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How do I show that as $N \to \infty$, that $$\sum_{i=N}^\infty{1} \to 0?$$ Don't know how to even start. Thanks..

Apparently this is wrong. But my teacher said that if $P_n f = \sum_{j=0}^n(f,w_j)w_j$, where $w_j$ is orthonormal basis of $L^2$, then $|P_n f- f|_{L^2} \to 0$. How can that be then? Because I thought $$|P_nf - f| = |\sum_{j=0}^n(f,w_j)w_j - \sum_{j=0}^\infty(f,w_j)w_j| = |\sum_{j={n+1}}^\infty(f,w_j)w_j| \leq \sum_{j={n+1}}^\infty|f|$$ where the last equality is by Cauchy Schwarz.

michael_faber
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3 Answers3

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The limit does not exist.

Consider the summation $\sum_{j=N}^{M-1} 1 = M- N$ for $M > N$. Roughly, you are after $\lim_{N \rightarrow \infty} \lim_{M \rightarrow \infty} M - N$.

This limit does not exist. You can reason this out formally by expanding out the meaning of a limit diverging to $\infty$. What I have below is most likely overkill for this problem.

Roughly, $\lim_{x \rightarrow \infty} f(x) = \infty$ means that $\forall K\ \exists x_0\ \forall x \geq x_0, f(x) > K$.

In this case, you want to show that

$ (\forall\ K \geq 0)\ (\exists N_0 \geq 0)\ (\forall N \geq N_0)\ (\exists M_0 \geq 0)\ (\forall M \geq M_0),\ M - N > K$.

The statement above follows by setting $N_0$ to be some fixed number and $M_0 \geq N + K+1$.

Sriram S
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  • $|P_n f - f|_{L2} \rightarrow 0$ only if $f$ is a $L_2$ function (square integrable). The function whose projection leaves an "infinite tail" of 1s is not square integrable. This follows from something like the Parseval's identity? – Sriram S Feb 09 '13 at 15:56
  • The last paragraph (beginning with This limit does not exist) is misleading. None of the cases $M=N^2$ and $M=N+c$ is relevant since one should only be concerned with the limit of $L_N$ when $N\to\infty$, where $L_N$ is the limit of $\sum\limits_N^M$ when $M\to\infty$. – Did Feb 10 '13 at 14:02
  • Thank you. I have made the argument entirely formal. Hope I did not mess up. :-) – Sriram S Feb 11 '13 at 01:54
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$$\lim\limits_{N\to\infty}\left(\sum_{i=N}^{+\infty}1\right)=+\infty$$

Did
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  • But my teacher said that if $P_n f = \sum_{j=0}^n(f,w_j)w_j$, where $w_j$ is orthonormal basis of $L^2$, then $|P_n f- f|_{L^2} \to 0$. How can that be then? – michael_faber Feb 09 '13 at 15:49
  • She probably mentioned "when the function is square integrable". – Did Feb 09 '13 at 15:57
  • @michael_faber: one theorem that is used to justify the interchange of limit and sum is the Dominated Convergence Theorem. The answer above shows that the limit fails. Suppose that $$ f_N(i)=\left{ \begin{array}{} 0&\text{if }i\lt N\ 1&\text{if }i\ge N \end{array} \right. $$ Why does the DCT fail to justify $$ \lim_{N\to\infty}\sum_{i=1}^\infty f_N(i)=\sum_{i=1}^\infty\lim_{N\to\infty}f_N(i) $$ ? – robjohn Feb 15 '13 at 19:23
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When you say $$\sum_{i=N}^\infty s_i$$ What you really have is a nice shorthand for: $$lim_{M\to\infty} \sum_{i=N}^Ms_i$$ You can't just extend the notion of a sum, which is a combination of two elements in a group to a third, to a sum of an infinite number without some sort of limiting process. So what you're looking for is a limit that doesn't exist.

guest196883
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