- #1
Doom of Doom
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Let S be a subspace of [tex]L^{2}(\left[0,1\right])[/tex] and suppose [tex]\left|f(x)\right|\leq K \left\| f \right\|[/tex] for all f in S.
Show that the dimension of S is at most [tex]K^{2}[/tex]
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The prof hinted us to use Bessel's inequality.
Namely, let [tex]\left\{ u_1,\dots, u_m \right\}[/tex] be a set of orthonormal vectors in [tex]L^{2}(\left[0,1\right])[/tex]. Then [tex]\left\| f \right\|^2 \geq \sum_{k=1}^m \left| \left\langle f, u_k \right\rangle\right|^2[/tex]
I just keep getting stuck, and getting things like
[tex]\left\|f \right\|^2 =\int_0^1 \left|f(x)\right|^2 dx \leq \int_0^1 K^2 \left\|f \right\|^2 dx = K^2 \left\|f \right\|^2[/tex]
and I can't figure out how to apply Bessel's inequality.
I guess the goal is to assume m linear independent vectors, and show that m is less than K^2. Help, please?
Show that the dimension of S is at most [tex]K^{2}[/tex]
---------
The prof hinted us to use Bessel's inequality.
Namely, let [tex]\left\{ u_1,\dots, u_m \right\}[/tex] be a set of orthonormal vectors in [tex]L^{2}(\left[0,1\right])[/tex]. Then [tex]\left\| f \right\|^2 \geq \sum_{k=1}^m \left| \left\langle f, u_k \right\rangle\right|^2[/tex]
I just keep getting stuck, and getting things like
[tex]\left\|f \right\|^2 =\int_0^1 \left|f(x)\right|^2 dx \leq \int_0^1 K^2 \left\|f \right\|^2 dx = K^2 \left\|f \right\|^2[/tex]
and I can't figure out how to apply Bessel's inequality.
I guess the goal is to assume m linear independent vectors, and show that m is less than K^2. Help, please?