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This is from my schools past examination. I'm stuck on $c$ part $i$. I'm guessing the answer is yes as they only differ by a finite number of points, but this is just from intuition. I'm trying to use part $b$ to solve it but to no avail. Any help please

DH.
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    You know that $S_n \to f$ in the $L^2$ norm, hence $S_n$ is Cauchy. Explicitly, $|S_n-S_m| \le |S_n-f|+|S_m-f|$. – copper.hat Apr 29 '15 at 15:04
  • @copper.hat I've never seen the $L^2$ norm, nor heard of it so this doesn't help me unfortunately. – DH. Apr 29 '15 at 15:05
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    $|f|= \sqrt{ \int_a^b f(x)^2 dx }$. – copper.hat Apr 29 '15 at 15:06
  • @copper.hat I see thanks, but we haven't proved that $S_n(x) \to f$ yet, so i'm not sure if this is the right way to go about it – DH. Apr 29 '15 at 15:08
  • @copper.hat no we haven't talked about completeness of an orthonormal system but your second comment gives me something to go on, thanks. – DH. Apr 29 '15 at 15:33

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Suppose $m\le n$, then we have $S_n-S_m = \sum_{k=m+1}^n a_k \phi_k$.

Then $(S_n(x)-S_m(x))^2 = \sum_{k_1=m+1}^n \sum_{k_2=m+1}^n a_{k_1} a_{k_2} \phi_{k_1}(x) \phi_{k_2}(x)$.

Since the $\phi_k$ are orthonormal, we have $\int_a^b \phi_{k_1}(x) \phi_{k_2}(x) dx = \delta_{k_1 k_2}$, which gives $\int_a^b ((S_n(x)-S_m(x))^2 dx = \sum_{k=m+1}^n a_k^2$.

If we can show that the $a_k$ are square summable then the desired result follows.

Since $f$ is Riemann integrable, we have that $f^2$ is also Riemann integrable (this result is not immediate) and Bessel's inequality gives $0 \le \int_a^b |f(x) -S_n(x)|^2 dx = \int_a^b f(x)^2 dx - \sum_{k=1}^n a_k^2$, from which we conclude that th $a_k$ are square summable.

Aside: To see that $f^2 $ is integrable, look at The product of two Riemann integrable functions is integrable for example.

Community
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copper.hat
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Hint: that is the definition of the sequence $S_n(x)$ being Cauchy with respect to the $L^2$ norm.

icurays1
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  • I've never seen the $L^2$ norm, nor heard of it so this doesn't help me unfortunately. – DH. Apr 29 '15 at 15:05