Finding the bound of a linear functional defined on $C[-1,1]$

Question

Define the linear functional:

$$f(x)=\int_{-1}^{0}x(t)dt-\int_0^1x(t)dt$$

On the normed space $C[-1,1]$ which consists of all contiuous functions on the interval. The norm is defined as: $\|x\|= \max_{t\in[-1,1]}|x(t)|$

Now, let $x(t)\in C[-1,1]$ be arbitrary

$$\left| f(x)\right|=\left| \int_{-1}^{0}x(t)dt-\int_0^1x(t)dt\right|$$

$$|f(x)|\leq \left|\int_{-1}^{0}x(t)dt\right|+\left|\int_0^1x(t)dt\right|$$

Also,

$$\left|\int_{-1}^{0}x(t)dt\right| \leq (0-(-1))\cdot \max_{\ t\in[-1,1]}|x(t)|=\|x\|$$

In a similar way for the other integral, one can conclude:

$$|f(x)|\leq 2\|x\|\implies \frac{|f(x)|}{\|x\|}\leq 2$$

Excluding the case where $x(t)$ is the zero function.

Define $\|f\|$: $$\|f\|=\sup_{x\in C[-1,1]\setminus\{0\}} \left( \frac{|f(x)|}{\|x\|}\right) $$

So $\|f\|\leq 2 $ since $x(t)$ was arbitrarily chosen and we've established an upper bound. But we can't conclude $\|f\| = 2 $ from this since it need not be the least upper bound.

It would be sufficient at this point to show $\frac{|f(x)|}{\|x\|}\geq 2$ for some specific $x(t)\in C[-1,1]$.

I'm having no luck finding this $x(t)$. What's clear is that it needs to be some anti-symmetric function on the interval that attains a maximum value of $1$.

Any suggestions would be appreciated.

oknp :) you can click edit and see how i did it so yours will look pretty next time. \left and \right constructs are floating in size and quite useful for such cases — gt6989b, Apr 12 '16 at 17:20
See also: Norm of the operator $Tf=\int_{-1}^0f(t)\ dt-\int_{0}^1f(t)\ dt$, Finding the bound of a linear functional defined on $C[-1,1]$, Find $\Vert f \Vert$ for a functional $f$ over $C[-1,1]$, find linear functional norm. — Martin Sleziak, Jan 05 '18 at 03:45

score 4 · Accepted Answer · answered Apr 12 '16 at 17:19

I don't think you can find $x$. the idea is that the perfect $x$ would be the function thqt has value -1 on $[-1,0[$ and 1 on $[0,1]$. The problem is that this function is not continuous...

Never mind! try to approximate it with well chosen partially linear functions. You don't need to find an $x\in C[-1,1]$ such that $\frac {\vert f(x)\vert}{\Vert x \Vert}\geqslant 2$.

It suffices to prove that, for any $\epsilon>0$, you can design a continuous function $x$ such that $\frac {\vert f(x)\vert}{\Vert x \Vert}\geqslant 2-\epsilon$.

score 3 · Answer 2 · answered Apr 12 '16 at 17:17

3

Hint: Connect the points $(-1,1),(-1/n,1), (1/n,-1),(1,-1)$ with line segments. Think of that as the graph of $x_n(t).$

answered Apr 12 '16 at 17:17

zhw.

105,693

I agree each $x_n(t)$ is piecewise continuous, so they are all in the space. But it converges to the step function - which isnt continuous. Is this not a problem ? – Walt van Amstel Apr 12 '16 at 17:27
But $C[-1,1]$ is complete... I'm pretty sure this sequence of functions is Cauchy. So that would mean the step function is in $C[-1,1]$. I find that a bit perplexing.. – Walt van Amstel Apr 12 '16 at 17:35
1

They converge point wise, but not uniformly, to sgn (x). They are not Cauchy in C[-1,1]. – zhw. Apr 12 '16 at 17:39

score 0 · Answer 3 · answered Apr 12 '16 at 17:20

Well, I don't know, that there exists a particular $x \in C[-1,1]$, which shows the sharpness of your inequality, but there is a non-continuous function in the closure of the named space, namely the function $-\text{sgn}$. It's enough, because if $x^*$ is close enough to it, the desired quotient is arbitrary close to 2.

Finding the bound of a linear functional defined on $C[-1,1]$

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