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Let $\mathcal{H}$ be the vector space of all complex-valued, absolutely continuous functions on $[0,1]$ such that $f(0)=0$ and $f^{'}\in L^2[0,1]$. Define an inner product on $\mathcal{H}$ by $$\langle f,g\rangle=\int_0^1f^{'}(x)\overline{g^{'}(x)}dx $$ for $f,g\in\mathcal{H}$.

If $0<x\leq 1$, define $L:\mathcal{H}\rightarrow \mathbb{C}$ by $L(f)=f(x)$. Show $L$ is a bounded linear functional and find $\|L\|$.

I was able to show $L$ is linear. That was easy. I am having trouble showing it is bounded and I cannot determine what $\|L\|$ is.

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There is a very similar exercise on Eidelman-Milman-Tsolomitis's book (click)(with solution). A nice remark: This functional $L$ is exactly a Dirac's delta concentrated at $x$. –  Giuseppe Negro Jun 10 '12 at 21:22
    
I cannot find the solution. –  john Jun 10 '12 at 21:29
    
You're right, Google preview trimmed it. You need pages 246-247. –  Giuseppe Negro Jun 10 '12 at 21:55

2 Answers 2

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Use Cauchy-Schwarz inequality $$ |L(f)|= \left|\int\limits_{0}^xf'(t)dt+f(0)\right|= \left|\int\limits_{0}^xf'(t)dt\right|\leq \left(\int\limits_{0}^x|f'(t)|^2dt\right)^{1/2} \left(\int\limits_{0}^x|1|^2dt\right)^{1/2}= $$ $$ \sqrt{x}\left(\int\limits_{0}^x|f'(t)|^2dt\right)^{1/2}\leq \sqrt{x}\left(\int\limits_{0}^1|f'(t)|^2dt\right)^{1/2} =\sqrt{x}\Vert f \Vert $$

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The operator $L_x$ evaluates the function at the point $x$, that is, $L_x(f)=f(x)$. As Norbert shows, the norm of this operator is less than or equal to $\sqrt{x}$.

For the piecewise linear function $f(y)={y\over \sqrt{x}}\wedge \sqrt{x}$, we see that this upper bound is achieved, and so deduce that $\|L_x\|=\sqrt{x}$.

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