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Analysis/differentiation help (scroll to post 8)

Trying to show that f(x, y) = xy is continuous using the epsilon delta definition. Take any point p = (a, b) in R^2 then

If ∣∣x−p∣∣<δ||\mathbf{x} -\mathbf{p}|| < \delta, then we need to show that

∣∣f(x,y)−f(a,b)∣∣=∣∣xy−ab∣∣<ϵ||f(x,y) - f(a,b)|| = ||xy - ab|| < \epsilon

Solutions say that ∣∣xy−ab∣∣<=∣x∣∣y−b∣+∣b∣∣x−a∣||xy - ab|| <= |x||y - b| + |b||x - a| which I understand

Then they say that ∣x∣∣y−b∣+∣b∣∣x−a∣<=(∣x∣+∣b∣)∣∣x−p∣∣|x||y - b| + |b||x - a| <= (|x| + |b|)||\mathbf{x} -\mathbf{p}||

Where is that last bit coming from? My understanding is that

∣∣x−p∣∣=(x−a)2+(y−b)2<δ||\mathbf{x} -\mathbf{p}|| = \sqrt{(x-a)^2 + (y - b)^2} < \delta

Can't see how they've thrown it in there.
(edited 12 years ago)
Reply 1
Avatar for nuodai
nuodai
17
Well the inequality x∣y−b∣+b∣x−a∣≤(∣x∣+∣b∣)(∣x−a∣+∣y−b∣)x|y-b| + b|x-a| \le (|x|+|b|)(|x-a|+|y-b|) is fairly clear. You need to try and show that, for r,s<1r,s <1 say, ∣r∣+∣s∣≤r2+s2|r|+|s| \le \sqrt{r^2+s^2}. [That is, ∥(r,s)∥1≤∥(r,s)∥2\lVert (r,s) \rVert_1 \le \lVert (r,s) \rVert_2, using Lp norms.]

EDIT: I'm being an idiot. Ignore this for the moment whilst I work out why I'm being an idiot and how to resolve it.
(edited 12 years ago)
Reply 2
Avatar for Swayum
Swayum
OP
18
Original post by nuodai
Well the inequality x∣y−b∣+b∣x−a∣≤(∣x∣+∣b∣)(∣x−a∣+∣y−b∣)x|y-b| + b|x-a| \le (|x|+|b|)(|x-a|+|y-b|) is fairly clear. You need to try and show that, for r,s<1r,s <1 say, ∣r∣+∣s∣≤r2+s2|r|+|s| \le \sqrt{r^2+s^2}. [That is, ∥(r,s)∥1≤∥(r,s)∥2\lVert (r,s) \rVert_1 \le \lVert (r,s) \rVert_2, using Lp norms.]


*Edit*

Missed your edit, sorry.

*/Edit*

Why would that inequality be true? Surely it's the case that

r2+s2<=r2+s2\sqrt{r^2 + s^2} <= \sqrt{r^2} + \sqrt{s^2}

If you square both sides you get

r2+s2<=r2+s2+2r2s2=r2+s2+2∣r∣∣s∣r^2 + s^2 <= r^2 + s^2 + 2\sqrt{r^2s^2} = r^2 + s^2 + 2|r||s|

i.e. ∣r∣∣s∣>=0|r||s| >= 0

Which is clearly true. So it must be that r2+s2<=r2+s2=∣r∣+∣s∣\sqrt{r^2 + s^2} <= \sqrt{r^2} + \sqrt{s^2} = |r| + |s| surely?
(edited 12 years ago)
Reply 3
Avatar for nuodai
nuodai
17
Original post by Swayum
*Edit*

Missed your edit, sorry.

*/Edit*

Why would that inequality be true? Surely it's the case that

r2+s2<=r2+s2\sqrt{r^2 + s^2} <= \sqrt{r^2} + \sqrt{s^2}

If you square both sides you get

r2+s2<=r2+s2+2r2s2=2∣r∣∣s∣r^2 + s^2 <= r^2 + s^2 + 2\sqrt{r^2s^2} = 2|r||s|

i.e. ∣r∣∣s∣>=0|r||s| >= 0

Which is clearly true. So it must be that r2+s2<=r2+s2=∣r∣+∣s∣\sqrt{r^2 + s^2} <= \sqrt{r^2} + \sqrt{s^2} = |r| + |s| surely?


Indeed; or, geometrically, the sum of the lengths of the two perpendicular sides of a right-angled triangle is greater than or equal to the length of the hypotenuse.

We could use the fact that ∥⋅∥1\lVert \cdot \rVert_1 and ∥⋅∥2\lVert \cdot \rVert_2 are Lipschitz equivalent, and hence your RHS becomes k(∣x∣+∣b∣)∥x−p∥k(|x|+|b|) \lVert \mathbf{x} - \mathbf{p} \rVert for some constant k>0k>0. I'm not sure how your proof continues from there but using this might work.
Reply 4
Avatar for Swayum
Swayum
OP
18
Original post by nuodai
We could use the fact that ∥⋅∥1\lVert \cdot \rVert_1 and ∥⋅∥2\lVert \cdot \rVert_2 are Lipschitz equivalent


No idea what that means :confused: Got any other tricks?
Reply 5
Avatar for Swayum
Swayum
OP
18
I think we can use that |x - a| <= ||x - p|| < d

And |y - b| <= ||x - p|| < d

So |x - a| + |y-b| <= 2||x - p||?
Reply 6
Avatar for DFranklin
DFranklin
18
Original post by Swayum
Trying to show that f(x, y) = xy is continuous using the epsilon delta definition. Take any point p = (a, b) in R^2 then

If ∣∣x−p∣∣<δ||\mathbf{x} -\mathbf{p}|| < \delta, then we need to show that

∣∣f(x,y)−f(a,b)∣∣=∣∣xy−ab∣∣<ϵ||f(x,y) - f(a,b)|| = ||xy - ab|| < \epsilon

Solutions say that ∣∣xy−ab∣∣<=∣x∣∣y−b∣+∣b∣∣x−a∣||xy - ab|| <= |x||y - b| + |b||x - a| which I understand

Then they say that ∣x∣∣y−b∣+∣b∣∣x−a∣<=(∣x∣+∣b∣)∣∣x−p∣∣|x||y - b| + |b||x - a| <= (|x| + |b|)||\mathbf{x} -\mathbf{p}||

Where is that last bit coming from? It follows because ∣y−b∣≤∣∣x−p∣∣|y-b| \leq ||\mathbf{x} -\mathbf{p}|| and ∣x−a∣≤∣∣x−p∣∣|x-a| \leq ||\mathbf{x} -\mathbf{p}||. So sub in for |y-b| and |x-a| on the LHS.

(Edit: looks like you found it yourself while I was struggling with the LaTeX...)
Reply 7
Avatar for Swayum
Swayum
OP
18
Original post by DFranklin
It follows because ∣y−b∣≤∣∣x−p∣∣|y-b| \leq ||\mathbf{x} -\mathbf{p}|| and ∣x−a∣≤∣∣x−p∣∣|x-a| \leq ||\mathbf{x} -\mathbf{p}||. So sub in for |y-b| and |x-a| on the LHS.

(Edit: looks like you found it yourself while I was struggling with the LaTeX...)


Thanks. In the interest of not clogging up the forum with another question, I'll post another one here (having a bad day I think...)

Suppose f: R --> R is differentiable on R and that f'(x) >= K for all x >= N, where K > 0 and N is some real number. Prove that f(x) tends to infinity as x tends to infinity.

Although I have a proof and I understand it, I'm confused as to what's going on. Surely you can have a bounded and increasing function that's differentiable everywhere which doesn't tend to infinity?

I'm struggling to think of an example, but I mean f(x) = 1 - 1/x for x > 0 is differentiable (not on R, but on the domain), increasing and bounded, so tends to 1.

Why does being differentiable on R change things? What's the main point here? Is it that f'(x) doesn't tend to 0 so you're not going to get convergence where as with f(x) = 1 - 1/x, f'(x) does tend to 0 so that eventually the function settles down?
(edited 12 years ago)
Reply 8
Avatar for DFranklin
DFranklin
18
Swayum
..
It's nothing (directly) to do with "differentiable on R" (as opposed to a subset of R).

The difference between your example and the given hypothesis is that they assume you're given K > 0 s.t. f'(x) >= K for all x>=N. This sets a lower bound for f'(x), so you know it can't tend to 0.

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