MHB Is there a contradiction in assuming that $J$ is linear over $C^1([a,b])$?

[evinda]

Hello! (Wave)

Is the following functional over $C^1([a,b])$ linear?

$$J(y)= \int_a^b (y')^2 dx+ G(y(b))$$

That's what I have thought:if $J$ would be linear it would have to hold:

$\forall y \in C^1([a,b]), \forall \lambda \in \mathbb{R}$

$J(\lambda y)=\lambda J(y)$ or equivalently

$\int_a^b (\lambda y')^2 dx+ G(\lambda y(b))=\lambda \left( \int_a^b (y')^2 dx+ G(y(b))\right)\\ \Rightarrow \lambda^2 \int_a^b (y')^2 dx+ G(\lambda y(b))= \lambda \int_a^b (y')^2 dx+ \lambda G(y(b)) $

For $\lambda=2, y=x$ the equality becomes:$4 \int_a^b 1 dx+ G(2b)=2 \int_a^b 1 dx+ 2 G(b) \Rightarrow 2(b-a)=2G(b)-G(2b) $

Can we take a specific function $G$ ? Or how else could we find a contradiction? (Thinking)

 

[I like Serena]

Hello! (Wave)

Is the following functional over $C^1([a,b])$ linear?

$$J(y)= \int_a^b (y')^2 dx+ G(y(b))$$

That's what I have thought:if $J$ would be linear it would have to hold:

$\forall y \in C^1([a,b]), \forall \lambda \in \mathbb{R}$

$J(\lambda y)=\lambda J(y)$ or equivalently

$\int_a^b (\lambda y')^2 dx+ G(\lambda y(b))=\lambda \left( \int_a^b (y')^2 dx+ G(y(b))\right)\\ \Rightarrow \lambda^2 \int_a^b (y')^2 dx+ G(\lambda y(b))= \lambda \int_a^b (y')^2 dx+ \lambda G(y(b)) $

For $\lambda=2, y=x$ the equality becomes:$4 \int_a^b 1 dx+ G(2b)=2 \int_a^b 1 dx+ 2 G(b) \Rightarrow 2(b-a)=2G(b)-G(2b) $

Can we take a specific function $G$ ? Or how else could we find a contradiction? (Thinking)

Hey! (Smile)

If it is a linear functional, there will have to be constraints on $G$.
Suppose we prove that whatever we pick for $G$, it won't be good enough?

Say, what if we pick the same example with y=2x.
Then what will we find as constraint for $G$? (Wondering)

 

[evinda]

Hey! (Smile)

If it is a linear functional, there will have to be constraints on $G$.
Suppose we prove that whatever we pick for $G$, it won't be good enough?

Say, what if we pick the same example with y=2x.
Then what will we find as constraint for $G$? (Wondering)

Then it will be as follows, right?

$$J(\lambda y)= \lambda J(y) \Rightarrow \int_a^b (\lambda y)' dx+ G(\lambda y(b))= \lambda \int_a^b (y')^2 dx+ \lambda G(y(b))$$

For $\lambda=2$ and $y(x)=2x$ we have:

$$4 \int_a^b 1 dx+ G(4b)=8 \int_a^b 1 dx+ 2 G(b) \Rightarrow G(4b)-2G(b)=4(b-a)$$

But how can we find constraint for $G$? (Thinking)

 

[I like Serena]

For $\lambda=2$ and $y(x)=2x$ we have:

$$4 \int_a^b 1 dx+ G(4b)=8 \int_a^b 1 dx+ 2 G(b) \Rightarrow G(4b)-2G(b)=4(b-a)$$

Shouldn't that be:
$$2^2 \int_a^b (2)^2 dx+ G(2\cdot 2b)=2 \int_a^b (2)^2 dx+ 2 G(2b) \Rightarrow 2G(2b)-G(4b)=8(b-a)$$
(Wondering)

But how can we find constraint for $G$? (Thinking)

From your first example, if we pick $\lambda=2, y=x, a=0, b=1$, we find $2G(b)-G(2b)=2(b-a) \Rightarrow \boxed{2G(1)-G(2)=2}$.

When we pick $\lambda=2, y=2x, a=0, b=\frac 12$, we find $2G(2b)-G(4b)=8(b-a) \Rightarrow \boxed{2G(1)-G(2)=4}$.

This is a contradiction. (Nerd)

 

[evinda]

Shouldn't that be:
$$2^2 \int_a^b (2)^2 dx+ G(2\cdot 2b)=2 \int_a^b (2)^2 dx+ 2 G(2b) \Rightarrow 2G(2b)-G(4b)=8(b-a)$$
(Wondering)

Oh yes, right... (Nod)

From your first example, if we pick $\lambda=2, y=x, a=0, b=1$, we find $2G(b)-G(2b)=2(b-a) \Rightarrow \boxed{2G(1)-G(2)=2}$.

When we pick $\lambda=2, y=2x, a=0, b=\frac 12$, we find $2G(2b)-G(4b)=8(b-a) \Rightarrow \boxed{2G(1)-G(2)=4}$.

This is a contradiction. (Nerd)

I see... But can we take two different $b$s, although the functional is defined over $C^1([a,b])$ where $a,b$ are fixed? (Thinking)

 

[I like Serena]

Oh yes, right... (Nod)
I see... But can we take two different $b$s, although the functional is defined over $C^1([a,b])$ where $a,b$ are fixed? (Thinking)

Oh yes, you're right... (Blush)

Then let's pick $\lambda = 2$ and $y=2x - b$.

Then we get:
$$2g(y(b)) - g(2y(b)) = 8(b-a) \Rightarrow \boxed{2g(b) - g(2b) = 8(b-a)}$$

This is a contradiction, assuming that $a\ne b$. (Thinking)

 

[evinda]

Oh yes, you're right... (Blush)

Then let's pick $\lambda = 2$ and $y=2x - b$.

Then we get:
$$2g(y(b)) - g(2y(b)) = 8(b-a) \Rightarrow \boxed{2g(b) - g(2b) = 8(b-a)}$$

This is a contradiction, assuming that $a\ne b$. (Thinking)

I see... Thanks a lot! (Mmm) (Smirk)

 

[evinda]

Oh yes, you're right... (Blush)

Then let's pick $\lambda = 2$ and $y=2x - b$.

Would we also get a contradiction picking an other $y$ for $\lambda=2$ or is it the only possibility? (Thinking)

 

[I like Serena]

Would we also get a contradiction picking an other $y$ for $\lambda=2$ or is it the only possibility? (Thinking)

Almost any other $y$ will also work, as long as $y(b)=b$ as you pointed out.
For instance $y=b$. (Wasntme)