Why do we deduce that the functional is continuous in respect to the other norm?

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Discussion Overview

The discussion revolves around the continuity of a functional \( J \) defined on the space \( V = C^1([a,b]) \) with respect to different norms, specifically \( ||y||_1 \) and \( ||y||_\infty \). Participants explore the implications of continuity in one norm for the other, as well as the conditions under which the inverse does not hold.

Discussion Character

  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • Some participants assert that if \( J \) is continuous with respect to the norm \( ||\cdot||_1 \), then it must also be continuous with respect to the norm \( ||\cdot||_\infty \).
  • Others argue that the converse is not true, providing examples to illustrate that continuity in \( ||\cdot||_\infty \) does not imply continuity in \( ||\cdot||_1 \).
  • A participant questions the reasoning behind the assertion that continuity implies boundedness, suggesting that if \( J \) is continuous, then there exists a constant \( M \) such that \( |J(y)| \leq M ||y||_\infty \) for all \( y \in V \).
  • Another participant clarifies that a linear functional is continuous if and only if it is bounded, reinforcing the relationship between continuity and boundedness.
  • There is a discussion about the specific case of \( J(y) = y'(0) \) and whether it can be expressed in terms of an integral functional.
  • Some participants express uncertainty about the conditions needed to demonstrate the failure of the converse claim regarding continuity.

Areas of Agreement / Disagreement

Participants generally agree on the implications of continuity in one norm leading to continuity in another, but there is disagreement regarding the converse. The discussion remains unresolved regarding the specific conditions needed to illustrate the failure of the converse claim.

Contextual Notes

Participants note that the definitions and assumptions regarding the continuity of functionals are crucial to the discussion, particularly the linearity of \( J \) and the implications of boundedness.

evinda
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Hello! (Wave)

Let $V=C^1([a,b])$. Show that if $J$ is a continuous functional in respect to the norm $||y||_1:=||y||_{\infty}+||y'||_{\infty}, y \in V$ then it is also continuous in respect to the norm $||y||:=||y||_{\infty}$.
Also, show that the inverse of the above claim does not hold.

Let $y_1, y_2 \in C^1([a,b])$. Then,

$$||y_1-y_2||_1=||y_1-y_2||_{\infty}+||y_1'-y_2'||_{\infty} \geq ||y_1-y_2||_{\infty} (\star)$$

and so from $(\star)$ we have that if $(y_n)_{n=1}^{\infty} \subset C^1([a,b])$ and $y \in C^1([a,b])$ with $||y_n-y||_1 \to 0$ then $||y_n-y||_{\infty} \to 0$.

So the continuity of $J$ in respect to the norm $|| \cdot ||_1$ ensures the continuity of $J$ in respect to the norm $||\cdot||_{\infty}$.

Why having shown that if $||y_n-y||_1 \to 0$ then $||y_n-y||_{\infty} \to 0$ do we deduce the continuity of $J$ in respect to the norm $||\cdot||_{\infty}$? (Thinking)The definition of a continuous functional is the following:

Let $(V, ||\cdot||)$ be a linear space with norm and let $J:V \to \mathbb{R}$ be a functional.
We say that the functional $J: V \to \mathbb{R}$ is continuous at $y_0 \in V$ if for all $\epsilon>0$ there exists a $\delta>0$ such that whenever $||y-y_0||< \delta$ for $y \in V$ then $|J(y)-J(y_0)|< \epsilon$
(or equivalently, if $y_n \in V, n=1,2, \dots$ with $||y_n-y_0|| \to 0$ then $|J(y_n)-J(y_0)| \to 0$).
 
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evinda said:
Hello! (Wave)

Let $V=C^1([a,b])$. Show that if $J$ is a continuous functional in respect to the norm $||y||_1:=||y||_{\infty}+||y'||_{\infty}, y \in V$ then it is also continuous in respect to the norm $||y||:=||y||_{\infty}$.
Also, show that the inverse of the above claim does not hold.

Let $y_1, y_2 \in C^1([a,b])$. Then,

$$||y_1-y_2||_1=||y_1-y_2||_{\infty}+||y_1'-y_2'||_{\infty} \geq ||y_1-y_2||_{\infty} (\star)$$

and so from $(\star)$ we have that if $(y_n)_{n=1}^{\infty} \subset C^1([a,b])$ and $y \in C^1([a,b])$ with $||y_n-y||_1 \to 0$ then $||y_n-y||_{\infty} \to 0$.

So the continuity of $J$ in respect to the norm $|| \cdot ||_1$ ensures the continuity of $J$ in respect to the norm $||\cdot||_{\infty}$.

Why having shown that if $||y_n-y||_1 \to 0$ then $||y_n-y||_{\infty} \to 0$ do we deduce the continuity of $J$ in respect to the norm $||\cdot||_{\infty}$? (Thinking)The definition of a continuous functional is the following:

Let $(V, ||\cdot||)$ be a linear space with norm and let $J:V \to \mathbb{R}$ be a functional.
We say that the functional $J: V \to \mathbb{R}$ is continuous at $y_0 \in V$ if for all $\epsilon>0$ there exists a $\delta>0$ such that whenever $||y-y_0||< \delta$ for $y \in V$ then $|J(y)-J(y_0)|< \epsilon$
(or equivalently, if $y_n \in V, n=1,2, \dots$ with $||y_n-y_0|| \to 0$ then $|J(y_n)-J(y_0)| \to 0$).
You seem to have this problem the wrong way round. In fact, if $J$ is continuous for the norm $\|\cdot\|_\infty$ then it is continuous for the norm $\|\cdot\|_1$. But the converse is false.

Recall that a linear functional is continuous if and only if it is bounded. So if $J$ is continuous for the norm $\|\cdot\|_\infty$ then there is a constant $M$ such that $|J(y)| \leqslant M\|y\|_\infty$ for all $y$ in $V$. From $(\star)$, $\|y\|_\infty \leqslant \|y\|_1$. It follows that $|Jy| \leqslant M\|y\|_1$ for all $y$ in $V$, so that $J$ is bounded (and hence continuous) for the $\|\cdot\|_1$-norm.

To see that the converse is false, you could take $V = C^1([-\pi,\pi])$ and $J(y) = y'(0)$. Then if for example $y_n(x) = \frac1n\sin(nx)$ you can check that $J$ is continuous for the $\|\cdot\|_1$-norm but not for the $\|\cdot\|_\infty$-norm (because $y_n\to0$ for the $\|\cdot\|_\infty$-norm, but $J(y)\not\to0$).
 
Opalg said:
You seem to have this problem the wrong way round. In fact, if $J$ is continuous for the norm $\|\cdot\|_\infty$ then it is continuous for the norm $\|\cdot\|_1$. But the converse is false.

Recall that a linear functional is continuous if and only if it is bounded. So if $J$ is continuous for the norm $\|\cdot\|_\infty$ then there is a constant $M$ such that $|J(y)| \leqslant M\|y\|_\infty$ for all $y$ in $V$. From $(\star)$, $\|y\|_\infty \leqslant \|y\|_1$. It follows that $|Jy| \leqslant M\|y\|_1$ for all $y$ in $V$, so that $J$ is bounded (and hence continuous) for the $\|\cdot\|_1$-norm.

Could you explain me further why if $J$ is continuous for the norm $\|\cdot\|_\infty$ then there is a constant $M$ such that $|J(y)| \leqslant M\|y\|_\infty$ for all $y$ in $V$? (Thinking)
 
evinda said:
Could you explain me further why if $J$ is continuous for the norm $\|\cdot\|_\infty$ then there is a constant $M$ such that $|J(y)| \leqslant M\|y\|_\infty$ for all $y$ in $V$? (Thinking)
The problem states that "J is a continuous functional". I am assuming this means that J is a continuous linear functional. (If $J$ is not linear then my comments are not relevant.)

If J is continuous then it is continuous at $0$. So given $\varepsilon>0$ there exists $\delta>0$ such that $|J(y)|<\varepsilon$ whenever $\|y\|<\delta.$ Now suppose that $x\in V$. Then $\left\|\dfrac{\delta\, x}{2\|x\|}\right\| <\delta$, so that $\left|J\Bigl(\dfrac {\delta\, x}{2\|x\|}\Bigr)\right| <\varepsilon$. If we now assume that $J$ is linear, then $J\Bigl(\dfrac{\delta\, x}{2\|x\|}\Bigr) = \dfrac{\delta}{2\|x\|}J(x)$. It follows that $|J(x)| <\dfrac{2\varepsilon}{\delta}\|x\|$. Let $M = \dfrac{2\varepsilon}{\delta}$, and you get $|J(x) \leqslant M\|x\|.$
 
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Opalg said:
The problem states that "J is a continuous functional". I am assuming this means that J is a continuous linear functional. (If $J$ is not linear then my comments are not relevant.)

If J is continuous then it is continuous at $0$. So given $\varepsilon>0$ there exists $\delta>0$ such that $|J(y)|<\varepsilon$ whenever $\|y\|<\delta.$ Now suppose that $x\in V$. Then $\left\|\dfrac{\delta\, x}{2\|x\|}\right\| <\delta$, so that $\left|J\Bigl(\dfrac {\delta\, x}{2\|x\|}\Bigr)\right| <\varepsilon$. If we now assume that $J$ is linear, then $J\Bigl(\dfrac{\delta\, x}{2\|x\|}\Bigr) = \dfrac{\delta}{2\|x\|}J(x)$. It follows that $|J(x)| <\dfrac{2\varepsilon}{\delta}\|x\|$. Let $M = \dfrac{2\varepsilon}{\delta}$, and you get $|J(x) \leqslant M\|x\|.$


I see... (Nod)

Recall that a linear functional is continuous if and only if it is bounded. So if $J$ is continuous for the norm $\|\cdot\|_\infty$ then there is a constant $M$ such that $|J(y)| \leqslant M\|y\|_\infty$ for all $y$ in $V$. From $(\star)$, $\|y\|_\infty \leqslant \|y\|_1$. It follows that $|Jy| \leqslant M\|y\|_1$ for all $y$ in $V$, so that $J$ is bounded (and hence continuous) for the $\|\cdot\|_1$-norm.

So does it hold that a functional $J$ is bounded iff there is a constant $M$ such that $|J(y)| \leq M ||y||$ ? (Thinking)
 
evinda said:
So does it hold that a functional $J$ is bounded iff there is a constant $M$ such that $|J(y)| \leq M ||y||$ ? (Thinking)
Yes, that is what is meant by bounded in this context.
 
Opalg said:
Yes, that is what is meant by bounded in this context.

A ok.. Nice... (Smile)
Opalg said:
To see that the converse is false, you could take $V = C^1([-\pi,\pi])$ and $J(y) = y'(0)$. Then if for example $y_n(x) = \frac1n\sin(nx)$ you can check that $J$ is continuous for the $\|\cdot\|_1$-norm but not for the $\|\cdot\|_\infty$-norm (because $y_n\to0$ for the $\|\cdot\|_\infty$-norm, but $J(y)\not\to0$).
Do we get $y'(0)$ from an integral of the form $\int_{-\pi}^{\pi} g(x,y,y') dx$ ? Because a functional should be of this form. Or am I wrong? (Thinking)

Also in order to show that the converse is false, do we have to show the following?

Whenever $\left| \left| \frac{1}{n} \sin{(n x)}\right| \right|_{\infty}+\left| \left| \left( \frac{1}{n} \sin{(n x)} \right)'\right| \right|_{\infty} \to 0 $ then $\left|J \left( \frac{1}{n} \sin{(n x)} \right) \right| \to 0$

and

whenever $\left| \left| \frac{1}{n} \sin{(n x)}\right| \right|_{\infty} \to 0$ then $\left|J \left( \frac{1}{n} \sin{(n x)} \right) \right| \not\to 0$ ? (Thinking)
 
Now I saw that in my book there is the hint that we can use the functional of arc length.

The functional is this: $J(y)= \int_a^b \sqrt{1+y'(x)^2} dx$, right?

I have tried the following:

If $y_1, y_2 \in V$ then:

$$|J(y_2)-J(y_1)|= \left| \int_a^b \sqrt{1+y_2'(x)^2} dx- \int_a^b \sqrt{1+y_1'(x)^2}dx \right| = \left| \int_a^b (\sqrt{1+y_2'(x)^2}- \sqrt{1+y_1'(x)^2}) dx \right|= \left| \int_a^b \frac{y_2'(x)^2-y_1'(x)^2}{\sqrt{1+y_2'(x)^2}+ \sqrt{1+y_1'(x)^2}} \right| \leq\int_a^b \frac{|y_2'(x)^2-y_1'(x)^2|}{|\sqrt{1+y_2'(x)^2}+ \sqrt{1+y_1'(x)^2}|} dx \leq \frac{1}{2} ||y_2'-y_1'||_{\infty} (b-a)$$

But $||y_1-y_2||_{\infty}+||y_1'-y_2'||_{\infty}$ didn't appear. So have I done something wrong? (Thinking)
 
We can bound $\frac{1}{2} ||y_2'-y_1'||_{\infty} (b-a)$ by $\frac{1}{2} (||y_2-y_1||_{\infty}+||y_2'-y_1'||_{\infty}) (b-a)$ and so we deduce that $J$ is continuous for the $||\cdot||_1$ norm, right?

But how could we show that $|J(y_2)-J(y_1)|$ is not bounded by $||y_1-y_2||_{\infty}$?
What counterexample could we consider? (Thinking)
 

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