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Consider a sequence of functionals $(f_n)$, $f_n(x)=\int_{-1}^1x(t)\cos(nt)dt,\ n\geq 1$, on the space $L_2(-1,1)$. I need to prove that $f_n(x)\to 0$, as $n\to\infty$, for all $x\in L_2(-1,1)$.

I know that $\int_{-1}^1\cos(nt)dt\to 0$, as $n\to\infty$, and I tried to extract this term as a multiplicand (applying Holder's inequality) but with no success.

Or maybe I should take into account that $L_2(-1,1)$ is a Hilbert space and somehow use the general form of continuous linear functionals there?

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    I guess I see what you are suggesting. I take a continuously differentiable function very close to $x$ and apply integration by parts, which I had totally forgotten of.2012-01-25

1 Answers 1

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Two ways (or more) to solve the problem:

First one: with analysis, using the fact that the set of continuously differentiable functions is dense in $L_2(-1,1)$. Then take $x\in L_2(-1,1)$, and $\{y_k\}\subset C^1(-1,1)$ which converges to $x$ in $L_2(-1,1)$. Then \begin{align*} |f_n(x)|&\leq |f_n(x-y_k)|+|f_n(y_k)| \\ &\leq \sqrt 2||x-y_k||_{L^2}+ \left|\frac{y_k(1)\sin(n)-y_k(-1)\sin(-n)}n\right|+\left|\int_{-1}^1y_k'(t) \frac{\sin(nt)}ndt\right|\\ &\leq \sqrt 2||x-y_k||_{L^2}+\frac{|y_k(1)|+|y_k(-1)|}n+\frac 2n\sup_{-1\leq t\leq 1}|y_k'(t)| \end{align*} and for all $k$: $\limsup_n|f_n(x)|\leq \sqrt 2||x-y_k||_{L^2}$ so you can conclude.

Second one: using only measure theory facts:

  1. Prove that $\mathcal B:=\left\{A\subset (-1,1),\lim_{n\to \infty}\int_A\cos(nt)dt=0\right\}$ is a $\sigma$-algebra.
  2. Conclude that for all Borel-measurable $B\subset (-1,1)$ we have $\lim_{n\to \infty}\int_B\cos(nt)dt=0$.
  3. Show that for all simple function $s$ we have $\lim_{n\to \infty}\int_{(-1,1)}s(t)\cos(nt)dt=0$.
  4. Conclude.
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    Thanks for the correction, since I changed involuntary when I edited (or these mistakes were already were). For the first term of the third line, I didn't compute it, I just bound it.2012-01-25