Doom of Doom
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Let S be a subspace of L^{2}(\left[0,1\right]) and suppose \left|f(x)\right|\leq K \left\| f \right\| for all f in S.
Show that the dimension of S is at most K^{2}
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The prof hinted us to use Bessel's inequality.
Namely, let \left\{ u_1,\dots, u_m \right\} be a set of orthonormal vectors in L^{2}(\left[0,1\right]). Then \left\| f \right\|^2 \geq \sum_{k=1}^m \left| \left\langle f, u_k \right\rangle\right|^2
I just keep getting stuck, and getting things like
\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
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 K^{2}
---------
The prof hinted us to use Bessel's inequality.
Namely, let \left\{ u_1,\dots, u_m \right\} be a set of orthonormal vectors in L^{2}(\left[0,1\right]). Then \left\| f \right\|^2 \geq \sum_{k=1}^m \left| \left\langle f, u_k \right\rangle\right|^2
I just keep getting stuck, and getting things like
\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
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?