1. ## True or false?

Which of the following statements are true or false? Give proofs or counterexamples to justify your answers.
i). If $\displaystyle g: \mathbb{R} \rightarrow \mathbb{R}$ is continuous and $\displaystyle f: \mathbb{R} \rightarrow \mathbb{R}$ is such that $\displaystyle f(x)=g(x) \ \forall \ x \in \mathbb{Q}$, then f is continuous.
I said that this statement was false and used the counterexample $\displaystyle g(x)=x$ and $\displaystyle f(x)= \begin{cases} x \ \ \ \ \ x \in \mathbb{Q}\\ 0 \ \ \ \ \ x \in \mathbb{R}- \mathbb{Q} \end{cases}$

ii). If $\displaystyle f: \mathbb{R} \rightarrow \mathbb{R}$ and $\displaystyle g: \mathbb{R} \rightarrow \mathbb{R}$ are both continuous and $\displaystyle f(x)=g(x) \ \forall \ x \in \mathbb{Q}$, then f=g.
I put true.

Since g is continuous I know that $\displaystyle \forall \ \epsilon_1>0 \exists \delta_1>0 \ s.t \ |x-c|< \delta \Rightarrow |f(x)-f(c)|< \epsilon_1$.

f is also continuous so $\displaystyle \forall \ \epsilon_2>0 \exists \delta_2>0 \ s.t \ |x-c|< \delta \Rightarrow |g(x)-g(c)|< \epsilon_2$.

$\displaystyle f(c)-\epsilon_1<f(x)<f(c)+\epsilon_1$
$\displaystyle g(c)-\epsilon_2<g(x)<g(c)+\epsilon_2$

$\displaystyle -|f(c)-g(c)|-\epsilon_1+\epsilon_2 \leq f(c)-\epsilon_1-g(c)+\epsilon_2<f(x)-g(x)<f(c)-g(c)+\epsilon_1-\epsilon_2 $$\displaystyle \leq |f(c)-g(c)|+\epsilon_1-\epsilon_2 Let \displaystyle \epsilon_3=|f(c)-g(c)| \displaystyle -\epsilon_3-\epsilon_1+\epsilon_2<f(x)-g(x)<\epsilon_3+\epsilon_1-\epsilon_2 Define \displaystyle \epsilon:= \epsilon_1-\epsilon_2+\epsilon_3. WLOG, let \displaystyle \epsilon_1+\epsilon_3> \epsilon_2. is this step valid? This gives \displaystyle |f(x)-g(x)|< \epsilon as required. Since my initial definitions only apply to a rational c, would I need to prove they are equal when c is not rational? iii). If \displaystyle f: \mathbb{R} \rightarrow \mathbb{R} is continuous at 0 with \displaystyle f(0)=0 and \displaystyle g: \mathbb{R} \rightarrow \mathbb{R} is bounded then the product fg is continuous at 0. I opted for true. f is continuous: \displaystyle \forall \ \epsilon>0 \ \exists \ \delta>0 \ s.t \ |x|< \delta \Rightarrow |f(x)-f(0)|< \epsilon. g bounded: \displaystyle \exists \ C \in \mathbb{R} \ s.t \ -C<g(x)<C where \displaystyle C= max \{|lower \ bound \ of \ g(x)|, \ |upper \ bound \ of \ g(x)| \}. \displaystyle f(0)-\epsilon<f(x)<f(0)+\epsilon \displaystyle -C<g(x)<C \displaystyle \Rightarrow \ -C(f(0)-\epsilon)<f(x)g(x)< C(f(0)+ \epsilon) Let \displaystyle \epsilon_1=max \{ |C(f(0)- \epsilon)|, \ |C(f(0)+ \epsilon)| \} \displaystyle \Rightarrow - \epsilon_1<f(x)g(x)< \epsilon_1 Since I am only interested in the behaviour of the multiple functions at 0, pick \displaystyle \delta_1:=\epsilon_1. can you do this? I need to restrict delta somehow so large values of delta can also be used. This gives that: \displaystyle \forall \ \epsilon_1>0 \ \exists \ \delta_1>0 \ s.t \ |x|< \delta_! \Rightarrow |f(x)g(x)-f(0)g(0)|< \epsilon_1 as required. If \displaystyle f:[-1,1] \rightarrow \mathbb{R} is such that the composition \displaystyle x \rightarrow \sin f(x) (this is a mapping, not a limit) is continuous on \displaystyle \mathbb{R} then f is continuous. I said this is true since the composition of two continuous functions is itself continuous. However, I cannot formulate a proof! If \displaystyle f:[-1,1] \rightarrow \mathbb{R} is such that the composition \displaystyle x \rightarrow f(\sin x) (mapping once again) is continuous on \displaystyle \mathbb{R} then f is continuous. I have a feeling this is false. Unfortunately I am unable to think of a counterexample. Any help with any of these questions would be appreciated. They get so much harder towards the end! 2. Originally Posted by Showcase_22 If \displaystyle f:[-1,1]\to\mathbb{R} is such that the composition \displaystyle x\mapsto\sin(f(x)) (this is a mapping, not a limit) is continuous on then f is continuous. I said this is true since the composition of two continuous functions is itself continuous. However, I cannot formulate a proof! If is such that the composition \displaystyle x\mapsto f(\sin x) (mapping once again) is continuous on then f is continuous. I have a feeling this is false. Unfortunately I am unable to think of a counterexample. Wrong way round! The first of those is false and the second is true. For the first one, look at the (discontinuous) function \displaystyle f(x) = \begin{cases}0 &\text{if } x<0,\\\pi&\text{if }x\geqslant0.\end{cases} For the second one, \displaystyle f(x) = f(\sin(\arcsin x)). This is the composition of two continuous functions and is therefore continuous. (I don't have time at the moment to comment in detail on problems (i)–(iii), but the answers to them are broadly on the right lines.) 3. For the second one, \displaystyle f(x) = f(\sin(\arcsin x)) . This is the composition of two continuous functions and is therefore continuous. Does this prove the statement? I feel like a proof should involve more \displaystyle \delta's and \displaystyle \epsilon's! (I don't have time at the moment to comment in detail on problems (i)–(iii), but the answers to them are broadly on the right lines.) YAY! I thought I was close!! =D Can you scrutinise my proofs? I want them to be perfect for my exam! P.S: Thanks for correcting my \displaystyle \rightarrow to \displaystyle \mapsto 4. Originally Posted by Showcase_22 I said that this statement was false and used the counterexample \displaystyle g(x)=x and \displaystyle f(x)= \begin{cases} x \ \ \ \ \ x \in \mathbb{Q}\\ 0 \ \ \ \ \ x \in \mathbb{R}- \mathbb{Q} \end{cases} I put true. Since g is continuous I know that \displaystyle \forall \ \epsilon_1>0 \exists \delta_1>0 \ s.t \ |x-c|< \delta \Rightarrow |f(x)-f(c)|< \epsilon_1. f is also continuous so \displaystyle \forall \ \epsilon_2>0 \exists \delta_2>0 \ s.t \ |x-c|< \delta \Rightarrow |g(x)-g(c)|< \epsilon_2. \displaystyle f(c)-\epsilon_1<f(x)<f(c)+\epsilon_1 \displaystyle g(c)-\epsilon_2<g(x)<g(c)+\epsilon_2 \displaystyle -|f(c)-g(c)|-\epsilon_1+\epsilon_2 \leq f(c)-\epsilon_1-g(c)+\epsilon_2<f(x)-g(x)<f(c)-g(c)+\epsilon_1-\epsilon_2$$\displaystyle \leq |f(c)-g(c)|+\epsilon_1-\epsilon_2$

Let $\displaystyle \epsilon_3=|f(c)-g(c)|$

$\displaystyle -\epsilon_3-\epsilon_1+\epsilon_2<f(x)-g(x)<\epsilon_3+\epsilon_1-\epsilon_2$

Define $\displaystyle \epsilon:= \epsilon_1-\epsilon_2+\epsilon_3$. WLOG, let $\displaystyle \epsilon_1+\epsilon_3> \epsilon_2$. is this step valid?

This gives $\displaystyle |f(x)-g(x)|< \epsilon$ as required.

Since my initial definitions only apply to a rational c, would I need to prove they are equal when c is not rational?

I opted for true.

f is continuous:
$\displaystyle \forall \ \epsilon>0 \ \exists \ \delta>0 \ s.t \ |x|< \delta \Rightarrow |f(x)-f(0)|< \epsilon$.

g bounded:
$\displaystyle \exists \ C \in \mathbb{R} \ s.t \ -C<g(x)<C$ where $\displaystyle C= max \{|lower \ bound \ of \ g(x)|, \ |upper \ bound \ of \ g(x)| \}$.

$\displaystyle f(0)-\epsilon<f(x)<f(0)+\epsilon$
$\displaystyle -C<g(x)<C$

$\displaystyle \Rightarrow \ -C(f(0)-\epsilon)<f(x)g(x)< C(f(0)+ \epsilon)$

Let $\displaystyle \epsilon_1=max \{ |C(f(0)- \epsilon)|, \ |C(f(0)+ \epsilon)| \}$

$\displaystyle \Rightarrow - \epsilon_1<f(x)g(x)< \epsilon_1$

Since I am only interested in the behaviour of the multiple functions at 0, pick $\displaystyle \delta_1:=\epsilon_1$. can you do this? I need to restrict delta somehow so large values of delta can also be used.

This gives that:

$\displaystyle \forall \ \epsilon_1>0 \ \exists \ \delta_1>0 \ s.t \ |x|< \delta_! \Rightarrow |f(x)g(x)-f(0)g(0)|< \epsilon_1$ as required.

I said this is true since the composition of two continuous functions is itself continuous. However, I cannot formulate a proof!

I have a feeling this is false. Unfortunately I am unable to think of a counterexample.

Any help with any of these questions would be appreciated. They get so much harder towards the end!
For (iii).
You want to prove :

For all ε>0 ,there exists a δ>0, such that:

for all ,x if |x|<δ,then |f(x)g(x)-f(0)g(0)|<ε,or |f(x)g(x)|<ε,since f(0)=0.

So ,let ε>0

SINCE the function g is bounded over THE REAL Nos R ,THERE exists a ,C>0 SUCH

For all ,x if xεR,then $\displaystyle |g(x)|\leq C$.......................................1

SINCE f is continouous at 0,then for all +ve Nos and thus for ε/C >0,

there exists a δ>0 such that:

for all ,x if |x|<δ. then |f(x)|< ε/C................................................. .2

So if |x|<δ ====> xε(-δ,+δ)====> xεR and using (1)

$\displaystyle |g(x)|\leq C$.................................................. ................3

Also using (2) we have :

|f(x)|<ε/C................................................. ..................................4

IN THE CASES where either |f(x)|=0 or |g(x)|= 0 we have :

|f(x)g(x)| = 0<ε

ΙN THE CASE where |f(x)|>0 and |g(x)|>0 by using (3) and (4)
again we have:

|f(x)g(x)| <ε

5. Originally Posted by Showcase_22

I put true.

Since g is continuous I know that $\displaystyle \forall \ \epsilon_1>0 \exists \delta_1>0 \ s.t \ |x-c|< \delta \Rightarrow |f(x)-f(c)|< \epsilon_1$.

f is also continuous so $\displaystyle \forall \ \epsilon_2>0 \exists \delta_2>0 \ s.t \ |x-c|< \delta \Rightarrow |g(x)-g(c)|< \epsilon_2$.

$\displaystyle f(c)-\epsilon_1<f(x)<f(c)+\epsilon_1$
$\displaystyle g(c)-\epsilon_2<g(x)<g(c)+\epsilon_2$

$\displaystyle -|f(c)-g(c)|-\epsilon_1+\epsilon_2 \leq f(c)-\epsilon_1-g(c)+\epsilon_2<f(x)-g(x)<f(c)-g(c)+\epsilon_1-\epsilon_2$$\displaystyle \leq |f(c)-g(c)|+\epsilon_1-\epsilon_2$

Let $\displaystyle \epsilon_3=|f(c)-g(c)|$

$\displaystyle -\epsilon_3-\epsilon_1+\epsilon_2<f(x)-g(x)<\epsilon_3+\epsilon_1-\epsilon_2$

Define $\displaystyle \epsilon:= \epsilon_1-\epsilon_2+\epsilon_3$. WLOG, let $\displaystyle \epsilon_1+\epsilon_3> \epsilon_2$. is this step valid?

This gives $\displaystyle |f(x)-g(x)|< \epsilon$ as required.

Since my initial definitions only apply to a rational c, would I need to prove they are equal when c is not rational?

!
Showcase i offer you the following proof to compare with your proof.

We want to prove that f=g.For that we must show that :

for all x belonging to the real Nos R ,f(x)=g(x).

To show that in proofs involving the use of the ε-δ variables we often use the following theorem:

for all a,b ,if for all ε>0,then |a-b|<ε that implies a=b............................................... .........................................1

And in our case we must prove for xεR ,FOR all ε>0 ,|f(x)-g(x)|<ε

SO,let xεR .

Also let ε>0

Νow we have the following theorem in analysis :

Let x be a real No.Given ε>0,there exists a rational No c such that |x-c|<ε.

Due to that theorem and because of the above assumptions and since f and g are continuous over R,

there exist ,$\displaystyle \delta_{1}>0,\delta_{2}>0$ and such that:

for all,x if $\displaystyle |x-c|<\delta_{1}$,then |f(x)-f(c)|<ε/2................................................. ....................................2

for all ,x if $\displaystyle |x-c|<\delta_{2}$,then |g(x)-g(c)|<ε/2................................................. ......................................3

Choose δ=min{$\displaystyle \delta_{1},\delta_{2}$}................................................. ....................................4

AND assuming:

|x-c|<δ

then by the use of (2),(3),and (4) |f(x)-f(c)|<ε/2 and |g(x)-g(c)|<ε/2................................................. .....................................5

Here now is a crucial point of the proof ,because at this point we cannot use YET theorem (1) neither the triangular inequality and conclude:

|f(x)-g(x)+g(c)-f(c)|<ε and finally f(x) = g(x) since |f(x)-g(x)|<ε, since g(c)-f(c)=0 because c is a rational No,

BECAUSE up to this point we have proved:

GIVEN ε>0 and

if |x-c|<δ,then |f(x)-f(c)|<ε/2 and |g(x)-g(c)|<ε/2................................................. ......................................6

Το get rid of the conditional (6) and be able to use theorem (1) and the triangular inequality we use the fact that c, belonging to R IS AN ACCUMULATION point and hence:

There exist an ,x such that 0<|x-c|<δ ,for all δ.

And now finally we can use theorem (1) and the triangular inequality and conclude :

for all xεR ,f(x) = g(x) and thus f=g