Banach space

1. A complex vector space \(X\) is said to be a normed linear space if to each \(x\in X\), there is associated a nonnegative real number \(\lVert x\rVert\), called the norm of \(x\), such that
   

1. \(\lVert x+y\rVert\leq\lVert x\rVert+\lVert y\rVert\) for all \(x\) and \(y\in X\),
   
2. \(\lVert ax\rVert=|a|\lVert x\rVert\) if \(x\in X\) and \(a\) is a scalar.
   
3. \(\lVert x\rVert=0\) implies \(x=0\).
   Every normed linear space may be regarded as a metric space, the distance between \(x\) and \(y\) being \(\lVert x-y\rVert\).
   A Banach space is a normed linear space which is complete in the metric defined by its norm. For instance, every Hilbert space is a Banach space, so is every \(L^p(\mu)\) normed by \(\lVert f\rVert_p\) if \(1\leq p\leq\infty\), and so is \(C_0(X)\) with the supremum norm.

2. Consider a linear transformation \(\Lambda\) from a normed linear space \(X\) into a normed linear space \(Y\), and define its norm by \[\lVert \Lambda\rVert=\sup\;\{\lVert \Lambda x\rVert:x\in X,\lVert x\rVert\leq1\}\] If \(\lVert \Lambda\rVert<\infty\), then \(\Lambda\) is called a bounded linear transformation.

3. For a linear transformation \(\Lambda\) of a normed linear space \(X\) into a normed linear space \(Y\), each of the following three conditions implies the other two:

1. \(\Lambda\) is bounded.
   
2. \(\Lambda\) is continuous.
   
3. \(\Lambda\) is continuous at one point of \(X\).
   Since \(\lVert \Lambda(x_1-x_2)\rVert\leq\lVert \Lambda\rVert\lVert x_1-x_2\rVert\), it is clear that (a) implies (b), and (b) implies (c). Suppose \(\Lambda\) is continuous at \(x_0\). To each \(\epsilon>0\) one can then find a \(\delta>0\) so that \(\lVert x-x_0\rVert<\delta\) implies \(\lVert \Lambda x-\Lambda x_0\rVert<\delta\). In other words, \(\lVert x\rVert\leq\delta\) implies \[\lVert \Lambda (x_0+x)-\Lambda x_0\rVert<\epsilon\] But then the linearity of \(\Lambda\) shows that \(\lVert\Lambda x\rVert<\epsilon\). Hence \(\lVert\Lambda\rVert\leq\frac{\epsilon}{\delta}\) and (c) implies (a).

4. Baire’s Theorem If \(X\) is a complete metric space, the intersection of every countable collection of dense open subsets of \(X\) is dense in \(X\). \(\Biggl(\) A subset \(A\) of a topological space \(X\) is called dense (in \(X\)) if every point \(x\) in \(X\) either belongs to \(A\) or is a limit point of \(A\); that is, the closure of \(A\) is constituting the whole set \(X\).\(\Biggl)\)
   Suppose \(V_1,V_2,V_3,\cdots\) are dense and open in \(X\). Let \(W\) be any open set in \(X\). We have to show that \(\bigcap V_n\) has a point in \(W\) if \(W\ne\varnothing\). Let \(\rho\) be the metric of \(X\); let us write \[S(x,r)=\{y\in X:\rho(x,y)<r\}\] and let \(\overline{S}(x,r)\) be the closure of \(S(x,r)\).
   Since \(V_1\) is dense, \(W\cap V_1\) is a nonempty open set, and we can therefore find \(x_1\) and \(r_1\) such that \[\overline{S}(x_1,r_1)\subset W\cap V_1,\quad 0<r_1<1\]
   If \(n\ge2\) and \(x_{n-1}\) and \(r_{n-1}\) are chosen, the denseness of \(V_n\) shows that \(V_n\cap S(x_{n-1},r_{n-1})\) is not empty and we can find \(x_n\) and \(r_n\) such that \[\overline{S}(x_n,r_n)\subset V_n\cap S(x_{n-1},r_{n-1}),\quad 0< r_n<\frac{1}{n}\]
   By induction, this process produces a sequence \(\{x_n\}\) in \(X\). If \(i>n\) and \(j>n\), the construction shows that \(x_i\) and \(x_j\) both lie in \(S(x_n,r_n)\), so that \(\rho(x_i,x_j)<2r_n<2/n\), and hence \(\{x_n\}\) is a Cauchy sequence. Since \(X\) is complete, there is a point \(x\in X\) such that \[x=\lim_{n\to\infty}x_n\]
   Since \(x_i\) lies in the closed set \(\overline{S}(x_n,r_n)\) if \(i>n\), it follows that \(x\) lies in each \(\overline{S}(x_n,r_n)\) and \[\overline{S}(x_n,r_n)\subset V_n\cap S(x_{n-1},r_{n-1}),\quad 0< r_n<\frac{1}{n}\] shows that \(x\) lies in each \(V_n\). By \[\overline{S}(x_1,r_1)\subset W\cap V_1,\quad 0<r_1<1\] \(x\in W\).
   In a complete metric space, the intersection of any countable collection of dense \(G_{\delta}\)’s is again a dense \(G_{\delta}\). This follows from the theorem, since every \(G_{\delta}\) is the intersection of a countable collection of open sets, and since the union of countably many countable sets is countable.

5. Baire’s theorem is sometimes called the Baire’s category Theorem.
   Call a set \(E\subset X\) nowhere dense if its closure \(\overline{E}\) contains no nonempty open subset of \(X\). Any countable union of nowhere dense sets is called a set of the first category; all other subsets of \(X\) are of the second category.
   Baire’s category Theorem is equivalent to the statement that no complete metric space is of the first category.

6. The Banach-Steinhaus Theorem (uniform boundedness principle) Suppose \(X\) is a Banach space, \(Y\) is a normed linear space, and \(\{\Lambda_a\}\) is a collection of bounded linear transformations of \(X\) into \(Y\), where \(a\) ranges over some index set \(A\). Then either there exists an \(M<\infty\) such that \[\lVert \Lambda_a\rVert\leq M,\quad \forall a\in A\] or \[\sup_{a\in A}\lVert \Lambda_ax\rVert=\infty\] for all \(x\) belonging to some dense \(G_{\delta}\) in \(X\).
   In geometric terminology: Either there is a ball \(B\) in \(Y\) (with radius \(M\) and center at \(0\)) such that every \(\Lambda_a\) maps the unit ball of \(X\) into \(B\), or there exist \(x\in X\) (in fact, a whole dense \(G_{\delta}\) of them) such that no ball in \(Y\) contains \(\Lambda_ax\) for all \(a\).
   For all \(y_1,y_2\in Y\), by the triangle inequality \[\lVert y_1\rVert\leq\lVert y_1-y_2\rVert+\lVert y_2\rVert\] \[\lVert y_1\rVert-\lVert y_2\rVert\leq\lVert y_1-y_2\rVert\] and if we interchange \(y_1\) and \(y_2\), we see \[|\lVert y_1\rVert-\lVert y_2\rVert|\leq\lVert y_1-y_2\rVert\] Thus \(y\to\lVert y\rVert\) is uniformly continuous mapping.
   Since each \(\lVert \Lambda_a\rVert\) is continuous and the norm of \(Y\) is a continuous function on \(Y\), each function \(x\to\lVert \Lambda_ax\rVert\) is continuous on \(X\). Hence \(\displaystyle\sup_{a\in A}\lVert \Lambda_ax\rVert\) is lower semicontinuous. Let \[V_n=\Bigl\{x:\sup_{a\in A}\lVert \Lambda_ax\rVert>n\Bigr\},\quad(n=1,2,3,\cdots)\] then each \(V_n\) is open.
   If one of these sets, say \(V_N\), fails to be dense in \(X\), then there exist a \(x_0\in X\) and a \(r>0\) such that \(\lVert x\rVert\leq r\) implies \(x_0+x\notin V_N\), this means that \[\sup_{a\in A}\lVert \Lambda_a(x_0+x)\rVert\leq N\] or \[\lVert \Lambda_a(x_0+x)\rVert\leq N\] for all \(a\in A\) and all \(x\) with \(\lVert x\rVert\leq r\). Since \(x=(x_0+x)-x_0\), then \[\lVert \Lambda_ax\rVert=\lVert \Lambda_a\rVert\lVert x\rVert\leq\lVert \Lambda_a(x_0+x)\rVert+\lVert \Lambda_ax_0\rVert\leq2N\] and then \[\lVert \Lambda_a\rVert\leq \frac{2N}{r}\] holds with \(M=\frac{2N}{r}\)
   The other possibility is that every \(V_n\) is dense in \(X\). In that case, by Baire’s theorem \(\bigcap V_n\) is a dense \(G_{\delta}\) in \(X\), and since \(\sup_{a\in A}\lVert \Lambda_ax\rVert=\infty\) for every \(x\in \bigcap V_n\), the proof is complete.

7. The Open Mapping Theorem Let \(U\) and \(V\) be the open unit balls of the Banach spaces \(X\) and \(Y\). To every bounded linear transformation \(\Lambda\) of \(X\) onto \(Y\) there corresponds a \(\delta>0\) so that \[\Lambda(U)\supset\delta V\]
   It follows from \(\Lambda(U)\supset\delta V\) and the linearity of \(\Lambda\) that the image of every open ball in \(X\), with center at \(x_0\), contains an open ball in \(Y\) with center at \(\Lambda x_0\). Hence the image of every open set is open. This explains the name of the theorem. Here is another way of stating: To every \(y\) with \(\lVert y\rVert<\delta\) there corresponds a \(x\) with \(\lVert x\rVert<1\) so that \(\Lambda x=y\).
   Given \(y\in Y\), there exist an \(x\in X\) such that \(\Lambda x=y\), if \(\lVert x\rVert<k\), it follows that \(y\in \Lambda(kU)\). Hence \(Y\) is the union of the sets \(\Lambda(kU),\quad k=1,2,3,\cdots\), since \(Y\) is complete, the Baire theorem implies that there is a nonempty open set \(W\) in the closure of some \(\Lambda(kU)\). This means that every point of \(W\) is the limit of a sequence \(\{\Lambda x_i\},\quad x_i\in kU\), from now on \(k\) and \(W\) are fixed.
   Choose \(y_0\in W, \eta>0\) so that \(y_0+y\in W\) if \(\lVert y\rVert<\eta\). For any such \(y\) there are sequences \(\{x_i'\}\), \(\{x_i''\}\) in \(kU\) such that \[\Lambda x_i'\to y_0,\quad \Lambda x_i''\to y_0+y\quad (i\to\infty)\] Setting \(x_i=x_i''-x_i'\), we have \(\lVert x_i\rVert\leq2k\) and \(\Lambda x_i\to y\). Since this holds for every \(y\) with \(\lVert y\rVert<\eta\), the linearity of \(\Lambda\) shows that the following is true,
   Since \(\lVert x\rVert<k\) and \(\lVert y\rVert<\eta\) then to each \(y\in Y\) and to each \(\epsilon>0\) there corresponds an \(x\in X\) such that \[\lVert x\rVert\leq\frac{2k}{\eta}\lVert y\rVert,\quad \lVert y-\Lambda x\rVert<\epsilon\] Fix \(y\in \frac{\eta}{2k} V\), and fix \(\epsilon>0\). By \(\lVert x\rVert\leq\frac{2k}{\eta}\lVert y\rVert,\quad \lVert y-\Lambda x\rVert<\epsilon\), there exists an \(x_1\) with \(\lVert x_1\rVert<1\) and \[\lVert y-\Lambda x_1\rVert<\frac{1}{2}\frac{\eta}{2k}\epsilon\] Suppose \(x_1,\cdots,x_n\) are chosen so that \[\lVert y-\Lambda x_1-\cdots-\Lambda x_n\rVert<2^{-n}\frac{\eta}{2k}\epsilon\] Use \(\lVert x\rVert\leq\frac{2k}{\eta}\lVert y\rVert,\quad \lVert y-\Lambda x\rVert<\epsilon\), with \(y\) replaced by the vector \(\lVert y-\Lambda x_1-\cdots-\Lambda x_n\rVert\), to obtain an \(x_{n+1}\) so that \[\lVert y-\Lambda x_1-\cdots-\Lambda x_n-\Lambda x_{n+1}\rVert<2^{-(n+1)}\frac{\eta}{2k}\epsilon\] and \[\lVert x_{n+1}\rVert\leq2^{-n}\epsilon\quad(n=1,2,3,\cdots)\] If we set \(s_n=x_1+\cdots+x_n\), \(s_n\) is a Cauchy sequence in \(X\). Since \(X\) is complete, there exists an \(x\in X\) so that \(s_n\to x\). The inequality \(\lVert x_1\rVert<1\), together with \(\lVert x_{n+1}\rVert\leq2^{-n}\epsilon\quad(n=1,2,3,\cdots)\), shows that \(\lVert x\rVert<1+\epsilon\). Since \(\Lambda\) is continuous, \(\Lambda s_n\to \Lambda x\). By \[\lVert y-\Lambda x_1-\cdots-\Lambda x_n\rVert<2^{-n}\frac{\eta}{2k}\epsilon\] \(\Lambda s_n\to y\). Hence \(\Lambda x=y\).
   Then \[\Lambda ((1+\epsilon)U)\supset\frac{\eta}{2k}V\] or \[\Lambda (U)\supset(1+\epsilon)^{-1}\frac{\eta}{2k}V\] for every \(\epsilon>0\). The union of the sets \((1+\epsilon)^{-1}\frac{\eta}{2k}V\), taken over all \(\epsilon>0\) is \(\frac{\eta}{2k}V\), then \[\Lambda(U)\supset\frac{\eta}{2k} V\]

8. If \(X\) and \(Y\) are Banach spaces and if \(\Lambda\) is a bounded linear transformation of \(X\) onto \(Y\) which is also one-to-one, then there is a \(\delta>0\) such that \[\lVert \Lambda x\rVert\ge\delta\lVert x\rVert\quad(x\in X)\] In other words, \(\Lambda^{-1}\) is a bounded linear transformation of \(Y\) onto \(X\).
   Since \(\lVert y\rVert=\lVert \Lambda x\rVert<\delta\) implies \(\lVert x\rVert<1\). Hence \(\lVert x\rVert\ge1\) implies \(\lVert y\rVert=\lVert \Lambda x\rVert\ge\delta\) then \[\lVert \Lambda x\rVert\ge\delta\lVert x\rVert\quad(x\in X)\]
   The transformation \(\Lambda^{-1}\) is defined on \(Y\) by the requirement that \(\Lambda^{-1}y=x\) if \(y=\Lambda x\). A trivial verification shows that \(\Lambda^{-1}\) is linear, and \(\lVert \Lambda x\rVert\ge\delta\lVert x\rVert\quad(x\in X)\) implies that \(\lVert \Lambda^{-1}\rVert\leq\frac{1}{\delta}\)

9. The \(n\)th partial sum of the Fourier series of \(f\) at the point \(x\) is given by \[s_n(f;x)=\frac{1}{2\pi}\int_{-\pi}^{\pi}f(t)D_n(x-t)dt\quad(n=0,1,2,\cdots)\] where \[D_n(t)=\sum_{k=-n}^{n}e^{ikt}\] For every \(f\in C(T)\) and for every real \(x\), whether the limit \[\lim_{n\to\infty}s_n(f;x)=f(x)?\]
   Put \[s^*(f;x)=\sup_{n}|s_n(f;x)|\] Take \(x=0\) and define \[\Lambda_nf=s_n(f;0)\quad(f\in C(T);n=1,2,3,\cdots)\] We know that \(C(T)\) is a Banach space, relative to the supremum norm \(\lVert f\rVert_{\infty}\). It follows from \[s_n(f;x)=\frac{1}{2\pi}\int_{-\pi}^{\pi}f(t)D_n(x-t)dt\quad(n=0,1,2,\cdots)\] that each \(\Lambda_n\) is a bounded linear functional on \(C(T)\), of norm \[\lVert \Lambda_n\rVert\leq\frac{1}{2\pi}\int_{-\pi}^{\pi}|D_n(t)dt|=\lVert D_n\rVert_1\] We claim that \[\lVert \Lambda_n\rVert\to\infty\quad\text{as }n\to\infty\] This will be proved by showing that equality holds in \[\lVert \Lambda_n\rVert\leq\frac{1}{2\pi}\int_{-\pi}^{\pi}|D_n(t)dt|=\lVert D_n\rVert_1\] and that \[\lVert D_n\rVert_1\to\infty\quad\text{as }n\to\infty\]
   Multiply \[D_n(t)=\sum_{k=-n}^{n}e^{ikt}\] by \(e^{it/2}\) we get \[e^{it/2}D_n(t)=\sum_{k=-n}^{n}e^{ikt+it/2}\] Multiply \[D_n(t)=\sum_{k=-n}^{n}e^{ikt}\] by \(e^{-it/2}\) we get \[e^{-it/2}D_n(t)=\sum_{k=-n}^{n}e^{ikt-it/2}\] and subtract one of the resulting two equations from the other, to obtain \[D_n(t)=\frac{\sum_{k=-n}^{n}e^{ikt+it/2}-\sum_{k=-n}^{n}e^{ikt-it/2}}{e^{it/2}-e^{-it/2}}=\frac{\sin(n+\frac{1}{2})t}{\sin(t/2)}\]

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