A real function \(\varphi\) defined on a segment \((a,b),\quad -\infty\leq a<b\leq \infty\), is called convex if the inequality \[\varphi((1-\lambda)x+\lambda y)\leq(1-\lambda)\varphi(x)+\lambda\varphi(y)\] or equivalently \[\frac{\varphi(t)-\varphi(s)}{t-s}\leq\frac{\varphi(u)-\varphi(t)}{u-t},\quad a<s<t<u<b\] holds whenever \(a<x,y<b,\quad 0\leq\lambda\leq1\). If \(x<t<y\), then the point \((t,\varphi(t))\) should lie below or on the connecting the points \((x,\varphi(x))\) and \((y,\varphi(y))\) in the plane.
If \(\varphi\) is convex on \((a, b)\) then \(\varphi\) is continuous on \((a, b)\).
Suppose \(a<s<x<y<t<b\), write \(S\) for the point \((s,\varphi(s))\) in the plane, and deal similarly with \(x\), \(y\), and \(t\). Then \(X\) is on or below the line \(SY\), hence \(Y\) is on or above the line \(SX\); also \(Y\) is on or below \(XT\). As \(y\to x\), it follows that \(Y\to X\) then \(\varphi(y)\to\varphi(x)\). Left-hand limits are handled in the same manner.Jensen’s Inequality For a convex function \(\varphi(x)\), the line through two points \((x_1,\varphi(x_1))\) and \((x_2,\varphi(x_2))\) is everywhere at least as large as \(\varphi(x)\). Let \(\mu\) be a positive measure on a \(\sigma-algebra\) \(\mathfrak M\) in a set \(\Omega\), so that \(\mu(\Omega)=1\). If \(f\) is a real/unction in \(L^1(\mu)\), if \(a<f(x)<b,\quad x\in \Omega\) and if \(\varphi\) is convex on \((a,b)\), then \[\varphi\Biggl(\int_{\Omega}fd\mu\Biggr)\leq\int_{\Omega}(\varphi\circ f)d\mu\] or when \(f\) is a probability density function of a random variable \(X\), then \[\varphi(EX)\leq E\varphi(X)\] Put \(t=\int_{\Omega}fd\mu\), then \(a<t<b\). If \(\beta\) is the supremum of the quotients \[\beta\ge\frac{\varphi(t)-\varphi(s)}{t-s},\quad a<s<t<b\] then \(\beta\) is no larger than any of the quotients \[\frac{\varphi(u)-\varphi(t)}{u-t},\quad a<s<t<u<b\] It follows that \[\beta(t-s)\ge\varphi(t)-\varphi(s)\quad(a<s<b)\] \[\varphi\Biggl(\int_{\Omega}fd\mu\Biggr)\leq\int_{\Omega}(\varphi\circ f)d\mu\]
To give an example, take \(\varphi(x)=e^x\) Then \[\exp\Biggl[\int_{\Omega}fd\mu\Biggr]\leq\int_{\Omega}e^fd\mu\] If \(\Omega\) is finite set consisting of points \(p_1,\cdots,p_n\), and if \[\mu(\{p_i\})=1/n,\quad f(p_i)=x_i\] then \[\exp\Biggl[\frac{1}{n}(x_1+\cdots+x_n)\Biggr]\leq\frac{1}{n}(e^{x_1}+\cdots+e^{x_n})\] for real \(x_i\). or \[\exp\Biggl[\int_{\Omega}\log g\;d\mu\Biggr]\leq\int_{\Omega}g\;d\mu,\quad g=e^{x_i}\] or \[(y_1y_2\cdots y_n)^{1/n}\leq \frac{1}{n}(y_1+y_2+\cdots+y_n)\quad (y_i=e^{x_i})\] which is the familiar inequality between the arithmetic and geometric means of \(n\) positive numbers.
If we take \(\mu(\{p_i\})=a_i>0\), where \(\sum a_i=1\), then we obtain \[y_1^{a_1}y_2^{a_2}\cdots y_n^{a_n}\leq a_1y_1+a_2y_2+\cdots+a_ny_n\]If \(p\) and \(q\) are positive real numbers such that \(p + q = pq,\quad 1<p<\infty,1<q<\infty\), or \[\frac{1}{p}+\frac{1}{q}=\frac{p+q}{pq}=1,\quad 1<p<\infty,1<q<\infty\] then we call \(p\) and \(q\) a pair of conjugate exponents.
Let \(p\) and \(q\) be conjugate exponents, \(1<p<\infty\). Let \(X\) be a measure space, with measure \(\mu\). Let \(f\) and \(g\) be measurable functions on \(X\), with range in \([0,\infty]\).
Then the Holder’s Inequality \[\int_Xfg\;d\mu\leq\Biggl[\int_Xf^p\;d\mu\Biggr]^{1/p}\Biggl[\int_Xg^q\;d\mu\Biggr]^{1/q}\] and if \(p=q=2\) the Schwarz Inequality \[\int_Xfg\;d\mu\leq\Biggl[\int_Xf^2\;d\mu\Biggr]^{1/2}\Biggl[\int_Xg^2\;d\mu\Biggr]^{1/2}\]
If \(\Biggl[\int_Xf^p\;d\mu\Biggr]^{1/p}=0\) if \(A_n=\{x\in X:f^p(x)>\frac{1}{n}\},n=1,2,3,\cdots\) then \[\frac{1}{n}\mu(A_n)\leq\int_{A_n}f^p\;d\mu\leq\int_{X}f^p\;d\mu=0\] so that \(\mu(A_n)=0\). Since \(\{x\in X:f^p(x)>0\}=\bigcup A_n=0\) then \(f^p=0\) and \(f=0\). hence \(fg=0\) so Holder’s Inequality holds. If \(\Biggl[\int_Xf^p\;d\mu\Biggr]^{1/p}>0\) and \(\Biggl[\int_Xg^q\;d\mu\Biggr]^{1/q}=\infty\) Holder’s Inequality holds again. If \(0<\Biggl[\int_Xf^p\;d\mu\Biggr]^{1/p}<\infty\) and \(0<\Biggl[\int_Xg^q\;d\mu\Biggr]^{1/q}<\infty\), \[\int_X\Biggl(\frac{f}{\Bigl[\int_Xf^p\;d\mu\Bigr]^{1/p}}\Biggr)^pd\mu=\int_X\frac{f^p}{\Bigl[\int_Xf^p\;d\mu\Bigr]}d\mu=1\] and \[\int_X\Biggl(\frac{g}{\Bigl[\int_Xg^q\;d\mu\Bigr]^{1/q}}\Biggr)^qd\mu=\int_X\frac{g^q}{\Bigl[\int_Xg^q\;d\mu\Bigr]}d\mu=1\] If \(x\in X\) is such that \[0<\frac{f}{\Bigl[\int_Xf^p\;d\mu\Bigr]^{1/p}}<\infty\] and \[0<\frac{g}{\Bigl[\int_Xg^q\;d\mu\Bigr]^{1/q}}<\infty\] there are real numbers \(s\) and \(t\) such that \(\frac{f}{\Bigl[\int_Xf^p\;d\mu\Bigr]^{1/p}}=e^{s/p},\quad \frac{g}{\Bigl[\int_Xg^q\;d\mu\Bigr]^{1/q}}=e^{t/q}\) Since \(1/p+1/q=1\), the convexity of the exponential function implies that \[e^{s/p}e^{t/q}=e^{s/p+t/q}\leq\frac{1}{p}e^s+\frac{1}{q}e^t\] It follows that \[\frac{f}{\Bigl[\int_Xf^p\;d\mu\Bigr]^{1/p}}\frac{g}{\Bigl[\int_Xg^q\;d\mu\Bigr]^{1/q}}\leq\frac{1}{p}\Biggl(\frac{f}{\Bigl[\int_Xf^p\;d\mu\Bigr]^{1/p}}\Biggr)^p+\frac{1}{q}\Biggl(\frac{g}{\Bigl[\int_Xg^q\;d\mu\Bigr]^{1/q}}\Biggr)^q\] for every \(x\in X\) then integration yields \[\int_X\frac{f}{\Bigl[\int_Xf^p\;d\mu\Bigr]^{1/p}}\frac{g}{\Bigl[\int_Xg^q\;d\mu\Bigr]^{1/q}}d\mu\leq\frac{1}{p}+\frac{1}{q}=1\] then \[\int_Xfg\;d\mu\leq\Biggl[\int_Xf^p\;d\mu\Biggr]^{1/p}\Biggl[\int_Xg^q\;d\mu\Biggr]^{1/q}\]Let \(p\) and \(q\) be conjugate exponents, \(1<p<\infty\). Let \(X\) be a measure space, with measure \(\mu\). Let \(f\) and \(g\) be measurable functions on \(X\), with range in \([0,\infty]\).
Then the Minkowski’s Inequality \[\Biggl[\int_X(f+g)^p\;d\mu\Biggr]^{1/p}\leq\Biggl[\int_Xf^p\;d\mu\Biggr]^{1/p}+\Biggl[\int_Xg^p\;d\mu\Biggr]^{1/p}\]
We write \[(f+g)^p=f\cdot(f+g)^{p-1}+g\cdot(f+g)^{p-1}\] HOlder’s inequality gives \[\int f\cdot(f+g)^{p-1}\leq\Biggl[\int f^p\Biggr]^{1/p}\Biggl[\int (f+g)^{(p-1)q}\Biggr]^{1/q}\] and \[\int g\cdot(f+g)^{p-1}\leq\Biggl[\int g^p\Biggr]^{1/p}\Biggl[\int (f+g)^{(p-1)q}\Biggr]^{1/q}\] then \[\int (f+g)^p\leq \Biggl[\int f^p\Biggr]^{1/p}\Biggl[\int (f+g)^{(p-1)q}\Biggr]^{1/q}+\Biggl[\int g^p\Biggr]^{1/p}\Biggl[\int (f+g)^{(p-1)q}\Biggr]^{1/q}\\ =\Biggl[\int (f+g)^{(p-1)q}\Biggr]^{1/q}\Biggl(\Biggl[\int f^p\Biggr]^{1/p}+\Biggl[\int g^p\Biggr]^{1/p}\Biggr)\\ =\Biggl[\int (f+g)^{p}\Biggr]^{1/q}\Biggl(\Biggl[\int f^p\Biggr]^{1/p}+\Biggl[\int g^p\Biggr]^{1/p}\Biggr)\] and The convexity of the function \(t^p,\quad 0<t<\infty\) shows that \[\Biggl(\frac{f+g}{2}\Biggl)^p\leq\frac{1}{2}(f^p+g^p)\] then \[0<\Bigl[\int (f+g)^{p}\Bigr]^{1/q}<\infty\] then \[\frac{\int (f+g)^p}{\Bigl[\int (f+g)^{p}\Bigr]^{1/q}}=\Bigl[\int (f+g)^{p}\Bigr]^{1/p}\leq \Biggl[\int f^p\Biggr]^{1/p}+\Biggl[\int g^p\Biggr]^{1/p}\]\(X\) is an arbitrary measure space with a positive measure \(\mu\). If \(0<p<\infty\) and if \(f\) is a complex measurable function on \(X\), define \[\lVert f\rVert_p=\Biggl[\int_X|f|^pd\mu\Biggr]^{1/p}\] and let \(L^p(\mu)\) consist of all \(f\) for which \[\lVert f\rVert_p<\infty\] We call \(\lVert f\rVert_p\) the \(L^p-norm\) of \(f\).
If \(\mu\) is Lebesgue measure on \(R^k\), we write \(L^p(R^k)\) instead of \(L^p(\mu)\). If \(\mu\) is the counting measure on a set \(A\), it is customary to denote the corresponding \(L^p\)-space by \(\ell^p(A)\). An element of \(\ell^p(A)\) may be regarded as a complex sequence \(x=\{a_n\}\) and \[\lVert x\rVert_p=\Biggl(\sum_{n=1}^{\infty}|a_n|^p\Biggr)^{1/p}\]Suppose \(g:X\to[0,\infty]\) is measurable. Let \(S\) be the set of all real \(a\) such that \[\mu(g^{-1}((a,\infty]))=0\] If \(S=\varnothing\), put \(\beta=\infty\). If \(S\ne\varnothing\), put \(\beta=\inf S\). Since \[g^{-1}((\beta,\infty])=\overset{\infty}{\underset{n=1}{\bigcup}}g^{-1}\Biggl(\Bigl(\beta+\frac{1}{n},\infty\Bigr]\Biggr)\] and since the union of a countable collection of sets of measure \(0\) has measure \(0\), so \(\beta\in S\). We call \(\beta\) the essential supremum of \(g\).
If \(f\) is a complex measurable function on \(X\), we define \(\lVert f\rVert_{\infty}\) to be the essential supremum of \(|f|\), and we let \(L^{\infty}(\mu)\) consist of all \(f\) for which \(\lVert f\rVert_{\infty}<\infty\). The members of \(L^{\infty}(\mu)\) are sometimes called essential bounded measurable functions on \(X\).
The inequality \(|f(x)|\leq \lambda\) holds for almost all \(x\) if and only if \(\lambda\ge\lVert f\rVert_{\infty}\)If \(p\) and \(q\) are conjugate exponents, \(1\leq p\leq\infty\), and if \(f\in L^p(\mu)\) and \(g\in L^q(\mu)\), then \[fg\in L^1(\mu)\] and \[\lVert fg\rVert_{1}\leq\lVert f\rVert_{p}\lVert g\rVert_{q}\]
For \(1<p<\infty\), the HOlder’s inequality applied to \(|f|\) and \(|g|\) is \[\lVert fg\rVert_{1}\leq\lVert f\rVert_{p}\lVert g\rVert_{q}\] If \(p=\infty,\quad q=1\), since \[|fg|\leq|f||g|\leq\lVert f\rVert_{\infty}\lvert g\rvert\] for almost all \(x\), integration will obtain \[\lVert fg\rVert_{1}\leq\lVert f\rVert_{p}\lVert g\rVert_{q}\] If \(q=\infty,\quad p=1\) the same argument applies.Suppose \(1\leq p\leq\infty\) and \(f\in L^p(\mu), g\in L^p(\mu)\) Then \(f+g\in L^p(\mu)\) and \[\lVert f+g\rVert_{p}\leq\lVert f\rVert_{p}+\lVert g\rVert_{p}\]
For \(1<p<\infty\), since \[\Biggl(\int_X|f+g|^pd\mu\Biggr)^{1/p}\leq\Biggl(\int_X(|f|+|g|)^pd\mu\Biggl)^{1/p}\] then \[\lVert f+g\rVert_{p}\leq\lVert f\rVert_{p}+\lVert g\rVert_{p}\] follows from Minkowski’s inequality. If \(p=1\) or \(p=\infty\), since \[|f+g|\leq|f|+|g|\] then \[\lVert f+g\rVert_{p}\leq\lVert f\rVert_{p}+\lVert g\rVert_{p}\] Since \[\lVert af\rVert_{p}=|a|\lVert f\rVert_{p}\] Then \(L^p(\mu)\) is a complex vector space.Since \[\lVert f-h\rVert_{p}\leq\lVert f-g\rVert_{p}+\lVert g-h\rVert_{p}\] is we define the distance \(d(f,g)\) in \(L^p(\mu)\) space is \(\lVert f-g\rVert_{p}\), then \(0\leq d(f,g)<\infty\) and \(d(f,g)=0, \; precisely\) when \(f(x)=g(x)\) for almost all \(x\).
We write \(f\sim g\) if and only if \(d(f,g)=0\).If \(\{f_n\}\) is a sequence in \(L^p(\mu)\), if \(f\in L^p(\mu)\) and if \[\lim_{n\to\infty}\lVert f_n-f\rVert_{p}=0\] we say that \(\{f_n\}\) converges to \(f\) in \(L^p(\mu)\). If there is a integer \(N\) such that \[\lVert f_n-f_m\rVert_{p}<\epsilon,\quad \forall \epsilon>0,\quad n,m>N\] we call \(\{f_n\}\) a Cauchy sequence in \(L^p(\mu)\).
\(L^p(\mu)\) is a complete metric space, since every Cauchy sequence in \(L^p(\mu)\) converges to an element of \(L^p(\mu)\). (A metric space \(M\) is called complete (or a Cauchy space) if every Cauchy sequence of points in \(M\) has a limit that is also in \(M\) or, alternatively, if every Cauchy sequence in \(M\) converges in \(M\). Intuitively, a space is complete if there are no “points missing” from it (inside or at the boundary). For instance, the set of rational numbers is not complete, because \(\sqrt {2}\) is “missing” from it, even though one can construct a Cauchy sequence of rational numbers that converges to it. It is always possible to “fill all the holes”, leading to the completion of a given space).\(L^p(\mu)\) is a complete metric space, for \(1\leq p\leq\infty\) and for every positive measure \(\mu\). If \(1\leq p<\infty\), let \(\{f_n\}\) be a Cauchy sequence in \(L^p(\mu)\). There is a subsequence \(\{f_{n_i}\},\quad n_1,n_2,\cdots\) such that \[\lVert f_{n_i+1}-f_{n_i}\rVert_{p}<2^{-i}\quad(i=1,2,3,\cdots)\] Put \[g_k=\sum_{i=1}^{k}|f_{n_i+1}-f_{n_i}|,\quad g=\sum_{i=1}^{\infty}|f_{n_i+1}-f_{n_i}|\] Then \[\lVert g_k\rVert_{p}\leq\sum_{i=1}^{k}2^{-i}<1\] Hence an application of Fatou’s lemma to \(\{g_k^p\}\) gives \(\lVert g\rVert_{p}\leq1\). In particular, \(g(x)<\infty\) so that the series \[f_{n_1}(x)+\sum_{i=1}^{\infty}(f_{n_i+1}(x)-f_{n_i}(x))\] converges absolutely for almost every \(x\in X\). Denote \[f(x)=f_{n_1}(x)+\sum_{i=1}^{\infty}(f_{n_i+1}(x)-f_{n_i}(x))\] for those \(x\) at which \(f_{n_1}(x)+\sum_{i=1}^{\infty}(f_{n_i+1}(x)-f_{n_i}(x))\) converges; put \(f(x)=0\) on the remaining set of measure zero. Since \[f_{n_1}(x)+\sum_{i=1}^{k-1}(f_{n_i+1}(x)-f_{n_i}(x))=f_{n_k}\] then \[f(x)=\lim_{i\to\infty}f_{n_i}(x)\quad a.e.\]
Choose \(\epsilon>0\). There exists an \(N\) such that \(\lVert f_n-f_m\rVert_{p}<\epsilon,\quad n,m>N\). then \(f\) is the pointwise limit a.e. of \(\{f_{n_i}\}\). For every \(m>N\), Fatou’s lemma shows therefore that \[\int_X|f-f_m|^pd\mu\leq\lim_{i\to\infty}\inf\int_X|f_{n_i}-f_m|^pd\mu\leq\epsilon^p\] then \[\Biggl(\int_X|f-f_m|^pd\mu\Biggr)^{1/p}=\lVert f-f_m\rVert_{p}<\epsilon,\quad \forall m>N\] Then \(f-f_m\in L^p(\mu)\), since \(f=f-f_m+f_m,\quad f_m\in L^p(\mu)\), hence that \(f\in L^p(\mu)\) and finally that \(\lVert f-f_m\rVert_{p}\to0\) as \(m\to\infty\). Then \(f\) is the \(L^p(\mu)\) limit of \(\{f_n\}\).
If \(p=\infty\), suppose \(\{f_n\}\) is a Cauchy sequence in \(L^{\infty}(\mu)\), let sets \[A_k=\{k:|f_k|>\lVert f_k\rVert_{\infty}\},\quad k=1,2,3,\cdots\] and \[B_{m,n}=\{n,m:|f_n-f_m|>\lVert f_n-f_m\rVert_{\infty}\},\quad n,m=1,2,3,\cdots\] and let \(E\) be the union of these sets, then \(\mu(E)=0\), and on the complement of \(E\), the sequence \(\{f_n\}\) converges uniformly to a bounded function \(f\). Define \(f(x)=0,\quad x\in E\). Then \(f\in L^p(\mu)\) and \(\lVert f_n-f\rVert_{\infty}\to0,\quad n\to\infty\)If \(1\leq p\leq\infty\) and if \(\{f_n\}\) is a Cauchy sequence in \(L^p(\mu)\), with limit \(f\), then \(\{f_n\}\) has a subsequence which converges pointwise almost everywhere to \(f(x)\).
Let \(S\) be the class of all complex, measurable, simple functions on \(X\) such that \[\mu\Bigl(\{x:s(x)\ne0\}\Bigr)<\infty\] if \(1\leq p<\infty\), then \(S\) is dense in \(L^p(\mu)\). (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 \(X\). Informally, for every point in \(X\), the point is either in \(A\) or arbitrarily “close” to a member of \(A\)).
It is clear that \(S\subset L^p(\mu)\). Suppose \(f\ge0,f\in L^p(\mu)\) and let simple measurable functions \(\{s_n\}\) such that \[0\leq s_1\leq s_2\leq\cdots\leq f\] and as \[n\to\infty,\quad s_n(x)\to f(x),\quad \forall x\in X\] Since \(0\leq s_n\leq f\), we have \(s_n\in L^p(\mu)\), hence \(s_n\in S\). Since \(|f-s_n|^p\leq f^p\), the dominated convergence theorem shows that as \[n\to\infty,\quad \lVert f-s_n\rVert_{p}\to0\] Thus \(f\) is in the \(L^p\)-closure of \(S\).Let \(X\) be a locally compact Hausdorff space, and let \(\mu\) be a measure on a \(\sigma-algebra\) \(\mathfrak M\) in \(X\), which contains all Borel sets in \(X\).
For \(1\leq p<\infty\), \(C_c(X)\) is dense in \(L^p(\mu)\).
Define \(S\) as in last Theorem. If \(s\in S, \epsilon>0\), there exists a \(g\in C_c(X)\) such that \(g(x)=s(x)\) except on a set of measure \(<\epsilon\), and \(|g|\leq \lVert s\rVert_{\infty}\) (Lusin’s theorem: An almost-everywhere finite function is measurable if and only if it is a continuous function on nearly all its domain). Hence \[\lVert g-s\rVert_{p}\leq\lVert g\rVert_{p}+\lVert s\rVert_{p}\leq2\epsilon^{1/p}\lVert s\rVert_{\infty}\] Since \(S\) is dense in \(L^p(\mu)\), this completes the proof. If \(1\leq p<\infty\), \(C_c(R^k)\) is dense in \(L^p(R^k)\) and \(L^p(R^k)\) is complete. Thus \(L^p(R^k)\) is the completion of the metric space which is obtained by endowing \(C_c(R^k)\) with the \(L^p\)-metric.A complex function \(f\) on a locally compact Hausdorff space \(X\) is said to vanish at infinity if to every \(\epsilon>0\) there exists a compact set \(K\subset X\) such that \(|f(x)|<\epsilon, \forall x\notin K\). The class of all continuous \(f\) on \(X\) which vanish at infinity is called \(C_0(X)\).
It is clear that \(C_c(X)\subset C_0(X)\), and that the two classes coincide if \(X\) is compact.If \(X\) is a locally compact Hausdorff space, then \(C_0(X)\) is the completion of \(C_c(X)\), relative to the metric defined by the supremum norm \[\lVert f\rVert=\sup_{x\in X}|f(x)|\]
\(C_0(X)\) satisfies the axioms of a metric space if the distance between \(f\) and \(g\) is taken to be \(\lVert f-g\rVert\). We have to show that (a) \(C_c(X)\) is dense in \(C_0(X)\) and (b) \(C_0(X)\) is a complete metric space.
Given \(f\in C_0(X),\;\epsilon>0\), there is a compact set \(K\) so that \(|f(x)|<\epsilon, x\notin K\). Urysohn’s lemma gives us a function \(g\in C_c(X)\), and \(0\leq g\leq1\),\(g(x)=1,\quad x\in K\). Then \(fg\in C_c(X)\) and \(\lVert f-fg\rVert<|f(x)||1-g|<\epsilon\), then \(C_c(X)\) is dense in \(C_0(X)\).
Let \(\{f_n\}\) be a Cauchy sequence in \(C_0(X)\), assume that \(\{f_n\}\) converges uniformly. Then its pointwise limit function \(f\) is continuous. There exists an \(n\) so that \(\lVert f_n-f\rVert\leq\frac{\epsilon}{2},\quad\epsilon>0\) and there is a compact set \(K\) so that \(|f_n(x)|\leq\frac{\epsilon}{2},\quad x\notin K\). Hence \(|f(x)|=|f+f_n-f|<|f|+|f_n-f|<\epsilon,x\notin K\) and then \(f\) vanishes at infinity. Then \(C_0(X)\) is a complete metric space.
BIBLIOGRAPHY
1. Rudin W. Real and complex analysis. Tata McGraw-hill education; 2006.