# A sequence of continuous functions on $[0,\infty)$ converging uniformly such that each $f_n$ has a zero but the limit function is nowhere zero.

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.

• 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. – Daniel Fischer Nov 15 '16 at 16:47
• @DanielFischer Thanks. How can I begin to think like you? :D – Error 404 Nov 15 '16 at 16:56
• Collect a few decades of experience. :P – Daniel Fischer Nov 15 '16 at 16:59
• @DanielFischer Well I am in a hurry. :P – Error 404 Nov 15 '16 at 17:06

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.

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

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.

• Are you taking $g_n f = g_n \circ f$ here? If yes, then if we take $f(x)= \dfrac {1}{1+x}$, then $(g_nf)(n)= \dfrac {\left(\dfrac {1}{1+n}-n \right)^2}{1+ \left ({\dfrac {1}{1+n}}-n \right)^2} \neq 0.$ – Error 404 Nov 17 '16 at 5:14
• No, $g_nf$ is $g_n$ times $f.$ – zhw. Nov 17 '16 at 6:55
• 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 $\text{sup} \{|g_n(x)-g(x)| : x \in [0,\infty) \}=\text{sup}\left\{ \left|\dfrac {1}{1+(x-n)^2}\right| : x \in [0,\infty) \right\} =1 \nrightarrow 0$. – Error 404 Nov 17 '16 at 7:19
• That's true, but $g_nf$ does converge uniformly to $f.$ That's because $f(x) \to 0$ at $\infty.$ – zhw. Nov 17 '16 at 7:21
• @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 $n\ge N.$ – zhw. Nov 18 '16 at 16:20

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

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.