Proving uniform continuity of $f(x) = sqrt{1-x^2}$ on $[-1,1]$
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I want to prove that $f(x) = sqrt{1-x^2}$ is uniform continuous on the interval $[-1,1]$. Let $f(x) = sqrt{1 - x^2}$. Then I need to show:
$forall epsilon > 0 enspace exists delta > 0 enspace forall x,y in [-1,1] enspace |x-y|<delta implies |f(x)-f(y)| < epsilon$.
Steps:
$|f(x)-f(y)| = |sqrt{1-x^2} - sqrt{1-y^2}|
= bigrlvertfrac{(sqrt{1-x^2}-sqrt{1-y^2})(sqrt{1-x^2}+sqrt{1-y^2})}{(sqrt{1-x^2}+sqrt{1-y^2})} bigrrvert = bigrlvert frac{1-x^2-(1-y^2)}{sqrt{1-x^2}+sqrt{1-y^2}} bigrrvert = bigrlvert frac{x^2-y^2}{sqrt{1-x^2}+sqrt{1-y^2}} bigrlvert = frac{|x+y||x-y|}{sqrt{1-x^2}+sqrt{1-y^2}} < frac{2delta}{sqrt{1-x^2}+sqrt{1-y^2}} leq frac{2delta}{sqrt{2 - x^2 - y^2}} $
The problem is that I can't bound this by constant as $sqrt{2-x^2-y^2} rightarrow 0$ as $x,y rightarrow1$. I feel I'm running out of tricks. How can one solve this?
analysis uniform-continuity
edited Jan 17 at 21:49
RRL
53.7k52675
asked Jan 17 at 20:09
TedTed
858
$endgroup$
add a comment |
$begingroup$
I want to prove that $f(x) = sqrt{1-x^2}$ is uniform continuous on the interval $[-1,1]$. Let $f(x) = sqrt{1 - x^2}$. Then I need to show:
$forall epsilon > 0 enspace exists delta > 0 enspace forall x,y in [-1,1] enspace |x-y|<delta implies |f(x)-f(y)| < epsilon$.
Steps:
$|f(x)-f(y)| = |sqrt{1-x^2} - sqrt{1-y^2}|
= bigrlvertfrac{(sqrt{1-x^2}-sqrt{1-y^2})(sqrt{1-x^2}+sqrt{1-y^2})}{(sqrt{1-x^2}+sqrt{1-y^2})} bigrrvert = bigrlvert frac{1-x^2-(1-y^2)}{sqrt{1-x^2}+sqrt{1-y^2}} bigrrvert = bigrlvert frac{x^2-y^2}{sqrt{1-x^2}+sqrt{1-y^2}} bigrlvert = frac{|x+y||x-y|}{sqrt{1-x^2}+sqrt{1-y^2}} < frac{2delta}{sqrt{1-x^2}+sqrt{1-y^2}} leq frac{2delta}{sqrt{2 - x^2 - y^2}} $
The problem is that I can't bound this by constant as $sqrt{2-x^2-y^2} rightarrow 0$ as $x,y rightarrow1$. I feel I'm running out of tricks. How can one solve this?
analysis uniform-continuity
edited Jan 17 at 21:49
RRL
53.7k52675
asked Jan 17 at 20:09
TedTed
858
$endgroup$
2
$begingroup$
There is a general result that you can use here: any continuous function on a compact interval is uniformly continuous.
$endgroup$
– GReyes
Jan 17 at 20:12
add a comment |
$begingroup$
I want to prove that $f(x) = sqrt{1-x^2}$ is uniform continuous on the interval $[-1,1]$. Let $f(x) = sqrt{1 - x^2}$. Then I need to show:
$forall epsilon > 0 enspace exists delta > 0 enspace forall x,y in [-1,1] enspace |x-y|<delta implies |f(x)-f(y)| < epsilon$.
Steps:
$|f(x)-f(y)| = |sqrt{1-x^2} - sqrt{1-y^2}|
= bigrlvertfrac{(sqrt{1-x^2}-sqrt{1-y^2})(sqrt{1-x^2}+sqrt{1-y^2})}{(sqrt{1-x^2}+sqrt{1-y^2})} bigrrvert = bigrlvert frac{1-x^2-(1-y^2)}{sqrt{1-x^2}+sqrt{1-y^2}} bigrrvert = bigrlvert frac{x^2-y^2}{sqrt{1-x^2}+sqrt{1-y^2}} bigrlvert = frac{|x+y||x-y|}{sqrt{1-x^2}+sqrt{1-y^2}} < frac{2delta}{sqrt{1-x^2}+sqrt{1-y^2}} leq frac{2delta}{sqrt{2 - x^2 - y^2}} $
The problem is that I can't bound this by constant as $sqrt{2-x^2-y^2} rightarrow 0$ as $x,y rightarrow1$. I feel I'm running out of tricks. How can one solve this?
analysis uniform-continuity
edited Jan 17 at 21:49
RRL
53.7k52675
asked Jan 17 at 20:09
TedTed
858
$endgroup$
I want to prove that $f(x) = sqrt{1-x^2}$ is uniform continuous on the interval $[-1,1]$. Let $f(x) = sqrt{1 - x^2}$. Then I need to show:
$forall epsilon > 0 enspace exists delta > 0 enspace forall x,y in [-1,1] enspace |x-y|<delta implies |f(x)-f(y)| < epsilon$.
Steps:
$|f(x)-f(y)| = |sqrt{1-x^2} - sqrt{1-y^2}|
= bigrlvertfrac{(sqrt{1-x^2}-sqrt{1-y^2})(sqrt{1-x^2}+sqrt{1-y^2})}{(sqrt{1-x^2}+sqrt{1-y^2})} bigrrvert = bigrlvert frac{1-x^2-(1-y^2)}{sqrt{1-x^2}+sqrt{1-y^2}} bigrrvert = bigrlvert frac{x^2-y^2}{sqrt{1-x^2}+sqrt{1-y^2}} bigrlvert = frac{|x+y||x-y|}{sqrt{1-x^2}+sqrt{1-y^2}} < frac{2delta}{sqrt{1-x^2}+sqrt{1-y^2}} leq frac{2delta}{sqrt{2 - x^2 - y^2}} $
The problem is that I can't bound this by constant as $sqrt{2-x^2-y^2} rightarrow 0$ as $x,y rightarrow1$. I feel I'm running out of tricks. How can one solve this?
analysis uniform-continuity
analysis uniform-continuity
edited Jan 17 at 21:49
RRL
53.7k52675
asked Jan 17 at 20:09
TedTed
858
edited Jan 17 at 21:49
RRL
53.7k52675
asked Jan 17 at 20:09
TedTed
858
edited Jan 17 at 21:49
RRL
53.7k52675
edited Jan 17 at 21:49
RRL
53.7k52675
edited Jan 17 at 21:49
RRL
53.7k52675
53.7k52675
asked Jan 17 at 20:09
TedTed
858
asked Jan 17 at 20:09
TedTed
858
asked Jan 17 at 20:09
TedTed
858
858
2
$begingroup$
There is a general result that you can use here: any continuous function on a compact interval is uniformly continuous.
$endgroup$
– GReyes
Jan 17 at 20:12
add a comment |
2
$begingroup$
There is a general result that you can use here: any continuous function on a compact interval is uniformly continuous.
$endgroup$
– GReyes
Jan 17 at 20:12
2
2
$begingroup$
There is a general result that you can use here: any continuous function on a compact interval is uniformly continuous.
$endgroup$
– GReyes
Jan 17 at 20:12
$begingroup$
There is a general result that you can use here: any continuous function on a compact interval is uniformly continuous.
$endgroup$
– GReyes
Jan 17 at 20:12
add a comment |
1 Answer
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Hint:
$$|sqrt{1-x^2} - sqrt{1-y^2}|^2 leqslant |sqrt{1-x^2} - sqrt{1-y^2}||sqrt{1-x^2} + sqrt{1-y^2}| = |x^2 - y^2|$$
answered Jan 17 at 20:23
RRLRRL
53.7k52675
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add a comment |
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$begingroup$
Hint:
$$|sqrt{1-x^2} - sqrt{1-y^2}|^2 leqslant |sqrt{1-x^2} - sqrt{1-y^2}||sqrt{1-x^2} + sqrt{1-y^2}| = |x^2 - y^2|$$
answered Jan 17 at 20:23
RRLRRL
53.7k52675
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add a comment |
$begingroup$
Hint:
$$|sqrt{1-x^2} - sqrt{1-y^2}|^2 leqslant |sqrt{1-x^2} - sqrt{1-y^2}||sqrt{1-x^2} + sqrt{1-y^2}| = |x^2 - y^2|$$
answered Jan 17 at 20:23
RRLRRL
53.7k52675
$endgroup$
add a comment |
$begingroup$
Hint:
$$|sqrt{1-x^2} - sqrt{1-y^2}|^2 leqslant |sqrt{1-x^2} - sqrt{1-y^2}||sqrt{1-x^2} + sqrt{1-y^2}| = |x^2 - y^2|$$
answered Jan 17 at 20:23
RRLRRL
53.7k52675
$endgroup$
Hint:
$$|sqrt{1-x^2} - sqrt{1-y^2}|^2 leqslant |sqrt{1-x^2} - sqrt{1-y^2}||sqrt{1-x^2} + sqrt{1-y^2}| = |x^2 - y^2|$$
answered Jan 17 at 20:23
RRLRRL
53.7k52675
answered Jan 17 at 20:23
RRLRRL
53.7k52675
answered Jan 17 at 20:23
RRLRRL
53.7k52675
answered Jan 17 at 20:23
RRLRRL
53.7k52675
53.7k52675
add a comment |
add a comment |
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There is a general result that you can use here: any continuous function on a compact interval is uniformly continuous.
$endgroup$
– GReyes
Jan 17 at 20:12