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

[evinda]

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 $|J(y_n)-J(y_0)| \to 0$).

 

[Opalg]

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 $J(y)\not\to0$).

 

[evinda]

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)

 

[Opalg]

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 $|J(x) \leqslant M\|x\|.$

 

[evinda]

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)

 

[Opalg]

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.

 

[evinda]

Yes, that is what is meant by bounded in this context.

A ok.. Nice... (Smile)

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)

 

[evinda]

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)

 

[evinda]

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)