Linearity and continuity of maps

[mathmari]

Hey!

Let $C[0,1]$ and $C^1[0,1]$ be the space of continuous and continuously differentiable (respectively) functions $x:[0,1]\rightarrow \mathbb{R}$ with the supremum norm $\displaystyle{\|x\|=\sup_{t\in [0,1]}|x(t)|}$ and $T_0, T_1, T_2: C^1[0,1]\rightarrow C[0,1]$ maps with \begin{equation*}T_0(x)(t)=x'(t), \ \ T_1(x)(t)=\int_0^tsx(s)\, ds \ \ \text{ und } \ \ T_2(x)(t)=\int_0^tsx^2(s)\, ds\end{equation*}

1. Check the linearity and the continuity of $T_0, T_1, T_2$.
2. Calculate the norm $\displaystyle{\|T_1\|:=\sup_{\|x\|\leq 1}\|T_1(x)\|}$.

I have done the following:

1. Let $x,y:[0,1]\rightarrow \mathbb{R}$ and $a\in \mathbb{R}$.
   • \begin{align*}&T_0(x+y)(t)=(x+y)'(t)=x'(t)+y'(t)=T_0(x)(t)+T_0(y)(t)\\ &T_0(ax)(t)=(ax)'(t)=ax'(t)=aT_0(x)(t) \end{align*}
     So $T_0$ is linear.
   • \begin{align*}&T_1(x+y)(t)=\int_0^1s(x+y)(s)\, ds=\int_0^1s(x(s)+y(s))\, ds=\int_0^1sx(s)\, ds+\int_0^1sy(s)\, ds=T_1(x)(t)+T_1(y)(t)\\ &T_1(ax)(t)=\int_0^ts(ax)(s)\, ds=\int_0^tasx(s)\, ds=a\int_0^tsx(s)\, ds=aT_1(x)(t) \end{align*}
     So $T_1$ is linear.
   • \begin{align*}T_2(x+y)(t)&=\int_0^1s(x+y)^2(s)\, ds=\int_0^1s[x^2(s)+2x(s)y(s)+y^2(s)]\, ds \\ & =\int_0^1sx^2(s)\, ds+2\int_0^1sx(s)y(s)\, ds+\int_0^1sy^2(s)\, ds \\ & =T_2(x)(t)+2\int_0^1sx(s)y(s)\, ds+T_2(y)(t)\neq T_2(x)(t)+T_2(y)(t)\end{align*}
     So $T_2$ is not linear.
   
   Is everything correct? (Wondering)
   
   How can we check the continuity? (Wondering)
2. \begin{align*}&\\ & \|T_1(x)\|=\left \|\int_0^tsx(s)\, ds\right \|\leq \int_0^t\|sx(s)\|\, ds=\int_0^t|s|\|x(s)\|\, ds=\int_0^ts\|x(s)\|\, ds \\ &\Rightarrow \sup_{\|x\|\leq 1}\|T_1(x)\|\leq \sup_{\|x\|\leq 1}\int_0^ts\|x(s)\|\, ds=\int_0^ts\, ds=\left [\frac{s^2}{2}\right ]_0^t=\frac{t^2}{2}\end{align*}
   
   Is this correct? Or do we have to find an equality and not an inequality? (Wondering)

 

[S.G. Janssens]

1. For $T_0$ I would write: $x,y \in C^1[0,1]$ and for $T_{1,2}$ I would write: $x,y \in C[0,1]$. For general functions the operators may not be defined.

Remember to integrate from $0$ to $t$, not from $0$ to $1$. Apart from that:

Linearity of $T_0$: ok.
Linearity of $T_1$: ok.

Indeed, $T_2$ is clearly not linear because of the square, but I would perhaps prefer an explicit counter-example, with specific function(s).

I would worry about continuity of $T_0$: Uniformly close functions need not have uniformly close derivatives. (It is up to you to find an example and make this precise.)

For continuity of $T_1$ you can use question 2. (Why?) You can also directly estimate $\|T_1(x) - T_1(y)\|$.

For continuity of $T_2$ you could estimate $\|T_2(x) - T_2(y)\|$ for given $x, y \in C[0,1]$. Bring everything under one integral and see if you can extract $\|x - y\|$ from the integrand.

2. As you have written it, you start with $|T_1(x)(t)|$, not $\|T_1(x)\|$. This notation is confusing you in the second line.

First estimate $|T_1(x)(t)|$ for arbitrary $x \in C[0,1]$ and $t \in [0,1]$, as you started to do in the first line.Then (by taking the supremum over $t$) try to arrive at an estimate of the form $\|T_1(x)\| \le C \|x\|$ for all $x \in C[0,1]$, where $C$ is a numerical constant. Finally, deduce from this an estimate for $\|T_1\|$.

This gives you one inequality. Try to show that this inequality is in fact an equality by making an appropriate choice for $x$.

 

[S.G. Janssens]

As a clarifying (hopefully!) side note:

There is a common abuse of notation - also in your exercise - to denote both the function norm $\|x\|$ and the operator norm $\|T\|$ by the same symbol $\|\cdot\|$. This does not lead to confusion, as long as you make sure you are aware (in your head) of what norm you are actually using. This is important.

Some people avoid this by attaching subscripts to $\|\cdot\|$, but a lot of others do not do this because it often leads to very "heavy" notation.

 

[mathmari]

1. For $T_0$ I would write: $x,y \in C^1[0,1]$ and for $T_{1,2}$ I would write: $x,y \in C[0,1]$. For general functions the operators may not be defined.

Why is it like that? have we not that $T_0, T_1, T_2: C^1[0,1]\rightarrow C[0,1]$ ? (Wondering)

Remember to integrate from $0$ to $t$, not from $0$ to $1$.

Oh yes! (Blush)

Indeed, $T_2$ is clearly not linear because of the square, but I would perhaps prefer an explicit counter-example, with specific function(s).

Let $x(t)=t$ and $y(t)=1$. Then we have that \begin{align*}T_2(t+1)&=\int_0^ts(s+1)^2\, ds=\int_0^ts(s^2+2s+1)\, ds=\int_0^t(s^3+2s^2+s)\, ds \\ & =\left [\frac{s^4}{4}+\frac{2s^3}{3}+\frac{s^2}{2}\right ]_0^t=\frac{t^4}{4}+\frac{2t^3}{3}+\frac{t^2}{2}\end{align*} But \begin{equation*}T_2(t)+T_2(1)=\int_0^ts\cdot s^2\, ds+\int_0^ts\cdot 1^2\, ds=\int_0^t s^3\, ds+\int_0^ts\, ds=\left [\frac{s^4}{4}\right ]_0^t+\left [\frac{s^2}{2}\right ]_0^t=\frac{t^4}{4}+\frac{t^2}{2}\end{equation*} So we see that $T_2(t+1)\neq T_2(t)+T_2(1)$, right? (Wondering)

I would worry about continuity of $T_0$: Uniformly close functions need not have uniformly close derivatives. (It is up to you to find an example and make this precise.)

You mean that it can be that $x(t)$ is continuous but $x'(t)$ is discontinuous? (Wondering)

For continuity of $T_1$ you can use question 2. (Why?) You can also directly estimate $\|T_1(x) - T_1(y)\|$.

We have that \begin{equation*}\|T_1(x) - T_1(y)\|=\left \|\int_0^tsx(s)\, ds-\int_0^tsy(s)\, ds\right \|=\left \|\int_0^ts\left [x(s)-y(s)\right ]\, ds\right \|\end{equation*} We have that $t\in [0,1]$, can we maximize the integral with upper bound $1$ insstead of $1$ ? (Wondering)

2. As you have written it, you start with $|T_1(x)(t)|$, not $\|T_1(x)\|$. This notation is confusing you in the second line.

First estimate $|T_1(x)(t)|$ for arbitrary $x \in C[0,1]$ and $t \in [0,1]$, as you started to do in the first line.Then (by taking the supremum over $t$) try to arrive at an estimate of the form $\|T_1(x)\| \le C \|x\|$ for all $x \in C[0,1]$, where $C$ is a numerical constant. Finally, deduce from this an estimate for $\|T_1\|$.

This gives you one inequality. Try to show that this inequality is in fact an equality by making an appropriate choice for $x$.

I haven't really understood why we have to consider $|T_1(x)(t)|$ instead of $\|T_1(x)(t)\|$. Could you explain it further to me?
At the exercise statement we have that $\displaystyle{\|T_1\|:=\sup_{\|x\|\leq 1}\|T_1(x)\|}$, do you mean that it this is wrong and it should be $\displaystyle{\|T_1\|:=\sup_{\|x\|\leq 1}|T_1(x)|}$ ?

(Wondering)

 

[S.G. Janssens]

Why is it like that? have we not that $T_0, T_1, T_2: C^1[0,1]\rightarrow C[0,1]$ ? (Wondering)

Yes, that is fine, too, and you were right: This is the way it was formulated in your exercise. My issue was more that you should not write: "Let $x, y : [0,1] \to \mathbb{R}$..." because then you allow arbitrary functions, for which derivatives and integrals may not exist.

Let $x(t)=t$ and $y(t)=1$. Then we have that \begin{align*}T_2(t+1)&=\int_0^ts(s+1)^2\, ds=\int_0^ts(s^2+2s+1)\, ds=\int_0^t(s^3+2s^2+s)\, ds \\ & =\left [\frac{s^4}{4}+\frac{2s^3}{3}+\frac{s^2}{2}\right ]_0^t=\frac{t^4}{4}+\frac{2t^3}{3}+\frac{t^2}{2}\end{align*} But \begin{equation*}T_2(t)+T_2(1)=\int_0^ts\cdot s^2\, ds+\int_0^ts\cdot 1^2\, ds=\int_0^t s^3\, ds+\int_0^ts\, ds=\left [\frac{s^4}{4}\right ]_0^t+\left [\frac{s^2}{2}\right ]_0^t=\frac{t^4}{4}+\frac{t^2}{2}\end{equation*} So we see that $T_2(t+1)\neq T_2(t)+T_2(1)$, right? (Wondering)

Yes, except that I would start with $T_2(x + y)(t)$ instead of $T_2(t + 1)$. (For these kinds of problems, it helps when you are very strict about your notation, keeping functions distinct for the formal variable they happen to use.)

Similarly, I would start with $T_2(x)(t)$ and $T_2(y)(t)$.

You mean that it can be that $x(t)$ is continuous but $x'(t)$ is discontinuous? (Wondering)

No, since $T_0$ acts on $C^1[0,1]$, its range is going to consist of continuous functions. What I mean is that there exist $C^1$-functions that are close in the sup-norm but with derivatives (that are themselves continuous functions) that are not close in the sup-norm.

We have that \begin{equation*}\|T_1(x) - T_1(y)\|=\left \|\int_0^tsx(s)\, ds-\int_0^tsy(s)\, ds\right \|=\left \|\int_0^ts\left [x(s)-y(s)\right ]\, ds\right \|\end{equation*} We have that $t\in [0,1]$, can we maximize the integral with upper bound $1$ insstead of $1$ ? (Wondering)

In the second and third term you should take the supremum over all $t \in [0,1]$ and use $|\cdot|$instead of $\|\cdot\|$. You can see this, because as you write it now, there is no $t$ on the left, but the $t$ does appear on the right.

Yes, your idea is correct. Bring the absolute values inside the integral and maximize the integral of a non-negative integrand by taking the integration domain as large as possible.

I haven't really understood why we have to consider $|T_1(x)(t)|$ instead of $\|T_1(x)(t)\|$. Could you explain it further to me?

$|T_1(x)(t)|$ is the absolute value of a real number, namely, the value of the function $T_1(x)$ at the point $t$. On the other hand, $\|T_1(x)\|$ is the norm of the function $T_1(x)$ itself, defined by taking the supremum of $|T_1(x)(t)|$ over $t \in [0,1]$. Moreover, $\|T_1\|$ is the operator norm of $T_1$.

By "abuse of notation" in post #2 I meant that $\|\cdot\|$ plays two roles: It is the norm of elements of $C[0,1]$ or $C^1[0,1]$ (functions), but it is also the norm of the operator $T_1$ that acts on these functions.

At the exercise statement we have that $\displaystyle{\|T_1\|:=\sup_{\|x\|\leq 1}\|T_1(x)\|}$, do you mean that it this is wrong and it should be $\displaystyle{\|T_1\|:=\sup_{\|x\|\leq 1}|T_1(x)|}$ ?

(Wondering)

No, the statement in the exercise is fine. To distinguish between the two roles of $\|\cdot\|$, people sometimes write things like $\|x\|_{C[0,1]}$ and $\|T_1\|_{C[0,1] \to C[0,1]}$, but I would not do that since the exercise does not use this notation either.

 

[mathmari]

Yes, that is fine, too, and you were right: This is the way it was formulated in your exercise. My issue was more that you should not write: "Let $x, y : [0,1] \to \mathbb{R}$..." because then you allow arbitrary functions, for which derivatives and integrals may not exist.

So, you mean that we have to write $x, y : [0,1] \to C^1[0,1]$ ? (Wondering)

Yes, except that I would start with $T_2(x + y)(t)$ instead of $T_2(t + 1)$. (For these kinds of problems, it helps when you are very strict about your notation, keeping functions distinct for the formal variable they happen to use.)

Similarly, I would start with $T_2(x)(t)$ and $T_2(y)(t)$.

Ah ok, I see!

We could also take $x(t)=y(t)=1$, couldn't we?

\begin{align*}&T_2(x+y)(t)=T_2(1+1)=T_2(2)=\int_0^ts\cdot 2^2\, ds=4\cdot \int_0^ts\, ds =4\cdot \left [\frac{s^2}{2}\right ]_0^t=2t^2\\ & T_2(x)(t)+T_2(y)(t)=T_2(1)+T_2(1)=2\cdot T_2(1)=2\cdot \int_0^ts\cdot 1^2\, ds=2\cdot \int_0^ts\, ds =2\cdot \left [\frac{s^2}{2}\right ]_0^t=t^2\end{align*}

Is this maybe a better example? (Wondering)

 

[mathmari]

$|T_1(x)(t)|$ is the absolute value of a real number, namely, the value of the function $T_1(x)$ at the point $t$. On the other hand, $\|T_1(x)\|$ is the norm of the function $T_1(x)$ itself, defined by taking the supremum of $|T_1(x)(t)|$ over $t \in [0,1]$. Moreover, $\|T_1\|$ is the operator norm of $T_1$.

By "abuse of notation" in post #2 I meant that $\|\cdot\|$ plays two roles: It is the norm of elements of $C[0,1]$ or $C^1[0,1]$ (functions), but it is also the norm of the operator $T_1$ that acts on these functions.

Does it mean that we have the following?
\begin{align*}\|T_1(x)\|&=\sup_{t\in [0,1]}|T_1(x)(t)|=\sup_{t\in [0,1]}\left |\int_0^tsx(s)\, ds\right |\leq \sup_{t\in [0,1]}\int_0^t|sx(s)|\, ds \\ & =\int_0^1|sx(s)|\, ds=\int_0^1|s|\cdot |x(s)|\, ds=\int_0^1s\cdot |x(s)|\, ds\end{align*}
(Wondering)

 

[S.G. Janssens]

So, you mean that we have to write $x, y : [0,1] \to C^1[0,1]$ ? (Wondering)

No, we would write $x, y \in C^1[0,1]$ or, equivalently, $x, y : [0,1] \to \mathbb{R}$ are continuously differentiable. (This means that $x, y$ are differentiable and their derivative is continuous.)

Ah ok, I see!

We could also take $x(t)=y(t)=1$, couldn't we?

\begin{align*}&T_2(x+y)(t)=T_2(1+1)=T_2(2)=\int_0^ts\cdot 2^2\, ds=4\cdot \int_0^ts\, ds =4\cdot \left [\frac{s^2}{2}\right ]_0^t=2t^2\\ & T_2(x)(t)+T_2(y)(t)=T_2(1)+T_2(1)=2\cdot T_2(1)=2\cdot \int_0^ts\cdot 1^2\, ds=2\cdot \int_0^ts\, ds =2\cdot \left [\frac{s^2}{2}\right ]_0^t=t^2\end{align*}

Is this maybe a better example? (Wondering)

Sorry for being so much stuck in Freud's second stage, but I would write:
\[
T_2(x + y)(t) = T_2(1 + 1)(t) = T_2(2)(t),
\]
where, in fact, 1 really is the function $t \mapsto 1$, etc. Similarly, I would write
\[
T_2(x)(t) + T_2(y)(t) = T_2(1)(t) + T_2(1)(t).
\]
You can check this yourself: When you suppress $t$ on the left-hand sides, and it does appear on the right-hand side (in the upper limit of the integral), this cannot be correct.

The rest of the lines is ok.

Does it mean that we have the following?
\begin{align*}\|T_1(x)\|&=\sup_{t\in [0,1]}|T_1(x)(t)|=\sup_{t\in [0,1]}\left |\int_0^tsx(s)\, ds\right |\leq \sup_{t\in [0,1]}\int_0^t|sx(s)|\, ds \\ & =\int_0^1|sx(s)|\, ds=\int_0^1|s|\cdot |x(s)|\, ds=\int_0^1s\cdot |x(s)|\, ds\end{align*}
(Wondering)

Yes, very good.

 

[mathmari]

No, we would write $x, y \in C^1[0,1]$ or, equivalently, $x, y : [0,1] \to \mathbb{R}$ are continuously differentiable. (This means that $x, y$ are differentiable and their derivative is continuous.)

Ah I see! (Nerd)

Sorry for being so much stuck in Freud's second stage, but I would write:
\[
T_2(x + y)(t) = T_2(1 + 1)(t) = T_2(2)(t),
\]
where, in fact, 1 really is the function $t \mapsto 1$, etc. Similarly, I would write
\[
T_2(x)(t) + T_2(y)(t) = T_2(1)(t) + T_2(1)(t).
\]
You can check this yourself: When you suppress $t$ on the left-hand sides, and it does appear on the right-hand side (in the upper limit of the integral), this cannot be correct.

The rest of the lines is ok.

Ah ok! (Smile)

Yes, very good.

Then we have that \begin{equation*}\|T_1\|=\sup_{\|x\|\leq 1}\|T_1(x)\|\leq \sup_{\|x\|\leq 1}\int_0^1s\cdot |x(s)|\, ds\end{equation*} $\|x\|$ is not the same as $|x(s)|$, is it? (Wondering)

 

[S.G. Janssens]

Then we have that \begin{equation*}\|T_1\|=\sup_{\|x\|\leq 1}\|T_1(x)\|\leq \sup_{\|x\|\leq 1}\int_0^1s\cdot |x(s)|\, ds\end{equation*} $\|x\|$ is not the same as $|x(s)|$, is it? (Wondering)

No, you are right, it is not the same (the latter depends on $s$ while the former does not), but you can estimate $|x(s)|$ by $\|x\|$, namely,
\[
|x(s)| \le \sup_{\sigma \in [0,1]}{|x(\sigma)|} = \|x\|.
\]

 

[mathmari]

No, you are right, it is not the same (the latter depends on $s$ while the former does not), but you can estimate $|x(s)|$ by $\|x\|$, namely,
\[
|x(s)| \le \sup_{\sigma \in [0,1]}{|x(\sigma)|} = \|x\|.
\]

So, we have that \begin{align*}\|T_1(x)\|&=\sup_{t\in [0,1]}|T_1(x)(t)|=\sup_{t\in [0,1]}\left |\int_0^tsx(s)\, ds\right |\leq \sup_{t\in [0,1]}\int_0^t|sx(s)|\, ds \\ & =\int_0^1|sx(s)|\, ds=\int_0^1|s|\cdot |x(s)|\, ds=\int_0^1s\cdot |x(s)|\, ds \\ & \leq \int_0^1s\cdot \sup_{\sigma \in [0,1]}{|x(\sigma)|}\, ds=\int_0^1s\cdot \|x\|\, ds\end{align*} And therefore we get \begin{equation*}\|T_1\|=\sup_{\|x\|\leq 1}\|T_1(x)\|\leq \sup_{\|x\|\leq 1}\int_0^1s\cdot \|x\|\, ds=\int_0^1s\cdot 1\, ds=\int_0^1s\, ds=\left [\frac{s^2}{2}\right ]_0^1=\frac{1}{2}\end{equation*}
Is this correct? (Wondering)

But how can we get an equality for $\|T_1\|$ and not an inequality? (Wondering)

 

[mathmari]

For continuity of $T_1$ you can use question 2. (Why?) You can also directly estimate $\|T_1(x) - T_1(y)\|$.

Let $x, y \in C[0,1]$ then we have the following:
\begin{align*}\|T_1(x)-T_1(y)\|&=\sup_{t\in [0,1]}|T_1(x)(t)-T_1(y)(t)|=\sup_{t\in [0,1]}\left|\int_0^tsx(s)\, ds-\int_0^tsy(s)\, ds\right | \\ & = \sup_{t\in [0,1]}\left |\int_0^ts\left (x(s)-y(s)\right )\, ds\right |=\left |\int_0^1s\left (x(s)-y(s)\right )\, ds\right | \\ & \leq \int_0^1\left |s\left (x(s)-y(s)\right )\right |\, ds=\int_0^1\left |s\right |\cdot \left |x(s)-y(s)\right |\, ds \\ & = \int_0^1s\cdot \left |x(s)-y(s)\right |\, ds\leq \int_0^1s\cdot \sup_{\sigma\in[0,1]}\left |x(\sigma)-y(\sigma)\right |\, ds\\ & = \int_0^1s\cdot \|x-y\|\, ds= \|x-y\|\cdot \int_0^1s\, ds=\|x-y\|\cdot \left [\frac{s^2}{2}\right ]_0^1\\ & = \|x-y\|\cdot \frac{1}{2}=\frac{\|x-y\|}{2}\end{align*} This makes $T_1$ Lipschitz continuous, hence also continuous.

Is everything correct? (Wondering)

For continuity of $T_2$ you could estimate $\|T_2(x) - T_2(y)\|$ for given $x, y \in C[0,1]$. Bring everything under one integral and see if you can extract $\|x - y\|$ from the integrand.

\begin{align*}\|T_2(x)-T_2(y)\|&=\sup_{t\in [0,1]}|T_2(x)(t)-T_2(y)(t)|=\sup_{t\in [0,1]}\left|\int_0^tsx^2(s)\, ds-\int_0^tsy^2(s)\, ds\right | \\ & = \sup_{t\in [0,1]}\left |\int_0^ts\left (x^2(s)-y^2(s)\right )\, ds\right |=\left |\int_0^1s\left (x^2(s)-y^2(s)\right )\, ds\right | \\ & \leq \int_0^1\left |s\left (x^2(s)-y^2(s)\right )\right |\, ds=\int_0^1\left |s\right |\cdot \left |x^2(s)-y^2(s)\right |\, ds \\ & = \int_0^1s\cdot \left |x^2(s)-y^2(s)\right |\, ds\leq \int_0^1s\cdot \sup_{\sigma\in[0,1]}\left |x^2(\sigma)-y^2(\sigma)\right |\, ds\\ & = \int_0^1s\cdot \|x^2-y^2\|\, ds= \|x^2-y^2\|\cdot \int_0^1s\, ds=\|x^2-y^2\|\cdot \left [\frac{s^2}{2}\right ]_0^1\\ & = \|x^2-y^2\|\cdot \frac{1}{2}=\frac{\|x^2-y^2\|}{2}=\frac{\|(x-y)(x+y)\|}{2}\leq \frac{\|x-y\|\|x+y\|}{2}\end{align*} Does this mean that $T_2$ is Lipschitz continuous? Or do we have to bound $\|x+y\|$ ? (Wondering)

 

[S.G. Janssens]

Is this correct? (Wondering)

Yes, very good.

But how can we get an equality for $\|T_1\|$ and not an inequality? (Wondering)

You can get an equality by looking at the inequalities you wrote in the sequence of estimates. Can you find a function $x_1$ with $\|x_1\| \ge 1$ for which these inequalities are in fact equalities? Namely, in that case,you have
\[
\|T_1\| = \sup_{\|x\| \le 1}{\|T_1x\|} \ge \|T_1x_1\| \ge \frac{1}{2},
\]
by the definition of the supremum, and you are done.

 

[S.G. Janssens]

Let $x, y \in C[0,1]$ then we have the following:
\begin{align*}\|T_1(x)-T_1(y)\|&=\sup_{t\in [0,1]}|T_1(x)(t)-T_1(y)(t)|=\sup_{t\in [0,1]}\left|\int_0^tsx(s)\, ds-\int_0^tsy(s)\, ds\right | \\ & = \sup_{t\in [0,1]}\left |\int_0^ts\left (x(s)-y(s)\right )\, ds\right |=\left |\int_0^1s\left (x(s)-y(s)\right )\, ds\right | \\ & \leq \int_0^1\left |s\left (x(s)-y(s)\right )\right |\, ds=\int_0^1\left |s\right |\cdot \left |x(s)-y(s)\right |\, ds \\ & = \int_0^1s\cdot \left |x(s)-y(s)\right |\, ds\leq \int_0^1s\cdot \sup_{\sigma\in[0,1]}\left |x(\sigma)-y(\sigma)\right |\, ds\\ & = \int_0^1s\cdot \|x-y\|\, ds= \|x-y\|\cdot \int_0^1s\, ds=\|x-y\|\cdot \left [\frac{s^2}{2}\right ]_0^1\\ & = \|x-y\|\cdot \frac{1}{2}=\frac{\|x-y\|}{2}\end{align*} This makes $T_1$ Lipschitz continuous, hence also continuous.

Is everything correct? (Wondering)

Yes, I agree with everything except the 4th equality, but the estimate also stands when you remove that one, so this aside it is correct.

Alternatively, using your answer to question 2,
\[
\|T_1x - T_1y\| = \|T_1(x - y)\| \le \|T_1\| \cdot \|x - y\| = \frac{1}{2} \|x - y\|,
\]
for every $x, y \in C[0,1]$, so then you also have a Lipschitz estimate. (In fact, $T_1$ is a contraction.)

\begin{align*}\|T_2(x)-T_2(y)\|&=\sup_{t\in [0,1]}|T_2(x)(t)-T_2(y)(t)|=\sup_{t\in [0,1]}\left|\int_0^tsx^2(s)\, ds-\int_0^tsy^2(s)\, ds\right | \\ & = \sup_{t\in [0,1]}\left |\int_0^ts\left (x^2(s)-y^2(s)\right )\, ds\right |=\left |\int_0^1s\left (x^2(s)-y^2(s)\right )\, ds\right | \\ & \leq \int_0^1\left |s\left (x^2(s)-y^2(s)\right )\right |\, ds=\int_0^1\left |s\right |\cdot \left |x^2(s)-y^2(s)\right |\, ds \\ & = \int_0^1s\cdot \left |x^2(s)-y^2(s)\right |\, ds\leq \int_0^1s\cdot \sup_{\sigma\in[0,1]}\left |x^2(\sigma)-y^2(\sigma)\right |\, ds\\ & = \int_0^1s\cdot \|x^2-y^2\|\, ds= \|x^2-y^2\|\cdot \int_0^1s\, ds=\|x^2-y^2\|\cdot \left [\frac{s^2}{2}\right ]_0^1\\ & = \|x^2-y^2\|\cdot \frac{1}{2}=\frac{\|x^2-y^2\|}{2}=\frac{\|(x-y)(x+y)\|}{2}\leq \frac{\|x-y\|\|x+y\|}{2}\end{align*} Does this mean that $T_2$ is Lipschitz continuous? Or do we have to bound $\|x+y\|$ ? (Wondering)

As with your previous estimate, I don't agree with the 4th equality, but fortunately that does not invalidate your estimate as a whole, well done.

No, we don't have a Lipschitz estimate, but think about just continuity in $x$. For example, you have
\[
\|x + y\| \le \|x\| + \|y\| = \|x\| + \|y - x + x\| \le 2\|x\| + \|x - y\|,
\]
so
\[
\|T_2x - T_2y\| \le \frac{1}{2} \|x - y\| \left(2\|x\| + \|x - y\| \right).
\]
Can you now apply the usual $(\epsilon,\delta)$-definition to deduce continuity at $x$?

P.S. Sorry for taking a week to follow up, I don't always have the time. You are improving a lot and you take asking questions very seriously, so good job!