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I have a linear functional $$A: L_2[0,2] \to \mathbb R, Ax = \int_0^2(t^2+2)x(t)dt$$

I need to find $C$, trying to measure $C$ and $||Ax||$ to find it, but how can I do it in this problem?

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Is $C$ a bound for $A$? Is $C = \Vert A \Vert$? – Robert Lewis May 13 '14 at 8:09
@RobertLewis Exactly. – michaeluskov May 13 '14 at 8:38

2 Answers 2

up vote 3 down vote accepted

The Riesz representation theorem tells you that in an inner product space, the continuous linear functional $f_y(x) = \langle y,x\rangle$ has operator norm $||y||$. In this case, $A x= \langle t^2+2,x(t)\rangle$, so $||A|| = ||t^2+2||_2$.

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But why is $\int_0^2(t^2+2)x(t)dt$ scalar and equals to $<t^2+2,x(t)>$? – michaeluskov May 13 '14 at 9:01
In $L_2[0,2]$, $<f(t),g(t)> = \int_0^2 f(t) g(t) dt$. $<\cdot,\cdot>$ denotes the inner product. – Batman May 13 '14 at 9:04
But how to prove it? – michaeluskov May 13 '14 at 9:09
Prove the Riesz representation theorem?…. The other option is to use Cauchy Schwarz to note that $||A x|| = |<t^2+2,x(t)>| \leq ||t^2+2|| \quad ||x(t)||$ (so $||A|| \leq ||t^2+2||$) and find an $x(t)$ which causes equality (hint: what happens if $x(t) = t^2+2$?), showing that $||A|| = ||t^2+2||$. – Batman May 13 '14 at 9:14
no, how to prove that $A$ is equal to scalar in $L_2$? – michaeluskov May 13 '14 at 9:16

Here I'll play the role of Robin to Batman, and emulate my mentor's work in a more elementary manner, as befits a protoge of the master!

First of all, note that $y(t) = t^2 +2 \in L^2[0, 2]$; since the inner product $\langle u, v \rangle$ on $L^2[0, 2]$ is given by

$\langle u, v \rangle = \int_0^2 u(t)v(t) dt, \tag{1}$

and thus

$\langle t^2 + 2, t^2 + 2 \rangle = \int_0^2 (t^2 + 2)^2 dt = \int_0^2 (t^4 + 4t^2 + 4)^2 dt$ $= (\dfrac{t^5}{5} + \dfrac{4t^3}{3} + 4t \vert_0^2 = \dfrac{32}{5} + \dfrac{32}{3} + 8 = \dfrac{96}{15} + \dfrac{160}{15} + \dfrac{120}{15} = \dfrac{376}{15}, \tag{2}$

so that

$\Vert t^2 + 2 \Vert^2 = \langle t^2 + 2, t^2 + 2 \rangle = \dfrac{376}{15}, \tag{3}$


$\Vert t^2 + 2 \Vert = \sqrt{\dfrac{376}{15}}; \tag{4}$

since $\Vert t^2 + 2 \Vert$ exists (and is finite!), we have indeed have that $t^2 + 2 \in L_2[0, 2]$; thus we may apply the Cauchy-Schwarz inequality to $t^2 + 2$ and $x(t)$ to conclude that

$\vert \langle t^2 + 2, x(t) \rangle \vert \le \Vert t^2 + 2 \Vert \Vert x(t) \Vert; \tag{5}$

since $\langle t^2 + 2, x(t) \rangle$ is in fact the functional $Ax$, we see that (5) immediately shows that $Ax$ is bounded by $\Vert t^2 + 2 \Vert$;

$\vert A(x) \vert \le \Vert t^2 + 2 \Vert \Vert x(t) \Vert = \sqrt{\dfrac{376}{15}} \Vert x(t) \Vert, \tag{6}$


$\Vert A \Vert \le \sqrt{\dfrac{376}{15}}. \tag{7}$

Now taking

$x(t) = t^2 + 2 \tag{8}$

(2)-(4) show that

$\vert A(t^2 + 2) \vert = \Vert t^2 + 2 \Vert^2 = \sqrt{\dfrac{376}{15}} \Vert t^2 + 2 \Vert; \tag{9}$

since $\vert Ax(t) \vert$ in fact takes the value $\sqrt{376/15}\Vert x(t) \Vert$ for $x(t) = t^2 + 2$, we have that indeed

$\Vert A \Vert = \sqrt{\dfrac{376}{15}}. \tag{10}$

Hope this helps! Cheerio,

and as always,

Fiat Lux!!!

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