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I've got the following question and the very last bit is confusing me slightly.

Fix $0 < \eta < 1$. Prove that the series $\displaystyle\sum_{n=0}^{\infty} x^n cos(nx^2)$ is uniformly convergent on $[0,\eta]$. Deduce that the function $\displaystyle s$ defined by $s(x) = \displaystyle\sum_{n=0}^{\infty} x^n cos(nx^2)$ is continuous on $[0,\eta]$. Explain why $s$ is also continuous on $[0,1)$

I've used the Weierstrass M-Test as $|x^n cos(nx^2)| \leq x^n$ for all $x \in [0,\eta]$ and $p_n(x) = \displaystyle\sum_{k=0}^n x^k = \displaystyle\frac{1-x^{n+1}}{1-x}$ tends to $\displaystyle\frac{1}{1-x}$ for any $x \in [0,\eta]$

Thus as $\displaystyle\sum_{n=0}^{\infty}x^n$ converges, then by the Weierstrass M-Test $\displaystyle\sum_{n=0}^{\infty}x^n cos(nx^2)$ converges uniformly for all $x \in [0,\eta]$.

Also as $x^n$ and $cos(nx^2)$ are continuous for all $x \in [0,\eta]$ we have that $x^n cos(nx^2)$ is continuous for all $x \in [0,\eta]$ and therefore we have that $s(x)$ is continuous on $[0,\eta]$ due to uniform convergence preserving continuity.

But the very last point about explaining why $s$ is also continuous $[0,1)$ seems a bit too "obvious" to me, since we defined $0 < \eta < 1$ and showed $s$ is continuous $[0,\eta]$ does this not imply $s$ is continuous on $[0,\eta]\cup[\eta,1)$ or am I missing something? The next question asks the same thing at the end, so I'm concerned I'm missing something major (as I usually do when something seems obvious). Thanks for the help!

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1 Answer 1

up vote 1 down vote accepted

Continuity is something local (not global). Continuity on an interval is continuity of all the points in this interval.

$f$ continuous on $[0,1)\Leftrightarrow \forall x\in [0,1), f$ continuous at $x$


$f$ is continuous on $\left[0,\frac{1}{2}\right]$ by taking $\eta=\frac{1}{2}$ so it is continuous at $0$ since $0\in\left[0,\frac{1}{2}\right]$

Let $x\in(0,1)$

$f$ is continuous on $\left[0,x\right]$ by taking $\eta=x$ so it is continuous at $x$ since $x\in\left[0,x\right]$

So $\forall x\in[0,1)$, $f$ is continous at $x$

So $f$ is continous on $[0,1)$

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I understand this, I just don't really understand how to conclude that $s$ is continuous on $[0,1)$ given that it's continuous on $[0,\eta]$ where $0 < \eta < 1$. –  Noble. Feb 27 '13 at 12:56
    
@Noble. : I edited. –  xavierm02 Feb 27 '13 at 13:01
    
Ah, ok so it is fairly "obvious" it's just put a lot more rigorously in your edit than I would've explained. Thank you very much! –  Noble. Feb 27 '13 at 13:01

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