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Home / Let V Be the Vector Space of Continuous Fzunctions F R R Show That the Product Defined Below

Let V Be the Vector Space of Continuous Fzunctions F R R Show That the Product Defined Below

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Next: Linear Independence Up: Calculus Revisited Previous: Calculus Revisited

Vector Spaces

Recall the theorem about sums and multiples of continuous functions: If $ f$ and $ g$ are continuous on an interval $ [a,b]$ and $ c$ is any real number, then

$\displaystyle f+g {\text{ is a continuous function on }} [a,b]$

and
$\displaystyle cf {\text{ is a continuous function on }} [a,b]$

This theorem says that the set of functions continuous on $ [a,b]$ is closed under sums and constant multiples. Thus we have a situation which may yield a vector space. With this in mind, we define $ C[a,b]$ as the set of functions continuous on $ [a,b]$ along with addition of functions and constant multiplication of functions defined in the usual way. To see whether $ C[a,b]$ is a vector space, we must check the vector space axioms. Closure holds by the theorem stated above. The properties
are easily checked. The zero vector is the constant function (it is continuous) which is identically zero on $ [a,b]$. Given $ f$, the function $ -f$ is continuous and is the additive inverse of $ f$. Therefore, $ C[a,b]$ is a vector space.

Differentiable functions are another important set of functions in calculus. If we define $ C^1[a,b]$ as the set of functions $ f$ defined on $ [a,b]$ with $ f'\in C[a,b]$ (that is, $ f'$ is continuous) and give $ C^1[a,b]$ the same operations as $ C[a,b]$, then $ C^1[a,b]$ is a subspace of $ C[a,b]$. To see this, we need only check closure. Since a theorem from calculus tells us that the sum and constant multiples of differentiable functions are differentiable, we have the necessary closure. So $ C^1[a,b]$ is a vector space in its own right as well as a subspace of $ C[a,b]$.

Example 1 Give an example of a function in $ C[-1,1]$ which is not in $ C^1[-1,1]$. This shows that $ C^1[-1,1]$ is a proper subspace of $ C[-1,1]$.

Solution A continuous function which is not differentiable will be sufficient. The function $ y=\vert x\vert$ is continuous on $ [-1,1]$ but fails to have a derivative at $ x=0$.

Several functions from calculus are differentiable an infinite number of times. For example, $ y=e^x$, $ y=\sin x$, $ y=\cos x$, and all polynomials are infinitely differentiable over all $ \mathbb{R}$. Thus we can define vector spaces

$\displaystyle C^2[a,b], C^3[a,b],\ldots, C^n[a,b],\ldots$

and even

$\displaystyle C^\infty[a,b]$

Example 2 Give a example of a function which is in $ C^1[-1,1]$ but is not in $ C^2[-1,1]$.

Solution Let $ y=x^{5/3}$. Then $ y' = \frac53 x^{2/3}$ and $ y'' = \frac{10}9x^{-1/3}$. Although $ y'$ is continuous on $ [-1,1], y''$ is not, because $ y''(0)$ is undefined.

Examples 1 and 2 illustrate the subspace relationship

$\displaystyle C[a,b]\supset C^1[a,b]\supset C^2[a,b]\supset\cdots\supset C^\infty[a,b]$

among these vector spaces.

The vector space $ C[a,b]$ differs from the vector spaces we have been studying in that $ C[a,b]$ is not finite-dimensional. To see this, suppose that $ \dim(C[a,b]) = n$. Then a finite basis $ S = \{f_1(x), f_2(x),\ldots, f_n(x)\}$ of continuous functions in $ C[a,b]$ would exist, and any set of $ n+1$ functions would have to be linearly dependent. However, $ T = \{1,x,x^2,\ldots, x^n\}$ is a set of $ n+1$ functions in $ C[a,b]$ which are linearly independent. So it is impossible to have $ \dim(C[a,b]) = n$ for any $ n$. Since polynomials are infinitely differentiable, the same argument shows that $ C^k[a,b]$ is not finite-dimensional for any $ k$.

Although $ C[a,b]$ is not finite-dimensional, there are important finite-dimensional subspaces of $ C[a,b]$.

Example 3 Show that the set of all solutions to the equation3.3

$ y'=\alpha y$ is a one-dimensional subspace of

$ C(\mathbb{R})$

Solution We recall that the functions3.4 satisfying $ y'=\alpha y$ are $ y(x) = Ce^{\alpha x}$, where $ C$ is arbitrary. That is, $ W = \{y\vert y(x) = Ce^{\alpha x}\}$. To show that $ W$ is a subspace of $ C(\mathbb{R})$, we first note that since $ e^x$ is a continuous function, $ W\subseteq C(\mathbb{R})$. Now closure of $ W$ under the operations must be shown. Since $ c_1e^{\alpha x} + c_2e^{\alpha x} = (c_1+c_2)e^{\alpha x} = c_3e^{\alpha x}$ and $ c_1(c_2e^{\alpha x}) = (c_1c_2)e^{\alpha x}$, we have the closure, so $ W$ is a subspace of $ C(\mathbb{R})$. A basis for $ W$ is $ S = \{e^{\alpha x}\}$. Thus $ \dim W=1$.

Example 4 Show that the set of all solutions to the equation

$ y'' = -\alpha^2y$ is a two-dimensional subspace of

$ C(\mathbb{R})$.

Solution The solutions of the equation are of the form $ A\cos \alpha x+B \sin\alpha x$. Showing that $ W = \{y\vert y(x) = A\cos \alpha x + B\sin \alpha x\}$ is a subspace is done in the same way as Example 3. A basis for $ W$ is $ S = \{\cos \alpha x, \sin \alpha x\}$. That $ S$ spans $ W$ is easy. For the linear independence consider $ c_1 \cos \alpha x + c_2\sin\alpha x=0$. The equation must hold for all $ x$. Substitute $ x=0$ to find $ c_1=0$. Substitute $ x=\pi/(2\alpha)$ to obtain $ c_2=0$.

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Next: Linear Independence Up: Calculus Revisited Previous: Calculus Revisited

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Source: https://www.math.tamu.edu/~dallen/DistanceEd/Math640/chapter3/node20.html

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