ISI 2018 #3

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integrated_JRC

3465 posts

integrated_JRC

#1 h May 13, 2018, 10:58 AM • 3 Y

Y by ark2001, Adventure10, Mango247

Let $f:\mathbb{R}\to\mathbb{R}$ be a continuous function such that for all $x\in\mathbb{R}$ and for all $t\geqslant 0$ , $f(x)=f(e^tx)$ Show that $f$ is a constant function.

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alexheinis

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alexheinis

#2 h May 13, 2018, 11:03 AM • 4 Y

Y by adityaguharoy, skyline_yi, AYSA02, Adventure10

Too easy. First choose $a>0$ , then $f$ is constant on $[a,\infty)$ . Let $a\downarrow 0$ to see that $f$ is constant on ${\bf R^+}$ . By continuity $f$ is constant on $[0,\infty)$ . In the same way, with $a<0$ one shows that $f$ is constant on $(-\infty,0]$ . Hence $f$ is constant.

This post has been edited 2 times. Last edited by alexheinis, May 13, 2018, 11:04 AM

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Neo139

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Neo139

#3 h May 13, 2018, 12:49 PM • 2 Y

Y by Adventure10, Mango247

Hey can't it be done by differentiating both sides?

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TomMarvoloRiddle

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TomMarvoloRiddle

#5 h May 13, 2018, 1:17 PM • 2 Y

Y by Adventure10, Mango247

Neo139 wrote:

Hey can't it be done by differentiating both sides?

No. It is not given that $f$ is differentiable.

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vickyricky

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vickyricky

#6 h May 13, 2018, 1:51 PM • 3 Y

Y by ring_r, Adventure10, Mango247

By proving differentiability and then applying rolle's u will get infinitely many a such that f'(a)=0 Then function is constant .

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rayuga

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rayuga

#7 h May 13, 2018, 1:59 PM • 3 Y

Y by ihaveaproblem2017, amar_04, Adventure10

Solution

This post has been edited 8 times. Last edited by rayuga, May 13, 2018, 7:27 PM
Reason: Latex

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adityaguharoy

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adityaguharoy

#8 h May 13, 2018, 2:50 PM • 1 Y

Y by Adventure10

jrc1729 wrote:

Let $f:\mathbb{R}\to\mathbb{R}$ be a continuous function such that for all $x\in\mathbb{R}$ and for all $t\geqslant 0$ , $f(x)=f(e^tx)$ Show that $f$ is a constant function.

A generalization of this problem asks whether we can replace the $e^t$ by any varying continuous function of a specific class.

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Sayan

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Sayan

#9 h May 13, 2018, 4:43 PM • 11 Y

Y by integrated_JRC, adityaguharoy, Devarka, biomathematics, Vimath, gatterman, pankajsinha, Madhavi, EnigmaAlpha, Adventure10, Mango247

The equation is an overkill I think. Take $t=1$ . We have $f(x)=f(ex)$ for all $x\in\mathbb{R}$ . Fix $x_0\in\mathbb{R}$ . Then
$f(x_0)=f\left(\frac{x_0}{e}\right)=f\left(\frac{x_0}{e^2}\right)=\cdots=\lim_{n\to \infty} f\left(\frac{x_0}{e^n}\right) = f(0)$ The last one is true by continuity.
Thus $f(x)=f(0)$ for all $x\in \mathbb{R}$ . Hence $f$ is a constant.

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SPC4

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SPC4

#10 h May 13, 2018, 5:49 PM • 3 Y

Y by Math_Renreal..., Adventure10, Mango247

But Sayan, it is not given that limit of f exists. We need to prove that f is one-one. And then we can use the limit. But how to prove that?

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Sayan

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Sayan

#11 h May 13, 2018, 6:03 PM • 2 Y

Y by adityaguharoy, Adventure10

The existence of limit comes from continuity not one-one. $y_n \to 0$ implies $f(y_n) \to f(0)$

This post has been edited 1 time. Last edited by Sayan, May 13, 2018, 6:04 PM
Reason: Error

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sagnikndp16

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sagnikndp16

#12 h May 14, 2018, 4:09 AM • 2 Y

Y by Adventure10, Mango247

putting x=e^-t and x=1 we will get two values of f(1) and they are f(e^t) and f(e^-t)
as t>=0 e^t and e^-t covers the whole real number set which is the domain of f so f(1) is equal to f(r) that means it is an constant function

This post has been edited 1 time. Last edited by sagnikndp16, May 14, 2018, 4:09 AM

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Sayan

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Sayan

#13 h May 14, 2018, 9:49 AM • 4 Y

Y by rayuga, adityaguharoy, Adventure10, Mango247

sagnikndp16 wrote:

as t>=0 e^t and e^-t covers the whole real number set

They cover $(0,\infty)$ only

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polarLines

499 posts

polarLines

#14 h May 14, 2018, 10:10 AM • 5 Y

Y by integrated_JRC, stranger_02, Fibonacci1123_-, Adventure10, Mango247

Put $x=1$ to get $f(e^t)=f(1)=c$ (say). So, $f(x)=c \forall x \in [1,\infty)$ . Put $x=e^{-t}$ to get $f(e^{-t})=f(1)=c$ . So, $f(x)=c\forall x\in (0,1]$ . Hence, $f(x)=c\forall x\in (0,\infty)$ .

Put $x=-1$ to get $f(-e^t)=f(-1)=k$ (say). So, $f(x)=k \forall x \in (-\infty,-1]$ . Put $x=-e^{-t}$ to get $f(-e^{-t})=f(-1)=k$ . So, $f(x)=k\forall x\in [-1,0)$ . Hence, $f(x)=k\forall x\in (-\infty,0)$ .

By continuity, $f(0)=\lim_{x\to 0^-} f(x)=\lim_{x\to 0^+} f(x)\implies f(0)=c=k$

This post has been edited 2 times. Last edited by polarLines, Apr 18, 2019, 6:16 AM

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LordofAngaband

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LordofAngaband

#15 h May 15, 2018, 9:07 AM • 2 Y

Y by Adventure10, Mango247

it can be represented in the form of first principle of derivative then proving it to be 0

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integrated_JRC

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integrated_JRC

#16 h May 15, 2018, 9:43 AM • 2 Y

Y by Adventure10, Mango247

Ishaan600 wrote:

it can be represented in the form of first principle of derivative then proving it to be 0

No you can't. Because, it's not given if $f$ is differentiable or not.

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LordofAngaband

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LordofAngaband

#17 h May 15, 2018, 4:57 PM • 2 Y

Y by Adventure10, Mango247

i am not differentiating
first principle auto proves it
replace (e^t)x by x+k
since it is continuous take limit k to 0
then it is done

This post has been edited 2 times. Last edited by LordofAngaband, May 15, 2018, 5:46 PM

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bmwi8

2 posts

bmwi8

#18 h May 15, 2018, 6:01 PM • 2 Y

Y by Adventure10, Mango247

$f(x) = f(e^tx)$
Partially differentiate w/ respect to t to get:
$\ [f'(x)]'_{t} = e^t[f(e^tx)]_{t}'$
$\ \implies [f(e^tx)]'_t = lim_{\lambda \to 0} \frac{f(e^{t+\lambda}x) - f(x)}{\lambda}=0$
$\ \implies lim _{x_2 \to x_1} \frac{f(x_2)-f(x_1)}{ln(\frac{x_1}{x_1})}=0$ where $\ x_2 = x_1e^\lambda,$ and as $\ \lambda \to 0, e ^\lambda \to 1, x_2 \to x_1$
$\ lim_{x_2 \to x_1} \frac {\frac {f(x_2)-f(x_1)} {x_2-x_1}} {\frac {ln x_2 - ln x_1}{x_2-x_1}} = \frac{f'(x_1)}{\frac{1}{x_1}}=0 \implies f'(x_1) = 0 \forall x_1 \in R$ Q.E.D.

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LordofAngaband

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LordofAngaband

#19 h May 15, 2018, 8:04 PM • 2 Y

Y by Adventure10, Mango247

i think first principle is better since the function is given to be continuous hence limit is applicable differentiation at first step may be wrong

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ArijitSinha

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ArijitSinha

#20 h May 21, 2018, 5:30 PM • 2 Y

Y by Adventure10, Mango247

alexheinis wrote:

Too easy. First choose $a>0$ , then $f$ is constant on $[a,\infty)$ . Let $a\downarrow 0$ to see that $f$ is constant on ${\bf R^+}$ . By continuity $f$ is constant on $[0,\infty)$ . In the same way, with $a<0$ one shows that $f$ is constant on $(-\infty,0]$ . Hence $f$ is constant.

Yup. Now f(x) is continuous at x=0. So , eventually f(x)=k all over R.

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pratyushjain

2 posts

pratyushjain

#21 h May 26, 2018, 4:57 PM • 1 Y

Y by Adventure10

guys , i have done the question by writing a story, and i am not even sure i have proved what was asked or not . tell if there is any way i can get partial credits for this method :
solution : lets prove it by contradiction. Therefore acc to question f(x) = f(e^t x) = k for some pairs (x,t) and f(x) = f(e^t x) = m for atleast one pair (x,t) where k not equal to m if f is not a const. function . Lets prove it by induction . Without loss of generality , we will consider x const. and prove this result for any value of t and then keeping t const. and varying x with the help of induction on 2 variables, we will say that above result is true. consider x= const. now consider set t varying from say (0 p) , 0 included for any value of p satisfying f(x) = f(e^t x) = k . now consider p+e where e tends to 0+ ( e is very close to 0). acc. to question the above condition is satisfied for all t >= 0 and if it satisifies for set (0 p) , then it must satisfy for p+e as well. and hence by induction on real variables we can say that f(x) = f(e^t x) = k is true for all t >= 0 or there is no value of t for which f(x) = f(e^t x) = m where k not equal to m . similarly taking t = const. and taking set x to be say set of values ranging ( -q. q ) and proving arguments similarly by taking -q-e and q+e seperately , we get this for x as well and hence by induction on 2 variables we are done.

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LordofAngaband

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LordofAngaband

#22 h May 26, 2018, 7:42 PM • 2 Y

Y by Adventure10, Mango247

the problem is done away
the condition of continuity here forces differentiability

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pratyushjain

2 posts

pratyushjain

#23 h May 27, 2018, 4:44 AM • 2 Y

Y by Adventure10, Mango247

ishaan i didnt get what u mean . i understand the exact solution but i am asking abt mine .

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LordofAngaband

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LordofAngaband

#24 h May 27, 2018, 8:42 PM • 1 Y

Y by Adventure10

my solution is the most simple i have seen
we take f(x) to other side and divide by e^t-x for some t which we define in next step, we take limit t approaching 0
then write e^t(x) as x+k where k also approaches 0 now the side auto transforms to a limit which exists and is 0 for all x
and the limit is definition of first principle which exists (as the function is cont ) and vanishes at every x hence the function never changes ie a contant

This post has been edited 2 times. Last edited by LordofAngaband, May 27, 2018, 8:54 PM

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Crazywithmath

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Crazywithmath

#25 h May 29, 2018, 4:39 PM • 2 Y

Y by Adventure10, Mango247

It was my method too

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jjagmath

1016 posts

jjagmath

#27 h May 29, 2018, 6:51 PM • 2 Y

Y by Adventure10, Mango247

For $t \le 0, f(e^t x) = f(e^{-t} (e^{t} x)) = f(x)$ , so the condition is valid for all $t$ .
For $x>0$ , $f(x) = f(e^{\log(1/x)} x) = f(1)$ . $f$ is constant in $\mathbb{R}^+$ .
For $x<0$ , $f(x) = f(e^{\log(-1/x)} x) = f(-1)$ . $f$ is constant in $\mathbb{R}^-$ .
$f$ is continuous, so $f$ is constant.

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Butterfly

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Butterfly

#28 h May 30, 2018, 2:04 AM • 3 Y

Y by zetayh, Adventure10, Mango247

$\textbf{Proof}$

Take $t=1$ . Then for any fixed $r \in \mathbb{R}$ , it's clear that $f(r)=f \left(\frac{r}{e}\right)=f \left(\frac{r}{e^{2}}\right)=f \left(\frac{r}{e^{3}}\right)=\cdots=f \left(\frac{r}{e^{n}}\right).$ Take the limits of both sides as $n \to \infty$ . Notice that $\dfrac{r}{e^n} \to 0$ as $n \to \infty$ . By the continuity of $f(x)$ , we have $f(r)=\lim_{n \to \infty}f(r)=\lim_{n \to \infty}f \left(\frac{r}{e^{n}}\right)=f\left(\lim_{n \to \infty}\frac{r}{e^{n}}\right)=f(0).$ This shows that $f(x) \equiv f(0)$ .

This post has been edited 1 time. Last edited by Butterfly, May 30, 2018, 2:06 AM

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LordofAngaband

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LordofAngaband

#29 h May 30, 2018, 4:43 PM • 2 Y

Y by Adventure10, Mango247

the beauty i find in this question is that all different correct methods all have a similar underlying idea and proof

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ftheftics

651 posts

ftheftics

#30 h Sep 15, 2019, 3:15 PM • 1 Y

Y by Adventure10

Sayan wrote:

The equation is an overkill I think. Take $t=1$ . We have $f(x)=f(ex)$ for all $x\in\mathbb{R}$ . Fix $x_0\in\mathbb{R}$ . Then
$f(x_0)=f\left(\frac{x_0}{e}\right)=f\left(\frac{x_0}{e^2}\right)=\cdots=\lim_{n\to \infty} f\left(\frac{x_0}{e^n}\right) = f(0)$ The last one is true by continuity.
Thus $f(x)=f(0)$ for all $x\in \mathbb{R}$ . Hence $f$ is a constant.

This post has been edited 1 time. Last edited by ftheftics, Mar 21, 2020, 8:08 PM
Reason: T

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ftheftics

651 posts

ftheftics

#31 h Mar 21, 2020, 8:08 PM

Y by

ftheftics wrote:

Sayan wrote:

The equation is an overkill I think. Take $t=1$ . We have $f(x)=f(ex)$ for all $x\in\mathbb{R}$ . Fix $x_0\in\mathbb{R}$ . Then
$f(x_0)=f\left(\frac{x_0}{e}\right)=f\left(\frac{x_0}{e^2}\right)=\cdots=\lim_{n\to \infty} f\left(\frac{x_0}{e^n}\right) = f(0)$ The last one is true by continuity.
Thus $f(x)=f(0)$ for all $x\in \mathbb{R}$ . Hence $f$ is a constant.

I think you need to use the continuous condition to show that it works for $x<0$ as well.

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Severus

742 posts

Severus

#32 h Apr 23, 2020, 2:17 PM

Y by

adityaguharoy wrote:

jrc1729 wrote:

Let $f:\mathbb{R}\to\mathbb{R}$ be a continuous function such that for all $x\in\mathbb{R}$ and for all $t\geqslant 0$ , $f(x)=f(e^tx)$ Show that $f$ is a constant function.

A generalization of this problem asks whether we can replace the $e^t$ by any varying continuous function of a specific class.

I believe we can replace $e^t$ with any surjective function with domain either $(0,\infty), (-\infty,0)$ or best, $\mathbb{R}$ .

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lifeismathematics

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lifeismathematics

#35 h Apr 17, 2023, 2:07 AM

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{storage}

since it works/holds for all $t$ , we assume $t=1$ for sake of simplicity

then we have $f(x)=f(ex) \forall x \in \mathbb{R}$

now we fix $\Omega \in \mathbb{R}$ ( any arbitrary) for which we define $\{x_{n}\}_{n=0}^{\infty}$ s.t. $x_{n}=\frac{\Omega}{e^{n}}$ , then we have:
$f(x_{n})=f\left(\frac{\Omega}{e^{n-1}}\right)=f(\Omega)$

now since $f$ is continuous over $\mathbb{R}$ we take limits both sides with $n\rightarrow \infty$ and we know that for a continuous function over $\mathbb{R}$ we have a sequence in $\mathbb{R}$ say $\{y_{n}\}$ s.t. $\lim_{n\rightarrow \infty}f(y_{n})=f\left(\lim_{n\rightarrow \infty } y_{n}\right)$

so from this fact we get $f(\Omega)=f\left(\lim_{n\rightarrow \infty} \frac{\Omega}{x^{n-1}}\right)\implies f(\Omega)=f(0)$

no we had chosen $\Omega$ to be arbitrary we replace $\Omega \rightarrow x$

we get $\boxed{f(x)=f(0)}$ hence $f$ is constant function. $\blacksquare$

This post has been edited 5 times. Last edited by lifeismathematics, Apr 17, 2023, 11:17 AM

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MV2

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MV2

#37 h Apr 17, 2023, 5:10 AM

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Solution with Motivation.

Let $x_0 \in \mathbb{R}, x_0 > 0$ . Consider the interval $I = (x_0, \infty)$ . Now consider $y \in I$ . As $\frac{y}{x_0} > 1$ , we have, $y_{0} = e^{\ln\frac{y} {x_0}}$ . So, as $\ln\frac{y}{x_0} > 0$ , which satisfies the condition for $t$ , we have, $f(x_0) = f(y)$ . Therefore, if we consider any number $x_1$ s.t.
$0< x_1 < x_0$ we see that, $f(x_1) = f(y), \forall y \in (x_1, \infty)$ . Hence, it's meaningful to consider a sequence on either side of $0$ that converges to $0$ and simultaneously making use of the property that the function is continuous.

Therefore, let ${(x_n)}_{n=0}^{\infty}$ be a sequence converging to $0$ . Then ${(f(x_n))}_{n=0}^{\infty}$ converges to $f(0)$ . And as, all the terms in the sequence are equal, the function is a constant function. $\blacksquare$

This post has been edited 1 time. Last edited by MV2, Aug 5, 2023, 6:06 AM

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bCarbon

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bCarbon

#38 h Apr 21, 2023, 6:08 AM

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Observe the statement holds over

(-\infty, 0) and (0, \infty)

because (1) t at least 0 implies exp(t) at least 0 and (2) exp(t) is monotonically increasing.

But why should

f(0) = f(x), x \neq 0

Continuity, or course.

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amar_04

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amar_04

#39 h Apr 22, 2023, 10:44 PM • 2 Y

Y by kamatadu, HoRI_DA_GRe8

ISI 2018 P3 wrote:

Let $f:\mathbb{R}\to\mathbb{R}$ be a continuous function such that for all $x\in\mathbb{R}$ and for all $t\geqslant 0$ , $f(x)=f(e^tx)$ Show that $f$ is a constant function.

Let $x>y\ne 0$ and $t>0$ , then $e^tx=e^{t'}y\implies e^{t'}=e^t\left(\frac{x}{y}\right)\implies t'=t+\ln\left(\frac{x}{y}\right)$ since, $\frac{x}{y}>1\implies \ln\left(\frac{x}{y}\right)>0\implies t'=t+\ln\left(\frac{x}{y}\right)>0$ . Thus, we have $\forall y<x$ $f(x)=f(e^tx)=f\left(e^{t+\ln\left(\frac{x}{y}\right)}y\right)=f(y)$ Switch the roles of $x$ and $y$ we get $\forall y>x$ and $x,y\ne0$ we get $f(x)=f(y)$ .
Now consider the sequence, $\left(\frac{1}{n}\right)\rightarrow 0$ , since, $g$ is continuous, $\left(f\left(\frac{1}{n}\right)\right)=\left(f(x)\right)\rightarrow f(0)=f(x)$ . Thus all in all, $f(x)=f(y)$ $\forall y\in\mathbb{R}$ which proves $f$ is constant. $\qquad\blacksquare$

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anudeep

64 posts

anudeep

#40 h Today at 7:08 AM

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We will first show that $f$ is constant over $\mathbb{R}_{\ge 0}$ . Consider $a>0$ . As $ae^t$ is continuous and not bounded above, I can always find a suitable $t\ge 0$ such that $ae^t=b$ for any $b\ge a$ . Picking an $a$ arbitrarily close to $0$ assures that $f$ is a constant over $\mathbb{R}_{\ge 0}$ , similarly we can argue that $f(a)=f(0)$ for all $a<0$ , which is the required. $\square$

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