# Thread: how to show the following result using continuity defination

1. ## how to show the following result using continuity defination

Using an appropriate definition of continuity, show that the continuity of f (.) at x is equivalent to

lim h->0 sup|t|<h |f (x + t) ─ f (x)| = 0

2. ## Re: how to show the following result using continuity defination

Hey ujgilani.

Recall that the limit of lim x->a f(x) = f(a) so what you are doing is instead of considering the epsilon delta definition of |f(x) - l| < e where the limit is l, then this limit l is exactly f(a) = l for continuity. If f(x+t) -> f(x) then that expression should be zero since the limit approaches l = f(x) [Recall: the definition of continuity is that the limit is equal to the value of the function).

3. ## Re: how to show the following result using continuity defination

Is this the statement?

$\text{Prop: f is continuous at x } \Longleftrightarrow \lim_{h \to 0+} \left( \underset{|t| < h}{sup} \ \lvert f(x+t) - f(x) \rvert \right) = 0.$

If so, then proof is an $\epsilon - \delta$ proof. To get you started:

$\text{f is continuous at x means } \lim_{y \to x} f(y) = f(x),$

$\text{or equivalently } \lim_{y \to x} (f(y) - f(x)) = 0,$

$\text{or equivalently } \lim_{t \to 0} (f(x+t) - f(x)) = 0,$

$\text{or equivalently } \lim_{t \to 0} |f(x+t) - f(x)| = 0.$

Each of those requires an argument, but is very straightforward. So you can see that the final form there is almost identical to what you're trying to prove, except for that sup (supremum = least upper bound) part.

To prove the proposition, you'll need to prove two things, the $\Rightarrow$ direction, and the $\Leftarrow$ direction.

When you're into the guts of the proof, you'll want to choose (in one direction, though this works for both) the $\delta > 0$ so that

it makes the limit you know $\frac{\epsilon}{2}.$

----------------------------------------------
This won't make any sense until you try to prove it yourself (but perhaps then it will be helpful):

If everything in a set is less than $\frac{\epsilon}{2}$, then the supremum of that set will be less than or equal to $\frac{\epsilon}{2}$, which is less than $\epsilon.$

That will be your decisive observation when proving it in one direction.

In the other direction, you'll observe that if the supremeum of a set is less than $\epsilon$ (or you could use $\frac{\epsilon}{2}$ there too),

then any individual one of those things in that set is less than $\epsilon.$

4. ## Re: how to show the following result using continuity defination

thanks for replies guys .i got it. actually i wanted to post the proof as well but didn't find the way to write post in latex form or the way in which johnsomeone has posted.

5. ## Re: how to show the following result using continuity defination

hi thanks for your replay, but i'm not getting it, i tried solving it using epsilon-delta method but couldn't complete it another unclear thing is "sup". how to deal with it.
as i'm new to this stuff so i need little more help in solving it. I'll appreciate if you help me to understand it.
Originally Posted by johnsomeone
Is this the statement?

$\text{Prop: f is continuous at x } \Longleftrightarrow \lim_{h \to 0+} \left( \underset{|t| < h}{sup} \ \lvert f(x+t) - f(x) \rvert \right) = 0.$

If so, then proof is an $\epsilon - \delta$ proof. To get you started:

$\text{f is continuous at x means } \lim_{y \to x} f(y) = f(x),$

$\text{or equivalently } \lim_{y \to x} (f(y) - f(x)) = 0,$

$\text{or equivalently } \lim_{t \to 0} (f(x+t) - f(x)) = 0,$

$\text{or equivalently } \lim_{t \to 0} |f(x+t) - f(x)| = 0.$

Each of those requires an argument, but is very straightforward. So you can see that the final form there is almost identical to what you're trying to prove, except for that sup (supremum = least upper bound) part.

To prove the proposition, you'll need to prove two things, the $\Rightarrow$ direction, and the $\Leftarrow$ direction.

When you're into the guts of the proof, you'll want to choose (in one direction, though this works for both) the $\delta > 0$ so that

it makes the limit you know $\frac{\epsilon}{2}.$

----------------------------------------------
This won't make any sense until you try to prove it yourself (but perhaps then it will be helpful):

If everything in a set is less than $\frac{\epsilon}{2}$, then the supremum of that set will be less than or equal to $\frac{\epsilon}{2}$, which is less than $\epsilon.$

That will be your decisive observation when proving it in one direction.

In the other direction, you'll observe that if the supremeum of a set is less than $\epsilon$ (or you could use $\frac{\epsilon}{2}$ there too),

then any individual one of those things in that set is less than $\epsilon.$