Proving Continuity of a Function Using Epsilon-Delta Definition

[Polter19]

Homework Statement

Hey guys, I've been given the following $\epsilon - \delta$ proof question. The trouble I'm having is I'm not 100% sure what it is actually asking for, and how to go about it. The more common proofs involving two variable functions and a given limit are easy enough to do, but I'm stuck on this one.

Homework Equations

Suppose $f: \left( \mathbb R ^2 , \| . \|_2 \right) \rightarrow \left( \mathbb R , | .| \right)$ is continuous at $\left(a,b \right) \in \mathbb R^2$. Prove using the $\epsilon - \delta$ definition only, that if we define the function

$f_b : \left( \mathbb R,|.| \right) \rightarrow \left( \mathbb R,|.| \right)$ s.t. $f_b \left(x \right) = f \left( x,b \right),$

then $f_b$ is continuous at $x=a$.

The Attempt at a Solution

Well, considering I am not actually sure what the question is acting for I can't go very far. I was assuming you would set up the usual proof as,

Let $\epsilon > 0$
if $|(x,b) - (a,b)| < \delta$
then $|f(x,b) - (a,b)| < \epsilon$

Possibly then, $x - a < \delta$?

Any help would be appreciated, cheers.

 

[micromass]

Homework Statement

Hey guys, I've been given the following $\epsilon - \delta$ proof question. The trouble I'm having is I'm not 100% sure what it is actually asking for, and how to go about it. The more common proofs involving two variable functions and a given limit are easy enough to do, but I'm stuck on this one.

Homework Equations

Suppose $f: \left( \mathbb R ^2 , \| . \|_2 \right) \rightarrow \left( \mathbb R , | .| \right)$ is continuous at $\left(a,b \right) \in \mathbb R^2$. Prove using the $\epsilon - \delta$ definition only, that if we define the function

$f_b : \left( \mathbb R,|.| \right) \rightarrow \left( \mathbb R,|.| \right)$ s.t. $f_b \left(x \right) = f \left( x,b \right),$

then $f_b$ is continuous at $x=a$.

The Attempt at a Solution

Well, considering I am not actually sure what the question is acting for I can't go very far. I was assuming you would set up the usual proof as,

Let $\epsilon > 0$
if $|(x,b) - (a,b)| < \delta$
then $|f(x,b) - (a,b)| < \epsilon$

Possibly then, $x - a < \delta$?

Any help would be appreciated, cheers.

Take $\varepsilon >0$ fixed.

You know that f is continuous at (a,b), thus you know that there exists a $\delta>0$ such that for all (x,y) it holds that

$$\|(x,y)-(a,b)\|_2<\delta~\Rightarrow~|f(x,y)-f(a,b)|<\varepsilon$$

Now, what you must do is to show $f_b$ continuous at a. Thus you must find a $\delta>0$ such that for all x it holds that

$$|x-a|<\delta~\Rightarrow~|f(x,b)-f(a,b)|<\varepsilon$$

Can you proceed now?

 

[jj22vw25]

I am also interested in this question. I understand what is required for the proof, but I am just stuck as to how to manipulate:
|f(x,b) - f(a,b)|
into something that resembles δ??

We know that:
0 < ||(x,y) - (a,b) || < δ $\Rightarrow$ |f(x,y) - f(a,b)| < ε
as the limit is equal to the value of the function at (a,b) since it is continuous there, but how can we follow on from this to show that if y is restricted to equalling b, then:
0 < ||(x,b) - (a,b) || = ||x-a|| < δ $\Rightarrow$ |f(x,b) - f(a,b)| < ε
?? I'm really stuck and any help would be appreciated.

 

[HallsofIvy]

I am also interested in this question. I understand what is required for the proof, but I am just stuck as to how to manipulate:
|f(x,b) - f(a,b)|
into something that resembles δ??

We know that:
0 < ||(x,y) - (a,b) || < δ $\Rightarrow$ |f(x,y) - f(a,b)| < ε
as the limit is equal to the value of the function at (a,b) since it is continuous there, but how can we follow on from this to show that if y is restricted to equalling b, then:
0 < ||(x,b) - (a,b) || = ||x-a|| < δ $\Rightarrow$ |f(x,b) - f(a,b)| < ε
?? I'm really stuck and any help would be appreciated.

How is ||(x, y)- (a, b)|| defined? (there are several equivalent definitions- which are you using?)

 

[jj22vw25]

||(x,y) - (a,b)|| is a norm measuring the distance from (a,b) to (x,y), what do you mean how is it defined? Distance between points in R^2, euclidean norm.