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Thread: Continuity of vector space operations

  1. #1
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    Continuity of vector space operations

    Dear Colleagues,

    I need the solution of the following problem:

    Show that in a normed space $\displaystyle X$, vector addition and multiplication by scalars are continuous operations with respect to the norm; that is, the mappings defined by $\displaystyle (x,y)\longmapsto x+y$ and $\displaystyle (\alpha,x)\longmapsto \alpha x$ are continuous.

    This problem in "Introductory Functional Analysis with applications, Erwin Kreyszig" problem 4 page 70.

    Regards,

    Raed.
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  2. #2
    Super Member girdav's Avatar
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    Let $\displaystyle (x_0,y_0)\in X\times X$. We have $\displaystyle ||x+y -(x_0+y_0)||\leq ||x-x_0||+||y-y_0||$. Now you can use the definition of continuity.
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  3. #3
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    Quote Originally Posted by raed View Post
    Show that in a normed space $\displaystyle X$, vector addition and multiplication by scalars are continuous operations with respect to the norm; that is, the mappings defined by $\displaystyle (x,y)\longmapsto x+y$ and $\displaystyle (\alpha,x)\longmapsto \alpha x$ are continuous.
    Can you show that
    $\displaystyle \alpha x - \alpha _o x_0 = \alpha _0 \left( {x - x_0 } \right) + \left( {\alpha - \alpha _0 } \right)x_0 + \left( {\alpha - \alpha _0 } \right)\left( {x - x_0 } \right)~?$
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  4. #4
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    Quote Originally Posted by girdav View Post
    Let $\displaystyle (x_0,y_0)\in X\times X$. We have $\displaystyle ||x+y -(x_0+y_0)||\leq ||x-x_0||+||y-y_0||$. Now you can use the definition of continuity.
    Thank you for reply, but my problem is how $\displaystyle ||(x,y) -(x_0,y_0)||< \delta $ how can be interpreted using norm.

    Regards.
    Last edited by raed; Mar 21st 2011 at 03:42 AM.
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  5. #5
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    Quote Originally Posted by raed View Post
    Thank you for reply, but my problem is how $\displaystyle ||(x,y) -(x_0,y_0)||< \delta $ can be interpreted using norm.
    Can you please explain what mean by "can be interpreted using norm."

    If you are saying that you do not understand what a norm is or how to use the concepts then you should understand that this is not a tutorial service. So you need to sit down with an instructor.
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  6. #6
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    Quote Originally Posted by girdav View Post
    Let $\displaystyle (x_0,y_0)\in X\times X$. We have $\displaystyle ||x+y -(x_0+y_0)||\leq ||x-x_0||+||y-y_0||$. Now you can use the definition of continuity.
    Thank you for your reply. what does $\displaystyle ||(x,y) -(x_0,y_0)|| $ equal? and what its relation to
    $\displaystyle ||x-x_0||$ and to $\displaystyle ||y-y_0||$.

    Regards.
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  7. #7
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    Quote Originally Posted by raed View Post
    Thank you for your reply. what does $\displaystyle ||(x,y) -(x_0,y_0)|| $ equal? and what its relation to
    $\displaystyle ||x-x_0||$ and to $\displaystyle ||y-y_0||$.
    I will show you part #2. But first look at reply #3.
    Be sure that you can show that is true.

    We show that $\displaystyle (\alpha,x)\mapsto \alpha x$ is continuous.
    Suppose that $\displaystyle \alpha_0$ is a scalar and $\displaystyle x_0$ is a point in the normed space.
    We start the proof the way we start each continuity proof.
    Suppose that $\displaystyle \varepsilon > 0$. Now let $\displaystyle \delta= \min \left\{ {1,\frac{\varepsilon }{{4\left( {\left| {\alpha _0 } \right| + 1} \right)}},\frac{\varepsilon }{{4\left( {\left\| {x_0 } \right\| + 1} \right)}}} \right\}$
    If we have $\displaystyle |\alpha-\alpha_0|<\delta$ and $\displaystyle \|x-x_0\|<\delta$ then

    $\displaystyle \|\alpha x -\alpha _o x_0\|\le |\alpha _0|\| \left( {x - x_0 } \right)\| +$

    $\displaystyle |\left( {\alpha - \alpha _0 } \right)\|x_0\| + | {\alpha - \alpha _0 | \; \| {x - x_0 }\|< \varepsilon$
    Last edited by Plato; Mar 21st 2011 at 09:59 AM.
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