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Let $\mu(E)<\infty$. Let $C$ be a constant. A subspace $V\subset L^2(E)$ is defined such that $f\in V$ implies that $|f(x)|<C\|f\|_2$ for almost every $x\in E$. Let $\{f_1,\dots,f_n\}$ be an orthonormal set in $V$. Prove that $\sum\limits_{I=1}^n |f_i(x)|^2\leq C^2$.

  1. Is this question correct? I can prove that $\sum\limits_{I=1}^n |f_i(x)|^2\leq nC^2$. Is this also true that $\sum\limits_{I=1}^n |f_i(x)|^2\leq C^2$?

  2. I thought a way to prove it would be to prove that $f_i(x)f_j(x)=0$ for $i\neq j$. Is this along the right track? I couldn't find a way to prove this though.

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    For a.e. given $x\in E$, let $g=\sum_{k=1}^{n}\overline{f_k(x)}f_k\in V$. Then $\sum_{k=1}^n|f_k(x)|^2=|g(x)|\leq C|g|=C\left(\sum_{k=1}^n|\overline{f_k(x)}|^2\right)^{1/2}$. Squaring you get that $\sum_{k=1}^n|f_k(x)|^2\leq C^2$. –  May 04 '18 at 12:09
  • @totoro- I don't see how $|g|=(\sum\limits_{k=1}^n |\overline{f_k(x)}|^2)^{1/2}$. Could you explain that part please? –  May 04 '18 at 13:23
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    You have $g$ defined as a linear combination of orthonormal vectors. Its norm is the length of its vector of coordinates. This is a basic computation: $|g|^2=\langle\sum_{k=1}^{n}\overline{f_k(x)}f_k,\sum_{k=1}^{n}\overline{f_k(x)}f_k\rangle=\sum_{k=1}^{n}|\overline{f_k(x)}|^2$, where you use that $\langle f_i,f_j\rangle=1$ for $i=j$ and $=0$ for $i\neq j$. –  May 04 '18 at 13:27
  • @totoro- Thanks! That was really helpful. I've never seen that construction before. –  May 04 '18 at 14:32
  • @totoro you start with $x$ and define a function $g=g_x$. The inequality $|g_x (y)| \leq C ||g_x||_2$ holds for almost all $y$. You cannot put $y=x$ now, can you? – Kavi Rama Murthy May 09 '18 at 09:51
  • @KaviRamaMurthy You didn't get it. In your notation: I define $g=g_x(y)$. Then $g_x(x)=\sum_{k=1}^{n}|f_k(x)|^2$ is a real number equal to $|g_x(x)|$. By the definition of $V$, for a.e. $x$, this number satisfies $|g_x(x)|\leq C|g_x|2=C(\sum{k=1}^{n}|f_k(x)|^2)^{1/2}$. Now, put on the left-hand side $\sum_{k=1}^{n}|f_k(x)|^2$, divide by $(\sum_{k=1}^{n}|f_k(x)|^2)^{1/2}$ and square. You obtain $\sum_{k=1}^n|f_k(x)|^2\leq C^2$. This holds for all $x$ for which $|g_x(x)|\leq C|g_x|_2$, which are almost all. –  May 09 '18 at 15:14
  • @KaviRamaMurthy The other thing that you are confused about is that an inequality like $|f(x)|<C|f|_2$ holding for a.e. $x\in E$, with $f\in L^2(E)$, is really a red herring. The elements of $L^2(E)$ are equivalence classes of functions that are equal a.e. Therefore, the inequality might as well hold for every $x\in E$. Redefining $f(x)$ as $=0$ for the $x$ that didn't hold the inequality, gives you another representative of the same element of $L^2(E)$, for which the inequality holds for all $x\in E$. So, just ignore the "a.e." if it is confusing –  May 10 '18 at 16:00
  • @ totoro You are right. We may suppose that the inequality holds for all $x$ since only a finite number of functions are involved. – Kavi Rama Murthy May 11 '18 at 06:43

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