Proving the sequence $f_{n} = sqrt{x^{2} + 1/n}$ converges uniformly to $f(x) = |x|$ on $(-1, 1)$.
$begingroup$
I have the following exercise from my book:
For each $n in mathbb{N}$ and each $xin (-1, 1),$ define
$$f_{n}(x) = sqrt{x^{2} + frac{1}{n}}$$
and define $f(x) = |x|$. Prove that the sequence ${f_{n}}$ converges
uniformly on the open interval $(-1, 1)$ to the function $f$. Check
that each function $f_{n}$ is continuously differentiable, whereas the
limit function $f$ is not differentiable at $x = 0$.
My attempt:
First, we need to show $forall epsilon > 0$, there exists an index $N$ such that
$$|f(x) - f_{n}(x)| < epsilon $$
for all $n geq N$ and all $x in D$. So, we have
$$left||x| - sqrt{x^2 + 1/n}right|.$$
But since $x^{2} + 1/n > 0$ always, we have
$$left||x| - sqrt{x^2 + 1/n}right| = left||x| - |sqrt{x^{2} + 1/n}|right| leq left|x - sqrt{x^{2} + 1/n}right|,$$
where the last equality follows from the reverse triangle inequality. Then,
$$left|x - sqrt{x^{2} + 1/n}right| leq |x| < epsilon.$$
So, choose $N = lceil{epsilon}rceil + 1$. Does this choice of $N$ work? I don't really know. Can someone help me with the rest of the problem?
real-analysis sequences-and-series convergence uniform-convergence
asked Dec 10 '18 at 1:42
josephjoseph
489111
$endgroup$
add a comment |
$begingroup$
I have the following exercise from my book:
For each $n in mathbb{N}$ and each $xin (-1, 1),$ define
$$f_{n}(x) = sqrt{x^{2} + frac{1}{n}}$$
and define $f(x) = |x|$. Prove that the sequence ${f_{n}}$ converges
uniformly on the open interval $(-1, 1)$ to the function $f$. Check
that each function $f_{n}$ is continuously differentiable, whereas the
limit function $f$ is not differentiable at $x = 0$.
My attempt:
First, we need to show $forall epsilon > 0$, there exists an index $N$ such that
$$|f(x) - f_{n}(x)| < epsilon $$
for all $n geq N$ and all $x in D$. So, we have
$$left||x| - sqrt{x^2 + 1/n}right|.$$
But since $x^{2} + 1/n > 0$ always, we have
$$left||x| - sqrt{x^2 + 1/n}right| = left||x| - |sqrt{x^{2} + 1/n}|right| leq left|x - sqrt{x^{2} + 1/n}right|,$$
where the last equality follows from the reverse triangle inequality. Then,
$$left|x - sqrt{x^{2} + 1/n}right| leq |x| < epsilon.$$
So, choose $N = lceil{epsilon}rceil + 1$. Does this choice of $N$ work? I don't really know. Can someone help me with the rest of the problem?
real-analysis sequences-and-series convergence uniform-convergence
asked Dec 10 '18 at 1:42
josephjoseph
489111
$endgroup$
1
$begingroup$
Why is $|x|<epsilon$? Your choice of $epsilon$ can't depend on $x$. Also, if $x<0$, then $|x-sqrt{x^2+1/n}|>|x|.$ My professor taught me a cool method to solve this problem using the squeeze theorem for uniform convergence. Can you show $|x|leqsqrt{x^2+1/n}leq(|x|+1/n)?$
$endgroup$
– Melody
Dec 10 '18 at 1:50
$begingroup$
I cannot prove both inequalities. Can you help me?
$endgroup$
– joseph
Dec 10 '18 at 1:54
$begingroup$
Consider $|x|^2leq|x|^2+1/nleq|x|^2+2|x|/n+1/n^2$. Now taking square roots gives us the inequality I mentioned earlier.
$endgroup$
– Melody
Dec 10 '18 at 1:59
add a comment |
$begingroup$
I have the following exercise from my book:
For each $n in mathbb{N}$ and each $xin (-1, 1),$ define
$$f_{n}(x) = sqrt{x^{2} + frac{1}{n}}$$
and define $f(x) = |x|$. Prove that the sequence ${f_{n}}$ converges
uniformly on the open interval $(-1, 1)$ to the function $f$. Check
that each function $f_{n}$ is continuously differentiable, whereas the
limit function $f$ is not differentiable at $x = 0$.
My attempt:
First, we need to show $forall epsilon > 0$, there exists an index $N$ such that
$$|f(x) - f_{n}(x)| < epsilon $$
for all $n geq N$ and all $x in D$. So, we have
$$left||x| - sqrt{x^2 + 1/n}right|.$$
But since $x^{2} + 1/n > 0$ always, we have
$$left||x| - sqrt{x^2 + 1/n}right| = left||x| - |sqrt{x^{2} + 1/n}|right| leq left|x - sqrt{x^{2} + 1/n}right|,$$
where the last equality follows from the reverse triangle inequality. Then,
$$left|x - sqrt{x^{2} + 1/n}right| leq |x| < epsilon.$$
So, choose $N = lceil{epsilon}rceil + 1$. Does this choice of $N$ work? I don't really know. Can someone help me with the rest of the problem?
real-analysis sequences-and-series convergence uniform-convergence
asked Dec 10 '18 at 1:42
josephjoseph
489111
$endgroup$
I have the following exercise from my book:
For each $n in mathbb{N}$ and each $xin (-1, 1),$ define
$$f_{n}(x) = sqrt{x^{2} + frac{1}{n}}$$
and define $f(x) = |x|$. Prove that the sequence ${f_{n}}$ converges
uniformly on the open interval $(-1, 1)$ to the function $f$. Check
that each function $f_{n}$ is continuously differentiable, whereas the
limit function $f$ is not differentiable at $x = 0$.
My attempt:
First, we need to show $forall epsilon > 0$, there exists an index $N$ such that
$$|f(x) - f_{n}(x)| < epsilon $$
for all $n geq N$ and all $x in D$. So, we have
$$left||x| - sqrt{x^2 + 1/n}right|.$$
But since $x^{2} + 1/n > 0$ always, we have
$$left||x| - sqrt{x^2 + 1/n}right| = left||x| - |sqrt{x^{2} + 1/n}|right| leq left|x - sqrt{x^{2} + 1/n}right|,$$
where the last equality follows from the reverse triangle inequality. Then,
$$left|x - sqrt{x^{2} + 1/n}right| leq |x| < epsilon.$$
So, choose $N = lceil{epsilon}rceil + 1$. Does this choice of $N$ work? I don't really know. Can someone help me with the rest of the problem?
real-analysis sequences-and-series convergence uniform-convergence
real-analysis sequences-and-series convergence uniform-convergence
asked Dec 10 '18 at 1:42
josephjoseph
489111
asked Dec 10 '18 at 1:42
josephjoseph
489111
asked Dec 10 '18 at 1:42
josephjoseph
489111
asked Dec 10 '18 at 1:42
josephjoseph
489111
asked Dec 10 '18 at 1:42
josephjoseph
489111
489111
1
$begingroup$
Why is $|x|<epsilon$? Your choice of $epsilon$ can't depend on $x$. Also, if $x<0$, then $|x-sqrt{x^2+1/n}|>|x|.$ My professor taught me a cool method to solve this problem using the squeeze theorem for uniform convergence. Can you show $|x|leqsqrt{x^2+1/n}leq(|x|+1/n)?$
$endgroup$
– Melody
Dec 10 '18 at 1:50
$begingroup$
I cannot prove both inequalities. Can you help me?
$endgroup$
– joseph
Dec 10 '18 at 1:54
$begingroup$
Consider $|x|^2leq|x|^2+1/nleq|x|^2+2|x|/n+1/n^2$. Now taking square roots gives us the inequality I mentioned earlier.
$endgroup$
– Melody
Dec 10 '18 at 1:59
add a comment |
1
$begingroup$
Why is $|x|<epsilon$? Your choice of $epsilon$ can't depend on $x$. Also, if $x<0$, then $|x-sqrt{x^2+1/n}|>|x|.$ My professor taught me a cool method to solve this problem using the squeeze theorem for uniform convergence. Can you show $|x|leqsqrt{x^2+1/n}leq(|x|+1/n)?$
$endgroup$
– Melody
Dec 10 '18 at 1:50
$begingroup$
I cannot prove both inequalities. Can you help me?
$endgroup$
– joseph
Dec 10 '18 at 1:54
$begingroup$
Consider $|x|^2leq|x|^2+1/nleq|x|^2+2|x|/n+1/n^2$. Now taking square roots gives us the inequality I mentioned earlier.
$endgroup$
– Melody
Dec 10 '18 at 1:59
1
1
$begingroup$
Why is $|x|<epsilon$? Your choice of $epsilon$ can't depend on $x$. Also, if $x<0$, then $|x-sqrt{x^2+1/n}|>|x|.$ My professor taught me a cool method to solve this problem using the squeeze theorem for uniform convergence. Can you show $|x|leqsqrt{x^2+1/n}leq(|x|+1/n)?$
$endgroup$
– Melody
Dec 10 '18 at 1:50
$begingroup$
Why is $|x|<epsilon$? Your choice of $epsilon$ can't depend on $x$. Also, if $x<0$, then $|x-sqrt{x^2+1/n}|>|x|.$ My professor taught me a cool method to solve this problem using the squeeze theorem for uniform convergence. Can you show $|x|leqsqrt{x^2+1/n}leq(|x|+1/n)?$
$endgroup$
– Melody
Dec 10 '18 at 1:50
$begingroup$
I cannot prove both inequalities. Can you help me?
$endgroup$
– joseph
Dec 10 '18 at 1:54
$begingroup$
I cannot prove both inequalities. Can you help me?
$endgroup$
– joseph
Dec 10 '18 at 1:54
$begingroup$
Consider $|x|^2leq|x|^2+1/nleq|x|^2+2|x|/n+1/n^2$. Now taking square roots gives us the inequality I mentioned earlier.
$endgroup$
– Melody
Dec 10 '18 at 1:59
$begingroup$
Consider $|x|^2leq|x|^2+1/nleq|x|^2+2|x|/n+1/n^2$. Now taking square roots gives us the inequality I mentioned earlier.
$endgroup$
– Melody
Dec 10 '18 at 1:59
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Observe that $$left(vert xvert +frac{1}{sqrt{n}}right)^2=x^2+frac{1}{n}+2 vert x vert frac{1}{sqrt{n}}=f_n^2+2 vert x vert frac{1}{sqrt{n}} geq f_n^2 geq 0$$ so $$0 leq f_n -vert x vertleq frac{1}{sqrt{n}} longrightarrow 0$$
answered Dec 10 '18 at 1:55
Chinnapparaj RChinnapparaj R
5,5072928
$endgroup$
$begingroup$
But doesn't this show pointwise convergence and not uniform convergence?
$endgroup$
– joseph
Dec 10 '18 at 1:55
$begingroup$
it's uniform because it only depends on n
$endgroup$
– user29418
Dec 10 '18 at 1:56
$begingroup$
@Joseph: the convergence is uniform!
$endgroup$
– Chinnapparaj R
Dec 10 '18 at 1:57
$begingroup$
how'd you get the expression after "so"?
$endgroup$
– joseph
Dec 10 '18 at 2:04
1
$begingroup$
computing $f_n'$ is easy and observe each $f_n$ is differentiable and $f_n'$ is continuous, so $f_n$ is continuously differentiable.
$endgroup$
– Chinnapparaj R
Dec 10 '18 at 2:24
|
show 4 more comments
Your Answer
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1 Answer
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1 Answer
1
active
oldest
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active
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active
oldest
votes
$begingroup$
Observe that $$left(vert xvert +frac{1}{sqrt{n}}right)^2=x^2+frac{1}{n}+2 vert x vert frac{1}{sqrt{n}}=f_n^2+2 vert x vert frac{1}{sqrt{n}} geq f_n^2 geq 0$$ so $$0 leq f_n -vert x vertleq frac{1}{sqrt{n}} longrightarrow 0$$
answered Dec 10 '18 at 1:55
Chinnapparaj RChinnapparaj R
5,5072928
$endgroup$
$begingroup$
But doesn't this show pointwise convergence and not uniform convergence?
$endgroup$
– joseph
Dec 10 '18 at 1:55
$begingroup$
it's uniform because it only depends on n
$endgroup$
– user29418
Dec 10 '18 at 1:56
$begingroup$
@Joseph: the convergence is uniform!
$endgroup$
– Chinnapparaj R
Dec 10 '18 at 1:57
$begingroup$
how'd you get the expression after "so"?
$endgroup$
– joseph
Dec 10 '18 at 2:04
1
$begingroup$
computing $f_n'$ is easy and observe each $f_n$ is differentiable and $f_n'$ is continuous, so $f_n$ is continuously differentiable.
$endgroup$
– Chinnapparaj R
Dec 10 '18 at 2:24
|
show 4 more comments
$begingroup$
Observe that $$left(vert xvert +frac{1}{sqrt{n}}right)^2=x^2+frac{1}{n}+2 vert x vert frac{1}{sqrt{n}}=f_n^2+2 vert x vert frac{1}{sqrt{n}} geq f_n^2 geq 0$$ so $$0 leq f_n -vert x vertleq frac{1}{sqrt{n}} longrightarrow 0$$
answered Dec 10 '18 at 1:55
Chinnapparaj RChinnapparaj R
5,5072928
$endgroup$
$begingroup$
But doesn't this show pointwise convergence and not uniform convergence?
$endgroup$
– joseph
Dec 10 '18 at 1:55
$begingroup$
it's uniform because it only depends on n
$endgroup$
– user29418
Dec 10 '18 at 1:56
$begingroup$
@Joseph: the convergence is uniform!
$endgroup$
– Chinnapparaj R
Dec 10 '18 at 1:57
$begingroup$
how'd you get the expression after "so"?
$endgroup$
– joseph
Dec 10 '18 at 2:04
1
$begingroup$
computing $f_n'$ is easy and observe each $f_n$ is differentiable and $f_n'$ is continuous, so $f_n$ is continuously differentiable.
$endgroup$
– Chinnapparaj R
Dec 10 '18 at 2:24
|
show 4 more comments
$begingroup$
Observe that $$left(vert xvert +frac{1}{sqrt{n}}right)^2=x^2+frac{1}{n}+2 vert x vert frac{1}{sqrt{n}}=f_n^2+2 vert x vert frac{1}{sqrt{n}} geq f_n^2 geq 0$$ so $$0 leq f_n -vert x vertleq frac{1}{sqrt{n}} longrightarrow 0$$
answered Dec 10 '18 at 1:55
Chinnapparaj RChinnapparaj R
5,5072928
$endgroup$
Observe that $$left(vert xvert +frac{1}{sqrt{n}}right)^2=x^2+frac{1}{n}+2 vert x vert frac{1}{sqrt{n}}=f_n^2+2 vert x vert frac{1}{sqrt{n}} geq f_n^2 geq 0$$ so $$0 leq f_n -vert x vertleq frac{1}{sqrt{n}} longrightarrow 0$$
answered Dec 10 '18 at 1:55
Chinnapparaj RChinnapparaj R
5,5072928
edited Dec 10 '18 at 2:07
answered Dec 10 '18 at 1:55
Chinnapparaj RChinnapparaj R
5,5072928
answered Dec 10 '18 at 1:55
Chinnapparaj RChinnapparaj R
5,5072928
answered Dec 10 '18 at 1:55
Chinnapparaj RChinnapparaj R
5,5072928
5,5072928
$begingroup$
But doesn't this show pointwise convergence and not uniform convergence?
$endgroup$
– joseph
Dec 10 '18 at 1:55
$begingroup$
it's uniform because it only depends on n
$endgroup$
– user29418
Dec 10 '18 at 1:56
$begingroup$
@Joseph: the convergence is uniform!
$endgroup$
– Chinnapparaj R
Dec 10 '18 at 1:57
$begingroup$
how'd you get the expression after "so"?
$endgroup$
– joseph
Dec 10 '18 at 2:04
1
$begingroup$
computing $f_n'$ is easy and observe each $f_n$ is differentiable and $f_n'$ is continuous, so $f_n$ is continuously differentiable.
$endgroup$
– Chinnapparaj R
Dec 10 '18 at 2:24
|
show 4 more comments
$begingroup$
But doesn't this show pointwise convergence and not uniform convergence?
$endgroup$
– joseph
Dec 10 '18 at 1:55
$begingroup$
it's uniform because it only depends on n
$endgroup$
– user29418
Dec 10 '18 at 1:56
$begingroup$
@Joseph: the convergence is uniform!
$endgroup$
– Chinnapparaj R
Dec 10 '18 at 1:57
$begingroup$
how'd you get the expression after "so"?
$endgroup$
– joseph
Dec 10 '18 at 2:04
1
$begingroup$
computing $f_n'$ is easy and observe each $f_n$ is differentiable and $f_n'$ is continuous, so $f_n$ is continuously differentiable.
$endgroup$
– Chinnapparaj R
Dec 10 '18 at 2:24
$begingroup$
But doesn't this show pointwise convergence and not uniform convergence?
$endgroup$
– joseph
Dec 10 '18 at 1:55
$begingroup$
But doesn't this show pointwise convergence and not uniform convergence?
$endgroup$
– joseph
Dec 10 '18 at 1:55
$begingroup$
it's uniform because it only depends on n
$endgroup$
– user29418
Dec 10 '18 at 1:56
$begingroup$
it's uniform because it only depends on n
$endgroup$
– user29418
Dec 10 '18 at 1:56
$begingroup$
@Joseph: the convergence is uniform!
$endgroup$
– Chinnapparaj R
Dec 10 '18 at 1:57
$begingroup$
@Joseph: the convergence is uniform!
$endgroup$
– Chinnapparaj R
Dec 10 '18 at 1:57
$begingroup$
how'd you get the expression after "so"?
$endgroup$
– joseph
Dec 10 '18 at 2:04
$begingroup$
how'd you get the expression after "so"?
$endgroup$
– joseph
Dec 10 '18 at 2:04
1
1
$begingroup$
computing $f_n'$ is easy and observe each $f_n$ is differentiable and $f_n'$ is continuous, so $f_n$ is continuously differentiable.
$endgroup$
– Chinnapparaj R
Dec 10 '18 at 2:24
$begingroup$
computing $f_n'$ is easy and observe each $f_n$ is differentiable and $f_n'$ is continuous, so $f_n$ is continuously differentiable.
$endgroup$
– Chinnapparaj R
Dec 10 '18 at 2:24
|
show 4 more comments
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$begingroup$
Why is $|x|<epsilon$? Your choice of $epsilon$ can't depend on $x$. Also, if $x<0$, then $|x-sqrt{x^2+1/n}|>|x|.$ My professor taught me a cool method to solve this problem using the squeeze theorem for uniform convergence. Can you show $|x|leqsqrt{x^2+1/n}leq(|x|+1/n)?$
$endgroup$
– Melody
Dec 10 '18 at 1:50
$begingroup$
I cannot prove both inequalities. Can you help me?
$endgroup$
– joseph
Dec 10 '18 at 1:54
$begingroup$
Consider $|x|^2leq|x|^2+1/nleq|x|^2+2|x|/n+1/n^2$. Now taking square roots gives us the inequality I mentioned earlier.
$endgroup$
– Melody
Dec 10 '18 at 1:59