Does this dual space functional pairing = 0 imply functional = 0?

Question

If $V$ is a Hilbert space, is it true that if $\phi_1, \phi_2 \in C_c^\infty(0,T)$, $$\int_0^T \langle \phi_1(t)g +\phi_2(t) f, v \rangle_{V', V} = 0$$ for all $v \in V$, then $\phi_1g + \phi_2f \equiv 0$? How about without the integral?

Is this some Hahn-Banach type thing?

score 1 · Accepted Answer · answered Apr 27 '13 at 15:48

1

Note that $$\int_0^T\langle \phi_1(t)g+\phi_2(t)f,v\rangle dt=\langle\int_0^T\phi_1(t)dt\cdot g+\int_0^T\phi_2(t)dt\cdot f ,v\rangle,$$ so you can only conclude that $$\int_0^T\phi_1(t)dt\cdot g+\int_0^T\phi_2(t)dt\cdot f=0.$$

answered Apr 27 '13 at 15:48

Hu Zhengtang

3,757

Thanks, can you justify the last step? – michael_faber Apr 27 '13 at 17:37
@michael_faber: Denote the left hand side of the second equation by $w$, Note that the right hand side of the first equation is $\langle w,v\rangle$, and $v$ is arbitrary. In particular you can choose $v=w$, which gives you $\langle w,w\rangle=0$, i.e. $w=0$. – Hu Zhengtang Apr 27 '13 at 19:54
But $w \in V'$ not $V$. So we can't pick $v=w$ I believe. – michael_faber Apr 27 '13 at 20:46
@michael_faber: I understood $\langle,\rangle$ as the inner product of $V$, if so, $w\in V$. Otherwise, if $w$ is a linear functional on $V$, since $w(v)=0$ for every $v\in V$, by definition, $w$ must be the $0$ functional. Besides, since $V$ is a Hilbert space, by Riesz representation theorem, there is a natural isomorphism between $V$ and $V'$. – Hu Zhengtang Apr 27 '13 at 21:01
Yeah that's right the RRT solves it. Thanks – michael_faber Apr 27 '13 at 21:34
@michael_faber: You are welcome. – Hu Zhengtang Apr 28 '13 at 16:29

Does this dual space functional pairing = 0 imply functional = 0?

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