# Thread: Continuity of vector space operations

1. ## 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.

2. 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.

3. Originally Posted by raed
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
$\displaystyle \alpha x - \alpha _o x_0 = \alpha _0 \left( {x - x_0 } \right) + \left( {\alpha - \alpha _0 } \right)x_0 + \left( {\alpha - \alpha _0 } \right)\left( {x - x_0 } \right)~?$

4. Originally Posted by girdav
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.

5. Originally Posted by raed
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.

6. Originally Posted by girdav
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.

7. Originally Posted by raed
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

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

$\displaystyle |\left( {\alpha - \alpha _0 } \right)\|x_0\| + | {\alpha - \alpha _0 | \; \| {x - x_0 }\|< \varepsilon$

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