \) has a radius of convergence \(R\) given by Hadamard’s formula \(\frac {1}{R} = \limsup |c_n|^{\frac {1}{n}}\)

  • Proof. For any \(r<R\) and large \(n\) we have \(|c_n|^{\frac {1}{n}}< \frac {1}{r}\) so if \(|z|<r<R\) we have \(\sum _n^\infty |c_n|z^n < \sum _n^\infty \frac {z^n}{r^n}\) so we have absolute convergence. Similarly if \(|z|>r>R\) there are infinitely many terms so that \(|c_n|\geq \frac {1}{r^n}\) so the series diverges. □

  • Proposition 5.3. Every power series \(\sum _0^\infty c_nz^n\) is holomorphic in its radius of convergence, and its derivative, \(\sum _0^\infty nc_nz^{n-1}\) has the same radius of convergence.

  • Proof. First note by Proposition 5.2 since \(n^{1/n} \to 1\) as \(n \to \infty \) we have that \(\sum _0^\infty nc_nz^{n-1}\) has the same radius of convergence.

    Let \(z_0,z_0+h\) lie in radius \(r\), which is less than the radius of convergence.

    \[\left |\frac {\sum _0^\infty c_j(z_0+h)^j-\sum _0^\infty c_jz_0^j}{h}-\sum _0^\infty c_jjz_0^{j-1}\right |\]

    \[= |\ee _n(h)| + \left |\frac {\sum _{n+1}^\infty c_j(z_0+h)^j-\sum _{n+1}^\infty c_jz_0^j}{h}\right |+|\sum _{n+1}^\infty c_jjz_0^{j-1}|\]

    where \(\ee _n(h)\) is the error from the \(n^{th}\) partial sum’s approximation by its derivative. Since \(a^j-b^j = (a-b)(a^{j-1}+a^{j-2}b+\dots +b^{j-1})\) we have \(|\frac {(z_0+h)^j-z_0^j}{h}|\leq jr^{j-1}\). For any \(\ee >0\), we can choose \(n\) large enough so that \(\sum _{n+1}^\infty jc_jr^{j-1}<\ee \). Afterwards we can choose \(h\) small enough so \(\ee _n(h)<\ee \). Then continuing from above we get

    \[\leq \ee + \ee + \ee = 3\ee \]

    concluding the proof. □

  • Proposition 5.4 (Abel’s Theorem). If a power series \(f(x) = \sum _0^\infty c_ix^i\) converges at an endpoint of its radius of convergence \(R\), it converges uniformly and hence is continuous on \([0,R]\).

  • Proof. WLOG,\(R = 1\). Let \(A_n = \sum _0^{n-1}c_i\). We have then for \(x \in [0,1]\)

    \[ |\sum _n^mc_ix^i|=|\sum _n^mA_{i+1}x^i-\sum _n^mA_ix^i|=|\sum _n^m(f(1)-A_{i+1})x^i-\sum _n^m(f(1)-A_i)x^i|\]

    \[ = |\sum _{n+1}^{m+1}(f(1)-A_i)x^{i-1}-\sum _n^m(f(1)-A_i)x^i |\]

    \[= |\sum _{n+1}^{m}(f(1)-A_i)(x^{i-1}-x^i)-(f(1)-A_i)x^n+(f(1)-A_i)x^{m}| \]

    now choosing \(N\) large enough so \(|f(1)-A_i|<\ee \) for \(i>n\) and noting \(x \in [0,1]\)

    \[\leq \sum _{n+1}^{m}|(f(1)-A_i)|(x^{i-1}-x^i)+|(f(1)-A_i)|x^n+(f(1)-A_i)x^{m}|\]

    \[\leq (x^n-x^m)\ee +x^n\ee +x^m\ee \leq 3\ee \]

    and we are done as the partial sums are uniformly Cauchy. □

  • Proposition 5.5. If \(f(x) = \sum _0^\infty c_iz^i\) converges within radius \(R\), and \(a\) is within this radius, \(f(a+x)\) converges and is analytic within radius of \(R-|a|\).

  • Proof. Inside \(|z|<R-|a|\) the power series absolutely converges, and we would like to recenter around 0. WLOG \(a\) is real and non-negative. For \(0<x<R-|a|\) we have

    \[ \sum _0^\infty c_i(a+x)^i = \sum _0^\infty c_i \sum _{j=0}^i\binom i j a^{i-j}x^j \]

    We can then split the inner sum up into separate terms, still absolutely convergent as everything is non-negative, and then we can collect like powers of \(x\) to get

    \[\sum _0^\infty c_i(a+x)^i = \sum _0^\infty \Big (\sum _m^\infty c_i\binom i m a^{i-m}\Big )x^m\]

    so \(f(a+z)\) has a power series that converges on the interval \((0,R-|a|)\). Hence its radius of convergence is at least \(R-|a|\). □

  • Lemma 5.6. The composite of two \(\FF \)-analytic functions \(\FF \to \FF \) is analytic.

  • Proof. Real analytic functions are restrictions of complex analytic ones, and by the chain rule for holomorphic functions, the composite is analytic. □

  • Lemma 5.7. \((1+z)^a\) is holomorphic within radius \(1\) around the origin, its power series given by \(\sum _0^\infty \binom a j z^j\). It converges absolutely and uniformly on the interval \([-1,1]\) when \(a>0\).

  • Proof. That this is analytic follows from Lemma 5.6, and that it is holomorphic of radius \(1\) comes from Hadamard’s formula. Now when \(z \in [-1,1]\), for \(n>a>0\) we have \(n\bigg (1-\Big |\frac {\binom a {n+1}}{\binom a n}z\Big |\bigg ) \geq n\bigg (1-\Big |\frac {\binom a {n+1}}{\binom a n}\Big |\bigg ) = n\big (1-\frac {n-a}{n+1}\big ) = \frac {n(1+\alpha )}{n+1}\) which is larger than \(1\) for sufficiently large \(n\) so by Lemma 5.1 and Theorem 3.17 we have the last statement. □

  • Lemma 5.8. The function \(|x|\) can be uniformly approximated by polynomials in a bounded interval.

  • Proof. WLOG the interval is \([-1,1]\). By Lemma 5.7 \((1-t)^{1/2}\) is uniformly approximated by its Taylor expansion in this interval, and we can set \(t = 1-x^2\) we get \((x^2)^{1/2} = |x|\) is as well. □

  • Theorem 5.9 (Stone-Weierstrass). If \(S\) is compact, any \(\RR \)-subalgebra of \(\cC (S,\RR )\) that separates points is dense.

  • Proof. Let \(A\) be a \(\RR \)-subalgebra of \(\cC (S,\RR )\) that separates points. If \(f \in \bar {A}\), then \(|f| \in \bar {A}\) as \(f\) has a bounded image so by Lemma 5.8 \(|f|\) can be approximated uniformly by polynomials in \(f\), which are in \(\bar {A}\).

    As a consequence, \(\max (f,g), \min (f,g) \in \bar {A}\) whenever \(f,g\) are as they are linear combinations of \(f,g,|f|,|g|\).

    Now for any \(\ee >0,h \in \cC (S,\RR )\) we can approximate \(h\) up to \(\ee \) as follows: for each \(a \in S\) choose \(f_{a,b}\) for each \(b\) so that \(f(a) = h(a),f(b) = h(b)\). Then in a small neighborhood of \(b\), \(f_{a,b}(x)>h(x)-\ee \), but \(S\) is compact so finitely many such neighborhoods suffice to have \(f_a = \max _{b_i}f_{a,b_i}\) be at least \(h(x)-\ee \) everywhere. Now we can similarly choose small neighborhoods for each \(a\) so that \(f_a < h(x)+\ee \), and once again finitely many suffice so that \(\min _{a_i}{f_{a_i}}\) is our desired approximation. □

There is an analogous version for \(\cC (S,\CC )\).

  • Corollary 5.10. If \(S\) is compact, any \(C^*\)-subalgebra of \(\cC (S,\CC )\) that separates points is dense.

  • Proof. Let \(A\) again be such a subalgebra. As \(A\) is closed under conjugation, it contains the real and imaginary parts of any element \(f\). Then by Theorem 5.9 the real and imaginary parts of any function \(h\) are in \(\bar A\) so \(h\) is as well. □