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This is exercise 5.11 in Brezis's Functional Analysis, Sobolev Spaces, and PDEs.

Let $H$ be a Hilbert space, and let $M \subset H$ be a nonzero closed linear subspace. Let $f \in H$, $f \notin M^\perp$.
1. Prove that $m = \inf_{u \in M, \lvert u \rvert = 1} (f,u)$ is uniquely achieved.
2. Let $\varphi_1, \varphi_2, \varphi_3 \in H$ be given and let $E$ denote the linear space spanned by $\{\varphi_1, \varphi_2, \varphi_3\}$. Determine $m$ in the following cases: (i) $M = E$, and (ii) $M = E^\perp$.
3. Examine the case in which $H = L^2(0,1)$, $\varphi_1(t) = t$, $\varphi_2(t) = t^2$, $\phi_3(t) = t^3$.

My work thus far

I think I have a proof of part (1) using minimizing sequences, which doesn't really allow me to compute $m$ explicitly. I also have a pretty good geometric picture of what is going on, at least for finite-dimensions: essentially to find the minimizing argument $u$ one simply projects $f$ onto the subspace $M$ and then reverse the direction and scale.

But I don't know what the question wants for parts (2) and (3). For 2(i), can I write down anything better than

\begin{equation} \inf_{a,b,c \in \mathbb{R}} \frac{a(f,\varphi_1) + b(f,\varphi_2) + c(f,\varphi_3)}{\lvert a \varphi_1 + b \varphi_2 + c \varphi_3 \rvert} ? \end{equation}

My other idea is that we may suppose wlog that $\{\varphi_1, \varphi_2, \varphi_3\}$ is an orthonormal basis for the subspace $E$. In this case the expression simplifies slightly to \begin{equation} \inf_{a,b,c \in \mathbb{R}} \frac{a(f,\varphi_1) + b(f,\varphi_2) + c(f,\varphi_3)}{\sqrt{a^2 + b^2 + c^2}}, \end{equation} in which case I believe the solution is \begin{gather} (a,b,c) = -((f,\varphi_1),(f,\varphi_2),(f,\varphi_3)) \\ m = -\sqrt{(f,\varphi_1)^2 + (f,\varphi_2)^2 + (f,\varphi_3)^2}. \end{gather}

But then I tried using this for part (3), and produced a ugly mess of algebra and in the end I'm still left with integrals like $\int_0^1 tf(t) \, dt$ that I can't integrate.

And of course, none of this seems to apply in the case for $E^\perp$, which is likely infinite-dimensional.

Any hints, pointers, or suggestions? I would particularly like to know what form my final answers should take.

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1 Answer

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Hint: In your situation it is possible to write $f=P_M f+P_{M^\perp} f$, where $P_M f\in M$ and $P_{M^\perp} f\in M^\perp$. Note that $(f,u)=(P_M f,u)+(P_{M^\perp} f,u)=(P_M f,u)$ for $u\in M$. Now I think you can explicitly find the minimum (draw a picture).

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Thank you for your hint. So now I have $u = -\frac{P_M f}{\lvert P_M f \vert}$, and therefore $m = -\lvert P_M f \rvert$. This seems to correspond well to the half-results I have above. So would my answers to question 2 simply be $-\lvert P_E f \rvert$ and $-\lvert P_{E^\perp} f \rvert$? And is there a good way to explicitly compute the projection of $f$ onto $\{t, t^2, t^3\}$ in question (3)? –  j1123567 Feb 22 '13 at 14:13
    
Theorically, you answer is that, but you can make some manipulations to obtain a more explicitly one. Try to do it and if you get stucked, post here that I help you. –  Tomás Feb 22 '13 at 14:17
    
OK, I thought about this a bit more, but can't seem to get anywhere. If I make the assumption that $\{\varphi_1, \varphi_2, \varphi_3\}$ form an orthonormal basis for the space $E$, then I think I can write $-\lvert P_E f \rvert$ as $-\sqrt{(f,\varphi_1)^2 + (f,\varphi_2)^2 + (f,\varphi_3)^2}$. Also, it occured to me that since $f = P_E f + P_{E^\perp} f$, we have $-\lvert P_{E^\perp} f \vert = -\lvert f - P_E f \rvert$, so this gives me something for 2(ii). Other than this though, I don't see how to proceed further. Can you point me in the right direction? –  j1123567 Feb 22 '13 at 23:02
    
I think that's the best representation. You alread know it: $P_E f=(f,\phi_1)\phi_2+(f,\phi_2)\phi_2+(f,\phi_3)\phi_3$. I am assuming that $\phi_1,\phi_2,\phi_3$ are orthonormal. Note that you know $f,\phi_1,\phi_2,\phi_3$ and hence you have a explicilty expression to $P_E f$. For 3) try to find a orthonormal basis and then use this formula. –  Tomás Feb 22 '13 at 23:11
    
Thanks! I've accepted your answer. –  j1123567 Feb 22 '13 at 23:27
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