Ngô Quốc Anh

Tháng Ba 31, 2009

Mellin transform with some examples

Chuyên mục: Linh Tinh, Nghiên Cứu Khoa Học — Ngô Quốc Anh @ 13:11

In mathematics, the Mellin transform is an integral transform that may be regarded as the multiplicative version of the two-sided Laplace transform. This integral transform is closely connected to the theory of Dirichlet series, and is often used in number theory and the theory of asymptotic expansions; it is closely related to the Laplace transform and the Fourier transform, and the theory of the gamma function and allied special functions.

The Mellin transform of a function is

\mathcal{M}\left\{ {f\left( x \right)} \right\}(s) = \varphi (s) = \int_0^\infty {{x^s}} f(x)\frac{{dx}} {x}.

The inverse transform is

{\mathcal{M}^{ - 1}}\left\{ {\varphi \left( s \right)} \right\}(x) = f(x) = \frac{1} {{2\pi i}}\int_{c - i\infty }^{c + i\infty } {{x^{ - s}}} \varphi (s)\,ds.

The notation implies this is a line integral taken over a vertical line in the complex plane. Conditions under which this inversion is valid are given in the Mellin inversion theorem. The transform is named after the Finnish mathematician Hjalmar Mellin.

Importance of the fundamental strip

A Mellin transform should never be computed without its fundamental strip, which tells us where the image function converges. This strip is key to the Mellin inversion process, which arises in number theoretic applications of the transform and in the study of harmonic sums, frequently encountered in computer science. The basic idea is to compute the Mellin transform of a sum and invert it thereafter, thus obtaining an asymptotic expansion. However the Mellin inversion integral is computed over a line parallel to the imaginary axis that lies in the fundamental strip. Without knowing where the strip lies, the integral cannot be computed, more precisely, one does not know which residues contribute to its value.

The fundamental strip arises from the analysis of the convergence properties of the Mellin integral:

\int_0^\infty x^s f(x)\frac{dx}{x}.

We split the integral into two parts, as follows:

\left( \int_0^1 + \int_1^\infty \right) x^s f(x)\frac{dx}{x}.

Assuming is locally integrable along the positive real line, the first integral must remain bounded at zero, and the second, at infinity (understood as “in the limit” if e.g. Riemann integrability is used).

http://upload.wikimedia.org/wikipedia/en/c/c2/MellinDomains.pngFile:MellinStrip.png

Letting , we find that

\left| \int_0^1 x^s f(x)\frac{dx}{x} \right| \le \int_0^1 x^\sigma |f(x)| \frac{dx}{x}

and

\left| \int_1^\infty x^s f(x)\frac{dx}{x} \right| \le \int_1^\infty x^\sigma |f(x)| \frac{dx}{x}.

Now suppose at . The first bounding integral converges if

\sigma + u - 1 > -1 \quad \mbox{or} \quad \sigma > -u.

Furthermore suppose that at infinity. The second bounding integral converges if

\sigma + v - 1 < -1 \quad \mbox{or} \quad \sigma < -v.

These two constraints on s define two half planes, the first a left half plane and the second one a right half plane. The intersection of the two half planes is the fundamental strip, denoted \left\langle { - u, - v} \right\rangle. It frequently happens that the image function can be analytically continued to the whole plane, which makes it possible to compute the inversion integral by shifting the line of integration to the left or to the right. The original Mellin integral, however, remains restricted to the fundamental strip.

Summary: if is locally integrable along the positive real line, and

f(x)_{x\rightarrow 0+} = O(x^u) \quad \mbox{and} \quad f(x)_{x\rightarrow +\infty} = O(x^v)

then its Mellin transform converges in the fundamental strip \left\langle { - u, - v} \right\rangle and the corresponding Mellin inversion integral is taken along a line parallel to the imaginary axis in this strip.

Computing the fundamental strip

As an example, consider the transform pair

f(x) = \frac{1}{1+x} \quad \mbox{and} \quad \varphi(s) = \frac{\pi}{\sin \pi s}.

By inspection, we have f(x)_{x\rightarrow 0+} = O(1) = O(x^0) and f(x)_{x\rightarrow +\infty} \sim \frac{1}{x} = O(x^{-1}) and the fundamental strip is \left\langle { 0, 1} \right\rangle. This is illustrated in the first diagram at the beginning of this section.

As a second example, consider the transform pair

f(x) = \exp(-x) -1 + x \quad \mbox{and} \quad \varphi(s) = \Gamma(s).

We have the following series expansion around :

f(x) = \sum_{n\ge 2} \frac{(-1)^n \, x^n}{n!},

which implies that f(x)_{x\rightarrow 0+} = O(x^2). At infinity, we have  f(x)_{x\rightarrow +\infty} \sim x = O(x^1) so that the fundamental strip is \left\langle { - 2, - 1} \right\rangle. This is shown in the second diagram.

Some relation

Gamma function: Clearly,

\mathcal M\left\{ {{e^{ - ax}}u\left( x \right)} \right\}\left( s \right) = \int_0^\infty {{e^{ - ax}}{x^{s - 1}}dx} = {a^{ - s}}\Gamma \left( s \right).

Relation to Laplace Transform: By letting , the transform becomes

\mathcal M\left\{ {f\left( x \right)} \right\}\left( s \right) = \int_{ - \infty }^\infty {f\left( {{e^{ - t}}} \right){e^{ - st}}dt} = \mathcal L\left\{ {f\left( {{e^{ - t}}} \right)} \right\}\left( s \right).

Relation to Fourier Transform: By setting we obtain

M\left\{ {f\left( x \right)} \right\}\left( {s = \sigma + j2\pi \beta } \right) = L\left\{ {f\left( {{e^{ - t}}} \right){e^{ - \sigma t}}} \right\}\left( \beta \right).

Properties of Mellin Transform

Scaling Property

\mathcal{M}\left\{ {f\left( {ax} \right)} \right\}\left( s \right) = \int_{ - \infty }^\infty {f\left( {ax} \right){x^{s - 1}}dx} = {a^{ - s}}\underbrace {\int_{ - \infty }^\infty {f\left( x \right){x^{s - 1}}dx} }_{\mathcal{M}\left\{ {f\left( x \right)} \right\}\left( s \right)}.


Multiplication by

\mathcal{M}\left\{ {{x^a}f\left( x \right)} \right\}\left( s \right) = \int_{ - \infty }^\infty {f\left( x \right){x^{\left( {a + s} \right) - 1}}dx} = \mathcal{M}\left\{ {f\left( x \right)} \right\}\left( {s + a} \right).
Raising the Independent Variable to a Real Power

\mathcal{M}\left\{ {f\left( {{x^a}} \right)} \right\}\left( s \right) = \int_{ - \infty }^\infty {f\left( {{x^a}} \right){x^{s - 1}}dx} = \underbrace {\int_{ - \infty }^\infty {f\left( x \right){x^{\frac{s} {a} - \frac{1} {a}}}\left( {\frac{1} {a}{x^{\frac{1} {a} - 1}}dx} \right)} }_{\mathcal{M}\left\{ {f\left( x \right)} \right\}\left( {\frac{s} {a}} \right),\quad a > 0}

Inverse of Independent Variable

\mathcal{M}\left\{ {\frac{1} {x}f\left( {\frac{1} {x}} \right)} \right\}\left( s \right) = \mathcal{M}\left\{ {f\left( x \right)} \right\}\left( {1 - s} \right).

Multiplication by

\mathcal{M}\left\{ {\left( {\ln x} \right)f\left( x \right)} \right\}\left( s \right) = \frac{d} {{ds}}\mathcal{M}\left\{ {f\left( x \right)} \right\}\left( s \right).

Multiplication by a Power of

\mathcal{M}\left\{ {{{\left( {\ln x} \right)}^k}f\left( x \right)} \right\}\left( s \right) = \frac{{{d^k}}} {{d{s^k}}}\mathcal{M}\left\{ {f\left( x \right)} \right\}\left( s \right).

Derivative

\mathcal{M}\left\{ {\frac{{{d^k}}} {{d{x^k}}}f\left( x \right)} \right\}\left( s \right) = {\left( { - 1} \right)^k}\frac{{\Gamma \left( s \right)}} {{\Gamma \left( {s - k} \right)}}\mathcal{M}\left\{ {f\left( x \right)} \right\}\left( {s - k} \right).

Derivative Multiplied by Independent Variable

\mathcal{M}\left\{ {{x^k}\frac{{{d^k}}} {{d{x^k}}}f\left( x \right)} \right\}\left( s \right) = {\left( { - 1} \right)^k}\frac{{\Gamma \left( {s + k} \right)}} {{\Gamma \left( s \right)}}\mathcal{M}\left\{ {f\left( x \right)} \right\}\left( s \right).

Convolution

\mathcal{M}\left\{ {f\left( x \right)g\left( x \right)} \right\}\left( s \right) = \frac{1} {{2\pi i}}\int_{c - i\infty }^{c + i\infty } {\mathcal{M}\left\{ {f\left( x \right)} \right\}\left( z \right)\mathcal{M}\left\{ {g\left( x \right)} \right\}\left( {s - z} \right)dz} .

Multiplicative Convolution

\begin{gathered} \mathcal{M}\left\{ {f\left( x \right) \vee g\left( x \right)} \right\}\left( s \right) = \mathcal{M}\left\{ {f\left( x \right)} \right\}\left( s \right)\mathcal{M}\left\{ {g\left( x \right)} \right\}\left( s \right), \hfill \\ {\mathcal{M}^{ - 1}}\left( {\mathcal{M}\left\{ {f\left( x \right)} \right\}\left( s \right)\mathcal{M}\left\{ {g\left( x \right)} \right\}\left( s \right)} \right)\left( x \right) = \int_0^\infty {f\left( {\frac{x} {u}} \right)g\left( u \right)\frac{{du}} {u}.} \hfill \\ \end{gathered}

Examples of Mellin Transform

  • Characteristic function.

\mathcal{M}\left\{ {{x^a}{\chi _{\left[ {{x_0},\infty } \right)}}\left( x \right)} \right\}\left( s \right) = \int_{{x_0}}^\infty {{x^{{x_0} + s - 1}}dx} = - \frac{{x_0^{a + s}}} {{a + s}},\quad \Re e x < - a.

  • Fractional function.

    • \mathcal{M}\left\{ {\frac{1} {{1 + x}}} \right\}\left( s \right) = \int_0^\infty {\frac{{{x^{s - 1}}}} {{1 + x}}dx} .

    Setting we obtain and dt = \frac{{dx}} {{{{\left( {1 + x} \right)}^2}}}. Therefore,

    \mathcal{M}\left\{ {\frac{1} {{1 + x}}} \right\}\left( s \right) = \int_0^1 {\left( {1 - t} \right)\frac{{{t^{s - 1}}}} {{{{\left( {1 - t} \right)}^{s - 1}}}}\frac{{dt}} {{{{\left( {1 - t} \right)}^2}}}} = \underbrace {\int_0^1 {{t^{s - 1}}{{\left( {1 - t} \right)}^{ - s}}dt} }_{\frac{\pi } {{\sin \pi s}}}.

  • Polynomial function. From

    \int_0^1 {{{\left( {1 - u} \right)}^{m - 1}}{u^{s - 1}}du} = \frac{{\Gamma \left( m \right)\Gamma \left( s \right)}} {{\Gamma \left( {m + s} \right)}}, \quad \Re e s > 0, \Re e m > 0

    with the setting , we obtain

    \int_0^\infty {\frac{{{x^{s - 1}}}} {{{{\left( {1 + x} \right)}^{m + s}}}}dx} = \frac{{\Gamma \left( m \right)\Gamma \left( s \right)}} {{\Gamma \left( {m + s} \right)}}.

    Hence

    \mathcal{M}\left\{ {{{\left( {\frac{1} {{1 + x}}} \right)}^a}} \right\}\left( s \right) = \frac{{\Gamma \left( s \right)\Gamma \left( {a - s} \right)}} {{\Gamma \left( a \right)}}, \quad 0 < \Re e s < \Re e a.

Source

Tháng Ba 26, 2009

Solving initial value problem for heat equation via Fourier transform

Chuyên mục: Các Bài Tập Nhỏ, Nghiên Cứu Khoa Học — Ngô Quốc Anh @ 21:05

Followed by Solving initial value problem for wave equation via Fourier transform, in this topic I will show you how to solve initial value problem for heat equation via Fourier transform.

Following is the problem

\left\{ \begin{gathered} {u_t} = c^2 {u_{xx}}, \qquad x \in \mathbb{R}, \hfill \\ u\left( {x,0} \right) = \varphi \left( x \right). \hfill \\ \end{gathered} \right.

From the equation by taking Fourier transform to the both sides, we obtain

\widehat{{u_t}\left( {\eta ,t} \right)} - c^2 {\left( {i\eta } \right)^2}\widehat{u\left( {\eta ,t} \right)} = 0.

This is an ODE, the solution is given by

\widehat{u\left( {\eta ,t} \right)} = C(\eta)e^{-c^2 \eta^2 t}.

From the initial date we get \widehat{u\left( {\eta ,0} \right)} = \widehat{\varphi \left( \eta \right)} which implies that \widehat{\varphi \left( \eta \right)} = C(\eta). Thus, we obtain

\widehat{u\left( {\eta ,t} \right)} = \widehat{\varphi \left( \eta \right)}e^{-c^2 \eta^2 t}.

Finally, we obtain by taking the inverse Fourier transform

u\left( {x,t} \right) = \frac{1} {{\sqrt {2\pi } }}\int_{ - \infty }^\infty {{e^{i\eta x}}\widehatu\left( {\eta ,t} \right)d\eta } = \frac{1} {{\sqrt {2\pi } }}\int_{ - \infty }^\infty {{e^{i\eta x - {c^2}{\eta ^2}t}}\widehat\varphi \left( \eta \right)d\eta } .

Tháng Ba 13, 2009

Green’s function and differential equations

Chuyên mục: Linh Tinh, Nghiên Cứu Khoa Học — Ngô Quốc Anh @ 12:31

In mathematics, a Green’s function is a type of function used to solve inhomogeneous differential equations subject to boundary conditions. The term is also used in physics, specifically in quantum field theory, electrodynamics and statistical field theory, to refer to various types of correlation functions, even those that do not fit the mathematical definition; for this sense, see Correlation function (quantum field theory) and Green’s function (many-body theory).

Green’s functions are named after the British mathematician George Green, who first developed the concept in the 1830s. In the modern study of linear partial differential equations, Green’s functions are largely studied from the point of view of fundamental solutions instead.

Definition and uses

Technically, a Green’s function, , of a linear differential operator acting on distributions over a subset of the Euclidean space , at a point , is any solution of

L G(x,s) = \delta(x-s) \ \ \ \ (1)

where is the Dirac delta function. This technique can be used to solve differential equations of the form

If the kernel of is nontrivial, then the Green’s function is not unique. However, in practice, some combination of symmetry, boundary conditions and/or other externally imposed criteria will give a unique Green’s function. Also, Green’s functions in general are distributions, not necessarily proper functions.

Green’s functions are also a useful tool in condensed matter theory, where they allow the resolution of the diffusion equation, and in quantum mechanics, where the Green’s function of the Hamiltonian is a key concept, with important links to the concept of density of states. The Green’s functions used in those two domains are highly similar, due to the analogy in the mathematical structure of the diffusion equation and Schrödinger equation. As a side note, the Green’s function as used in Physics is usually defined with the opposite sign; that is, . This definition does not significantly change any of the properties of the Green’s function.

Motivation

Loosely speaking, if such a function can be found for the operator , then if we multiply the equation (1) for the Green’s function by , and then perform an integration in the variable, we obtain;

\int L G(x,s) f(s) ds = \int \delta(x-s)f(s) ds = f(x).

The right hand side is now given by the equation (2) to be equal to , thus:

Because the operator is linear and acts on the variable alone (not on the variable of integration ), we can take the operator outside of the integration on the right hand side obtaining;

Lu(x) = L\left(\int G(x,s) f(s) ds\right).

And this implies;

u(x) = \int G(x,s) f(s) ds . \ \ \ \ (3)

Thus, we can obtain the function through knowledge of the Green’s function in equation (1), and the source term on the right hand side in equation (2). This process has resulted from the linearity of the operator .

In other words, the solution of equation (2), , can be determined by the integration given in equation (3). Although is known, this integration cannot be performed unless is also known. The problem now lies in finding the Green’s function that satisfies equation (1). For this reason, the Green’s function is also sometimes called the fundamental solution associated to the operator .

Not every operator admits a Green’s function. A Green’s function can also be thought of as a right inverse of . Aside from the difficulties of finding a Green’s functions for a particular operator, the integral in equation (3), may be quite difficult to perform. However the method gives a theoretically exact result.

This can be thought of as an expansion of according to a Dirac delta function basis (projecting over and a superposition of the solution on each projection. Such an integral is known as a Fredholm integral equation, the study of which constitutes Fredholm theory.

Green’s functions for solving inhomogeneous boundary value problems

The primary use of Green’s functions in mathematics is to solve inhomogeneous boundary value problems. In modern theoretical physics, Green’s functions are also usually used as propagators in Feynman diagrams (and the phrase “Green’s function” is often used for any correlation function).

Framework

Let be the Sturm-Liouville operator, a linear differential operator of the form

L = {d \over dx}\left[ p(x) {d \over dx} \right] + q(x)

and let be the boundary conditions operator

Du = \left\{\begin{matrix} \alpha _1 u'(0) + \beta _1 u(0) \\ \alpha _2 u'(\ell) + \beta _2 u(\ell). \end{matrix}\right.

Let be a continuous function in . We shall also suppose that the problem

\begin{matrix}Lu = f \\ Du = 0 \end{matrix}

is regular, i.e. only the trivial solution exists for the homogeneous problem.

Theorem

There is one and only one solution which satisfies

\begin{matrix}Lu = f \\ Du = 0 \end{matrix}

and it is given by

u(x) = \int_0^\ell f(s) G(x,s) \, ds

where is a Green’s function satisfying the following conditions:

1. is continuous in and .
2. For , .
3. For , .
4. Derivative “jump”: G'(s_{ + 0}, s ) - G'(s_{ - 0}, s ) = \frac{1}{p(s)}.
5. Symmetry: .

Finding Green’s functions (eigenvalue expansions)

If a differential operator admits a set of eigenvectors (i.e. a set of functions and scalars such that ) that are complete, then it is possible to construct a Green’s function from these eigenvectors and eigenvalues.

Complete means that the set of functions satisfies the following completeness relation:

\delta(x - x') = \sum_{n=0}^\infty \Psi_n(x) \Psi_n(x').

Then the following holds:

G(x, x') = \sum_{n=0}^\infty \frac{\Psi_n(x) \Psi_n(x')}{\lambda_n}.

Applying the operator to each side of this equation results in the completeness relation, which was assumed true.

The general study of the Green’s function written in the above form, and its relationship to the function spaces formed by the eigenvectors, is known as Fredholm theory.

Green’s functions for the Laplacian

Green’s functions for linear differential operators involving the Laplacian may be readily put to use using the second of Green’s identities. To derive Green’s theorem, begin with the divergence theorem (otherwise known as Gauss’s law):

\int_V \nabla \cdot \hat A\ dV = \int_S \hat A \cdot d\hat\sigma.

Let \hat A = \phi\nabla\psi - \psi\nabla\phi and substitute into Gauss’ law. Compute and apply the chain rule for the operator:

\nabla\cdot\hat A = \nabla\cdot(\phi\nabla\psi - \psi\nabla\phi) = (\nabla\phi)\cdot(\nabla\psi) + \phi\nabla^2\psi - (\nabla\phi)\cdot(\nabla\psi) - \psi\nabla^2\phi = \phi\nabla^2\psi - \psi\nabla^2\phi.

Plugging this into the divergence theorem produces Green’s theorem:

\int_V (\phi\nabla^2\psi - \psi\nabla^2\phi) dV = \int_S (\phi\nabla\psi - \psi\nabla\phi)\cdot d\hat\sigma.

Suppose that the linear differential operator is the Laplacian, , and that there is a Green’s function for the Laplacian. The defining property of the Green’s function still holds:

L G(x,x') = \nabla^2 G(x,x') = \delta(x-x').

Let in Green’s theorem. Then:

\int_V \phi(x') \delta(x - x') - G(x,x') \nabla^2\phi(x')\ d^3x' = \int_S \left[\phi(x')\nabla' G(x,x') - G(x,x')\nabla'\phi(x')\right] \cdot d\hat\sigma'.

Using this expression, it is possible to solve Laplace’s equation or Poisson’s equation , subject to either Neumann or Dirichlet boundary conditions. In other words, we can solve for everywhere inside a volume where either (1) the value of is specified on the bounding surface of the volume (Dirichlet boundary conditions), or (2) the normal derivative of is specified on the bounding surface (Neumann boundary conditions).

Suppose the problem is to solve for inside the region. Then the integral \int\limits_V {\phi(x')\delta(x-x')\ d^3x'} reduces to simply due to the defining property of the Dirac delta function and we have:

\phi(x) = \int_V G(x,x') \rho(x')\ d^3x' + \int_S \left[\phi(x')\nabla' G(x,x') - G(x,x')\nabla'\phi(x')\right] \cdot d\hat\sigma'.

This form expresses the well-known property of harmonic functions that if the value or normal derivative is known on a bounding surface, then the value of the function inside the volume is known everywhere. In electrostatics, is interpreted as the electric potential, as electric charge density, and the normal derivative \nabla\phi(x')\cdot d\hat\sigma' as the normal component of the electric field. If the problem is to solve a Dirichlet boundary value problem, the Green’s function should be chosen such that vanishes when either or is on the bounding surface; conversely, if the problem is to solve a Neumann boundary value problem, the Green’s function is chosen such that its normal derivative vanishes on the bounding surface. Thus only one of the two terms in the surface integral remains.

With no boundary conditions, the Green’s function for the Laplacian (Green’s function for the three-variable Laplace equation) is G(\hat x, \hat x') = \frac{1}{|\hat x - \hat x'|}. Supposing that the bounding surface goes out to infinity, and plugging in this expression for the Green’s function, this gives the familiar expression for electric potential in terms of electric charge density (in the CGS unit system) as \phi(\hat x) = \int_V \frac{\rho(x')}{|\hat x - \hat x'|} \ d^3x'.

Example

Given the problem

\begin{matrix}Lu\end{matrix} = u ' ' + u = f( x ) \\ u(0) = 0,\\ u\left(\frac{\pi}{2}\right) = 0.

Find the Green’s function.

First step: The Green’s function for the linear operator at hand is defined as the solution to If , then the delta function gives zero, and the general solution is For , the boundary condition at implies g(0,s) = c_1 \cdot 1 + c_2 \cdot 0 = 0, \quad c_1 = 0. The equation of is skipped because if and . For , the boundary condition at implies
g\left(\frac{\pi}{2},s\right) = c_3 \cdot 0 + c_4 \cdot 1 = 0, \quad c_4 = 0. The equation of is skipped for similar reasons. To summarize the results thus far:

g(x,s)=\left\{\begin{matrix} c_2 \sin x, \;\; x < s \\ c_3 \cos x, \;\; s < x \end{matrix}\right.

Second step: The next task is to determine and . Ensuring continuity in the Green’s function at implies One can also ensure proper discontinuity in the first derivative by integrating the defining differential equation from to and taking the limit as goes to zero c_3 \cdot [ - \sin s ] - c_2 \cdot \cos s = 1. The two (dis)continuity equations can be solved for and to obtain c_2 = - \cos s \quad ; \quad c_3 = - \sin s. So the Green’s function for this problem is:

g(x,s)=\left\{\begin{matrix} -\cos s \cdot \sin x, \;\; x < s, \\ - \sin s \cdot \cos x, \;\; s < x. \end{matrix}\right.

Further examples
* Let and let the subset be all of . Let be . Then, the Heaviside step function is a Green’s function of at .

* Let and let the subset be the quarter-plane and be the Laplacian. Also, assume a Dirichlet boundary condition is imposed at and a Neumann boundary condition is imposed at . Then the Green’s function is

G(x, y, x_0, y_0)=\frac{1}{2\pi}\left[\ln\sqrt{(x-x_0)^2+(y-y_0)^2}-\ln\sqrt{(x+x_0)^2+(y-y_0)^2}\right] +\frac{1}{2\pi}\left[\ln\sqrt{(x-x_0)^2+(y+y_0)^2}-\ln\sqrt{(x+x_0)^2+(y+y_0)^2}\right]

which completes this example.

Source

Tháng Ba 4, 2009

Show that the closure of a convex set also is convex

Chuyên mục: Các Bài Tập Nhỏ, Giải tích 8 (MA5206), Linh Tinh — Ngô Quốc Anh @ 23:29

Today on the way, my Chinese friend and I have discussed the following question: Let denote a convex subset of a locally convex topological linear space . Show that the closure of also is convex.

I have suggested the following solution.

We define f: X \times X \times \mathbb{R} \to X as following f: \left( {x,y,\lambda } \right) \mapsto \lambda x + \left( {1 - \lambda } \right)y. Then is continuous and f\left( {K \times K \times \left[ {0,1} \right]} \right) \subset K since is convex. We then obtain that f\left( {\overline K \times \overline K \times \left[ {0,1} \right]} \right) \subset \overline K due to the continuity of , that is \lambda \overline K + \left( {1 - \lambda } \right)\overline K \in \overline K for every . Therefore is a convex set.

So what I am going to tell you is how correct the solution is? If no, what’s the problem, otherwise, what’s the main point? I will show you a little bit latter. I think I should go for sleep, it’s late now :(

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