3
$\begingroup$

We have a sequence of continuous functions $f_n : [0,\infty) \to \Bbb R$ such that $f_n \to f$ uniformly. Also each $f_n$ has a zero. But $f$ does not have any zero. Then what is the example of such a sequence of functions and their limit function.

I am trying to think about a sequence of functions such that their limit function is $f(x)=1 \quad \forall x \in [0,\infty)$. However, I am really stuck on this for a long time and completely clueless after trying many functions $f_n$.

Here is the link where I proved (after getting help) that if the domain is $[a,b]$ then $f$ will always have a zero because of sequential compactness of $[a,b]$. In case of $[0,\infty)$ it is not possible.

$\endgroup$
  • 1
    $\begingroup$ You can't have the limit $1$, $f$ must come arbitrarily close to $0$. The zero of $f$ would be at $+\infty$, but that's not in the domain. $\endgroup$ – Daniel Fischer Nov 15 '16 at 16:47
  • $\begingroup$ @DanielFischer Thanks. How can I begin to think like you? :D $\endgroup$ – Error 404 Nov 15 '16 at 16:56
  • 3
    $\begingroup$ Collect a few decades of experience. :P $\endgroup$ – Daniel Fischer Nov 15 '16 at 16:59
  • 1
    $\begingroup$ @DanielFischer Well I am in a hurry. :P $\endgroup$ – Error 404 Nov 15 '16 at 17:06
5
$\begingroup$

Consider $$f_n(x)=\frac{1}{x+1}-\dfrac{\sin x}{n}.$$ Note that every $f_n$ has at least a zero. But, its limit $f(x)=\dfrac{1}{x+1}$ has not zeroes.

$$f_n(x)=0\iff h(x)=(x+1)\sin x-n=0.$$ But $h(0)=-n$ and $h\left(\dfrac{(4n+1)\pi}{2}\right)=\dfrac{(4n+1)\pi}{2}+1-n>0.$ Since $h$ is continuous it must have a zero in the interval $\left(0,\dfrac{(4n+1)\pi}{2}\right).$

Finally, note that $$|f_n(x)-f(x)|=\dfrac{|\sin x|}{n}\le \dfrac 1n,$$ from where we can conclude that $f_n\to f$ uniformly.

$\endgroup$
  • $\begingroup$ Hey I just noticed that $h(\frac {(2n+1)\pi}2)=[\frac {(2n+1)\pi}{2} +1](-1)^n -n.$ So at $n=1, h(x) \lt 0$ $\endgroup$ – Error 404 Nov 15 '16 at 17:24
  • $\begingroup$ @VikrantDesai I have made a typo. It must be $\dfrac{(4n+1)\pi}{2}.$ I have fixed it. $\endgroup$ – mfl Nov 15 '16 at 17:30
  • $\begingroup$ Also we have '$x+1$' in RHS. So the corresponding term would be $\dfrac{(4n+1)\pi}{2} +1$ $\endgroup$ – Error 404 Nov 15 '16 at 17:32
  • 1
    $\begingroup$ @VikrantDesai Thank you for noticing it. $\endgroup$ – mfl Nov 15 '16 at 17:36
5
$\begingroup$

I'll mention a whole class of such functions, without going into details. Let

$$g_n(x) = \frac{(x-n)^2}{1+(x-n)^2}.$$

Now let $f$ be any positive continuous function on $[0,\infty)$ such that $\lim_{x\to \infty}f(x) = 0.$ Then $g_n f \to f$ uniformly on $[0,\infty).$ Because $(g_nf)(n) = 0,$ this gives many examples for the problem at hand.

I can go into this further if there are questions.

$\endgroup$
  • $\begingroup$ Are you taking $g_n f = g_n \circ f$ here? If yes, then if we take $f(x)= \dfrac {1}{1+x}$, then $(g_nf)(n)= \dfrac {\left(\dfrac {1}{1+n}-n \right)^2}{1+ \left ({\dfrac {1}{1+n}}-n \right)^2} \neq 0.$ $\endgroup$ – Error 404 Nov 17 '16 at 5:14
  • $\begingroup$ No, $g_nf$ is $g_n$ times $f.$ $\endgroup$ – zhw. Nov 17 '16 at 6:55
  • $\begingroup$ I see that $g_n$, the sequence of continuous functions on $[0,\infty)$, doesn't converge uniformly to $g(x)=1 \; \forall \; x \in [0,\infty)$. This is because $\text{sup} \{|g_n(x)-g(x)| : x \in [0,\infty) \}=\text{sup}\left\{ \left|\dfrac {1}{1+(x-n)^2}\right| : x \in [0,\infty) \right\} =1 \nrightarrow 0$. $\endgroup$ – Error 404 Nov 17 '16 at 7:19
  • $\begingroup$ That's true, but $g_nf$ does converge uniformly to $f.$ That's because $f(x) \to 0$ at $\infty.$ $\endgroup$ – zhw. Nov 17 '16 at 7:21
  • 1
    $\begingroup$ @VikrantDesai Basic idea: Verify that $g_n$ converges to $1$ uniformly on each $[0,R].$ This implies $g_nf \to f$ uniformly on each $[0,R].$ So let $\epsilon>0.$ Choose $R$ such that $f<\epsilon/2$ on $[R,\infty).$ Then on $[R,\infty),$ we have $|g_nf-f|\le 2f < \epsilon.$ If we then choose $N$ such that $|g_nf-f|<\epsilon$ on $[0,R]$ for $n\ge N,$ we have $|g_nf-f|<\epsilon$ everywhere in $[0,\infty)$ for $n\ge N.$ $\endgroup$ – zhw. Nov 18 '16 at 16:20
1
$\begingroup$

Hint: Use $f(x) =\frac{1}{1+x}$.

$\endgroup$
1
$\begingroup$

Take $f(x) = e^{-x}$, and let $f_n(x)$ equal $f(x)$ for $x \notin[n,n+1]$, but $f_n(x)$ takes the value computed by linearly interpolating the points $(n,f(n)), (n+{1 \over 2}, 0), (n+1, f(n+1))$. Then $\|f-f_n\|_\infty = f(n+{1 \over 2})$ hence converges uniformly, and $f_n(n+{1 \over 2}) = 0$. Since $f(x) \neq 0$, we are finished.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.