# Continuity if g(x) = f(x-c)

1. Mar 8, 2009

### H2Pendragon

Suppose $$f:D\rightarrow \Re, c \in \Re$$ and g(x) = f(x-c)

1) What's the Domain of g?

I think it's $$\Re$$, am I right?

2) Suppose that f is continuous at $$a \in D \Leftrightarrow$$ g is continuous at c + a

So far I have this:
($$\Rightarrow$$) Assume f is continuous. Then:
$$\forall \epsilon$$ > 0 $$\exists \delta$$ > 0 such that $$x \in D, 0 < |x-a| < \delta \Rightarrow |f(x) - f(a)| < \epsilon$$.
Let g(x) = f(x-c).

We must prove that $$\forall \epsilon'$$ > 0 $$\exists \delta'$$ > 0 such that $$x \in D, 0 < |x-c-a| < \delta' \Rightarrow |f(x-c) - f(a-c)| < \epsilon'$$

I see the obvious outcome here because we'll have |x - (c + a)| < $$\delta'$$, but I'm confused on how to prove the rest of it so that it solves the first part of the iff statement.

The way I've structured it, I'm pretty sure the ($$\Leftarrow$$) half with come naturally if I can figure out why this other part works.

Can anyone steer me towards the right wording?

Last edited: Mar 8, 2009
2. Mar 9, 2009

### Staff: Mentor

No. The domain of f is some set D. What does the graph of g look like relative to the graph of f?
I think the line immediately above should be: Assume f is continuous at a, where a is in D.
For the stuff above, your LaTeX stuff is somewhat garbled up, and it's difficult to decipher what you're trying to say. I would advise putting one whole statement within a pair of <tex> tags. The last part above should say:
$$x\in D and 0 < |x-c-a| < \delta' \Rightarrow |g(x) - g(a+c)| < \epsilon'$$

You will also need to clear up your confusion about the domains of f and g, which are different.

In the inequality above, with g(x) and g(a+c), use the relationship between f and g. You're given that g(x) = f(x - c), or equivalently, f(x) = g(x + c).

3. Mar 9, 2009

### H2Pendragon

I think you came in while I was fixing the <tex> stuff actually.

Exactly how do you know that f(x) = g(x+c)? I don't remember a lot of my early calculus tricks. And why would that help?

4. Mar 9, 2009

### lanedance

substitution
let u = x-c then x = u-c

g(x) = f(x-c) so
g(u+c) = f(u)

can just re-write this with x's, but may need to be a tiny bit careful which domain each variable comes from

5. Mar 9, 2009

### Staff: Mentor

Well, you know that g(x) = f(x - c), right? So if x = 3 and c = 1, for example, and assuming that 3 - 1 (= 2) is in D, then g(3) = f(3 - 1) = f(2).

In my example, an input to g is 1 larger than an input to f.

It works the same way with your problem. The input to g is c larger than the input to f, so g(x + c) = f(x), as long as all the numbers are in the domain for each function.

Another way to think about it is that the graph of g is the translation to the right (assuming c > 0) by c units of the graph of f. Alternatively, the graph of f is the translation to the left by c units of the graph of g.

You need to show that g is continuous at a + c, so you need to show that |g(x) - g(a + c)| < epsilon'. You need to use what you know about f being continuous at a, and the relationship above is how you can do that.

6. Mar 9, 2009

### H2Pendragon

A guy on the math forums wrote out the solution for me and it's obviously simple now that I see it:

$$\left| {x - a} \right| < \delta \, \Rightarrow \,\left| {\left( {x + c} \right) - \left( {a + c} \right)} \right| < \delta \, \Rightarrow \left| {f(x + c) - f(a + c)} \right|\, < \varepsilon \Rightarrow \,\,\left| {g(x) - g(a)} \right| < \varepsilon$$

but I would of had to have remembered the calculus trick so thank you for jogging that portion of my memory

I'm still confused why the domain of g isn't R if g(x) = f(x -c). Shouldn't that c cause g to be in R?

7. Mar 9, 2009

### lanedance

but the domain of f is D, which is a subset of $$\Re$$, but f is not defined outside this domain

with
g(u+c) = f(u)
the domain of g will be all u+c with $$u \in D$$

so the domain of g is a subset of $$\Re$$, but n t $$\Re$$, it is essentially D shifted by an constant amount c.