How do I find $text{sup}{|f_n(x)|}$?












0














Given $f_n : [0, infty) mapsto mathbb{R}$ a sequence of functions, defined by $$f_{n}(x) = frac{x}{n^2} e^{frac{-x}{n}} $$



for $n geq 1$



I am trying to prove that $f_n$ converges uniformly to the zero function by showing that $$delta_{n} := text{sup}{|f_n(x)| : x in [0, infty) }
longrightarrow 0$$
but I don't understand $delta_{n}$ very well. Does it depend on $n$? How do I actually find it?






real-analysis uniform-convergence







share|cite|improve this question






















  • You don't need to find $delta_n$ explicitly. It would suffice to find constants $C_n$ such that $C_n$ tends to zero and $delta_n leq C_n$ for each $n$
    – D_S
    Nov 24 at 21:41










  • So I am trying to find a sequence $a_{n}$ that tends to zero, such that $|f_n(x)| < a_{n}$ for all $x$ ?
    – maa'our
    Nov 24 at 21:47










  • Yes, that would do it for you, right?
    – D_S
    Nov 24 at 21:48










  • You can show using calculus that $f_n(x)$ has a global maximum at $x=n$, thus $suplimits_{xin [0,infty)} |f_n(x)|=suplimits_{xin [0,infty)} f_n(x)=frac{1}{ecdot n}$.
    – projectilemotion
    Nov 24 at 21:49


















0














Given $f_n : [0, infty) mapsto mathbb{R}$ a sequence of functions, defined by $$f_{n}(x) = frac{x}{n^2} e^{frac{-x}{n}} $$



for $n geq 1$



I am trying to prove that $f_n$ converges uniformly to the zero function by showing that $$delta_{n} := text{sup}{|f_n(x)| : x in [0, infty) }
longrightarrow 0$$
but I don't understand $delta_{n}$ very well. Does it depend on $n$? How do I actually find it?






real-analysis uniform-convergence







share|cite|improve this question






















  • You don't need to find $delta_n$ explicitly. It would suffice to find constants $C_n$ such that $C_n$ tends to zero and $delta_n leq C_n$ for each $n$
    – D_S
    Nov 24 at 21:41










  • So I am trying to find a sequence $a_{n}$ that tends to zero, such that $|f_n(x)| < a_{n}$ for all $x$ ?
    – maa'our
    Nov 24 at 21:47










  • Yes, that would do it for you, right?
    – D_S
    Nov 24 at 21:48










  • You can show using calculus that $f_n(x)$ has a global maximum at $x=n$, thus $suplimits_{xin [0,infty)} |f_n(x)|=suplimits_{xin [0,infty)} f_n(x)=frac{1}{ecdot n}$.
    – projectilemotion
    Nov 24 at 21:49
















0












0








0


1





Given $f_n : [0, infty) mapsto mathbb{R}$ a sequence of functions, defined by $$f_{n}(x) = frac{x}{n^2} e^{frac{-x}{n}} $$



for $n geq 1$



I am trying to prove that $f_n$ converges uniformly to the zero function by showing that $$delta_{n} := text{sup}{|f_n(x)| : x in [0, infty) }
longrightarrow 0$$
but I don't understand $delta_{n}$ very well. Does it depend on $n$? How do I actually find it?






real-analysis uniform-convergence







share|cite|improve this question













Given $f_n : [0, infty) mapsto mathbb{R}$ a sequence of functions, defined by $$f_{n}(x) = frac{x}{n^2} e^{frac{-x}{n}} $$



for $n geq 1$



I am trying to prove that $f_n$ converges uniformly to the zero function by showing that $$delta_{n} := text{sup}{|f_n(x)| : x in [0, infty) }
longrightarrow 0$$
but I don't understand $delta_{n}$ very well. Does it depend on $n$? How do I actually find it?






real-analysis uniform-convergence




real-analysis uniform-convergence






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Nov 24 at 21:38









maa'our

604




604












  • You don't need to find $delta_n$ explicitly. It would suffice to find constants $C_n$ such that $C_n$ tends to zero and $delta_n leq C_n$ for each $n$
    – D_S
    Nov 24 at 21:41










  • So I am trying to find a sequence $a_{n}$ that tends to zero, such that $|f_n(x)| < a_{n}$ for all $x$ ?
    – maa'our
    Nov 24 at 21:47










  • Yes, that would do it for you, right?
    – D_S
    Nov 24 at 21:48










  • You can show using calculus that $f_n(x)$ has a global maximum at $x=n$, thus $suplimits_{xin [0,infty)} |f_n(x)|=suplimits_{xin [0,infty)} f_n(x)=frac{1}{ecdot n}$.
    – projectilemotion
    Nov 24 at 21:49




















  • You don't need to find $delta_n$ explicitly. It would suffice to find constants $C_n$ such that $C_n$ tends to zero and $delta_n leq C_n$ for each $n$
    – D_S
    Nov 24 at 21:41










  • So I am trying to find a sequence $a_{n}$ that tends to zero, such that $|f_n(x)| < a_{n}$ for all $x$ ?
    – maa'our
    Nov 24 at 21:47










  • Yes, that would do it for you, right?
    – D_S
    Nov 24 at 21:48










  • You can show using calculus that $f_n(x)$ has a global maximum at $x=n$, thus $suplimits_{xin [0,infty)} |f_n(x)|=suplimits_{xin [0,infty)} f_n(x)=frac{1}{ecdot n}$.
    – projectilemotion
    Nov 24 at 21:49


















You don't need to find $delta_n$ explicitly. It would suffice to find constants $C_n$ such that $C_n$ tends to zero and $delta_n leq C_n$ for each $n$
– D_S
Nov 24 at 21:41




You don't need to find $delta_n$ explicitly. It would suffice to find constants $C_n$ such that $C_n$ tends to zero and $delta_n leq C_n$ for each $n$
– D_S
Nov 24 at 21:41












So I am trying to find a sequence $a_{n}$ that tends to zero, such that $|f_n(x)| < a_{n}$ for all $x$ ?
– maa'our
Nov 24 at 21:47




So I am trying to find a sequence $a_{n}$ that tends to zero, such that $|f_n(x)| < a_{n}$ for all $x$ ?
– maa'our
Nov 24 at 21:47












Yes, that would do it for you, right?
– D_S
Nov 24 at 21:48




Yes, that would do it for you, right?
– D_S
Nov 24 at 21:48












You can show using calculus that $f_n(x)$ has a global maximum at $x=n$, thus $suplimits_{xin [0,infty)} |f_n(x)|=suplimits_{xin [0,infty)} f_n(x)=frac{1}{ecdot n}$.
– projectilemotion
Nov 24 at 21:49






You can show using calculus that $f_n(x)$ has a global maximum at $x=n$, thus $suplimits_{xin [0,infty)} |f_n(x)|=suplimits_{xin [0,infty)} f_n(x)=frac{1}{ecdot n}$.
– projectilemotion
Nov 24 at 21:49












1 Answer
1






active

oldest

votes


















2














You need to show that the supremum tends to $0$, not to necessarily find the supremum exactly. Namely, it is sufficient to find $(a_n)_n$ such that
$$
forall ngeq 1,qquad sup_{xin[0,infty)} lvert f_n(x)rvert leq a_n
$$
and $lim_{ntoinfty} a_n =0$.



Therefore, let's try and do that here: for all $xgeq 0$ and $ngeq 1$,
$$
0 leq frac{x}{n^2} e^{-frac{x}{n}} = frac{1}{n}cdot frac{x}{n} e^{-frac{x}{n}} leq frac{1}{n}cdot sup_{tin[0,infty) } t e^{-t} stackrel{rm def}{=} a_n
$$
(can you see why?) But it is easy to study the function $h(t) = t e^{-t}$: it is differentiable, and $h'(t) = e^{-t}(1-t)$, and it is near-immediate to conclude it achieves its maximum at $t=1$, where $h(1) = 1/e$. Therefore, in light of the above you have
$$
forall ngeq 1,qquad sup_{xin[0,infty)} lvert f_n(x)rvert leq frac{1}{en}
$$
and you can conclude.




Note: in this case, it so happens that the $a_n$ we found is actually equal to $sup_{xin[0,infty)} lvert f_n(x)rvert$.






share|cite|improve this answer





















  • Thank you! that's really helpful :)
    – maa'our
    Nov 24 at 21:56










  • @maa'our You're welcome -- glad this helped!
    – Clement C.
    Nov 24 at 21:57











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1 Answer
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1 Answer
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active

oldest

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active

oldest

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2














You need to show that the supremum tends to $0$, not to necessarily find the supremum exactly. Namely, it is sufficient to find $(a_n)_n$ such that
$$
forall ngeq 1,qquad sup_{xin[0,infty)} lvert f_n(x)rvert leq a_n
$$
and $lim_{ntoinfty} a_n =0$.



Therefore, let's try and do that here: for all $xgeq 0$ and $ngeq 1$,
$$
0 leq frac{x}{n^2} e^{-frac{x}{n}} = frac{1}{n}cdot frac{x}{n} e^{-frac{x}{n}} leq frac{1}{n}cdot sup_{tin[0,infty) } t e^{-t} stackrel{rm def}{=} a_n
$$
(can you see why?) But it is easy to study the function $h(t) = t e^{-t}$: it is differentiable, and $h'(t) = e^{-t}(1-t)$, and it is near-immediate to conclude it achieves its maximum at $t=1$, where $h(1) = 1/e$. Therefore, in light of the above you have
$$
forall ngeq 1,qquad sup_{xin[0,infty)} lvert f_n(x)rvert leq frac{1}{en}
$$
and you can conclude.




Note: in this case, it so happens that the $a_n$ we found is actually equal to $sup_{xin[0,infty)} lvert f_n(x)rvert$.






share|cite|improve this answer





















  • Thank you! that's really helpful :)
    – maa'our
    Nov 24 at 21:56










  • @maa'our You're welcome -- glad this helped!
    – Clement C.
    Nov 24 at 21:57
















2














You need to show that the supremum tends to $0$, not to necessarily find the supremum exactly. Namely, it is sufficient to find $(a_n)_n$ such that
$$
forall ngeq 1,qquad sup_{xin[0,infty)} lvert f_n(x)rvert leq a_n
$$
and $lim_{ntoinfty} a_n =0$.



Therefore, let's try and do that here: for all $xgeq 0$ and $ngeq 1$,
$$
0 leq frac{x}{n^2} e^{-frac{x}{n}} = frac{1}{n}cdot frac{x}{n} e^{-frac{x}{n}} leq frac{1}{n}cdot sup_{tin[0,infty) } t e^{-t} stackrel{rm def}{=} a_n
$$
(can you see why?) But it is easy to study the function $h(t) = t e^{-t}$: it is differentiable, and $h'(t) = e^{-t}(1-t)$, and it is near-immediate to conclude it achieves its maximum at $t=1$, where $h(1) = 1/e$. Therefore, in light of the above you have
$$
forall ngeq 1,qquad sup_{xin[0,infty)} lvert f_n(x)rvert leq frac{1}{en}
$$
and you can conclude.




Note: in this case, it so happens that the $a_n$ we found is actually equal to $sup_{xin[0,infty)} lvert f_n(x)rvert$.






share|cite|improve this answer





















  • Thank you! that's really helpful :)
    – maa'our
    Nov 24 at 21:56










  • @maa'our You're welcome -- glad this helped!
    – Clement C.
    Nov 24 at 21:57














2












2








2






You need to show that the supremum tends to $0$, not to necessarily find the supremum exactly. Namely, it is sufficient to find $(a_n)_n$ such that
$$
forall ngeq 1,qquad sup_{xin[0,infty)} lvert f_n(x)rvert leq a_n
$$
and $lim_{ntoinfty} a_n =0$.



Therefore, let's try and do that here: for all $xgeq 0$ and $ngeq 1$,
$$
0 leq frac{x}{n^2} e^{-frac{x}{n}} = frac{1}{n}cdot frac{x}{n} e^{-frac{x}{n}} leq frac{1}{n}cdot sup_{tin[0,infty) } t e^{-t} stackrel{rm def}{=} a_n
$$
(can you see why?) But it is easy to study the function $h(t) = t e^{-t}$: it is differentiable, and $h'(t) = e^{-t}(1-t)$, and it is near-immediate to conclude it achieves its maximum at $t=1$, where $h(1) = 1/e$. Therefore, in light of the above you have
$$
forall ngeq 1,qquad sup_{xin[0,infty)} lvert f_n(x)rvert leq frac{1}{en}
$$
and you can conclude.




Note: in this case, it so happens that the $a_n$ we found is actually equal to $sup_{xin[0,infty)} lvert f_n(x)rvert$.






share|cite|improve this answer












You need to show that the supremum tends to $0$, not to necessarily find the supremum exactly. Namely, it is sufficient to find $(a_n)_n$ such that
$$
forall ngeq 1,qquad sup_{xin[0,infty)} lvert f_n(x)rvert leq a_n
$$
and $lim_{ntoinfty} a_n =0$.



Therefore, let's try and do that here: for all $xgeq 0$ and $ngeq 1$,
$$
0 leq frac{x}{n^2} e^{-frac{x}{n}} = frac{1}{n}cdot frac{x}{n} e^{-frac{x}{n}} leq frac{1}{n}cdot sup_{tin[0,infty) } t e^{-t} stackrel{rm def}{=} a_n
$$
(can you see why?) But it is easy to study the function $h(t) = t e^{-t}$: it is differentiable, and $h'(t) = e^{-t}(1-t)$, and it is near-immediate to conclude it achieves its maximum at $t=1$, where $h(1) = 1/e$. Therefore, in light of the above you have
$$
forall ngeq 1,qquad sup_{xin[0,infty)} lvert f_n(x)rvert leq frac{1}{en}
$$
and you can conclude.




Note: in this case, it so happens that the $a_n$ we found is actually equal to $sup_{xin[0,infty)} lvert f_n(x)rvert$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Nov 24 at 21:48









Clement C.

49.2k33785




49.2k33785












  • Thank you! that's really helpful :)
    – maa'our
    Nov 24 at 21:56










  • @maa'our You're welcome -- glad this helped!
    – Clement C.
    Nov 24 at 21:57


















  • Thank you! that's really helpful :)
    – maa'our
    Nov 24 at 21:56










  • @maa'our You're welcome -- glad this helped!
    – Clement C.
    Nov 24 at 21:57
















Thank you! that's really helpful :)
– maa'our
Nov 24 at 21:56




Thank you! that's really helpful :)
– maa'our
Nov 24 at 21:56












@maa'our You're welcome -- glad this helped!
– Clement C.
Nov 24 at 21:57




@maa'our You're welcome -- glad this helped!
– Clement C.
Nov 24 at 21:57


















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