Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

My Question

Let $\{g_k\}$ be a sequence of continuous real-valued functions on $[0,1]$. Assume that there is a number $M$ such that $|g_{k}(x)|\leq M$ for every integer $k$ and every $x\in [0,1]$ and also that there is continuous real-valued funtion $g$ on $[0,1]$ such that

$$\int_0^1 g_k(x)p(x) \ dx \rightarrow \int_0^1 g(x)p(x) \ dx$$ as $k \rightarrow \infty$ for every polynomial $p$. Proved that $|g(x)|\leq M$ for every $x\in[0,1]$.

Remarks

(Note that the two bounds are the same and is $M$)

This is what I have done so far.

Suppose it is not true, that is there exists a $x_0 \in [0,1]$ such that $|g(x_0)|>M$, then by the continuity of $g$ at $x_0$, there will be an interval $(x_0-\delta, x_0+\delta)$ such that $|g(x)|>M$ for all $x\in (x_0-\delta, x_0+\delta)$. Then take note that it is possible to have a non-negative continuous function $\phi$ such that it is zero outside the interval $(x_0-\delta, x_0+\delta)$ and the integral $\int_{x_0-\delta}^{x_0+\delta} \phi(x)\ dx$ is one.

Then by Weierstrass Approximation Theorem, we know that
$$\int_0^1 g_k(x)\phi(x) \ dx \rightarrow \int_0^1 g(x)\phi(x) \ dx$$ as $k \rightarrow \infty$ and I managed to obtained a contradiction by assuming $g(x_0)>0$, but I am lost with all the inequalities for the other case. I hope someone can help me with this, perhaps we do not need any Weierstrass approximation theorem, for that I am not sure.

Thanks

share|improve this question
    
If $g(x_0)<0$, you replace each $g_k$ by $-g_k$ and $g$ by $-g$. –  Brian M. Scott May 20 '12 at 22:40
    
Thanks, it works. Is this a general method that always apply or just for this question? –  KWO May 20 '12 at 23:02
    
It’s a fairly common trick. It works here because (a) the integral respects the sign change, and (b) the condition $|g_k(x)|\le M$ is the same for $g_k$ and $-g_k$. Such circumstances are not unusual, so there are lots of situations in which such a trick can work. –  Brian M. Scott May 20 '12 at 23:06
    
Thanks a lot for your help. –  KWO May 20 '12 at 23:21
add comment

1 Answer 1

up vote 1 down vote accepted

In the interests of getting this off the Unanswered list, I’m turning my comment into an answer.

If $g(x_0)<0$, you can reduce the problem to the positive case by replacing each $g_k$ by $-g_k$ and $g$ by $-g$, since the condition that $|g_k(x)|\le M$ applies equally to $g_k$ and $-g_k$, and the integral respects the sign change. This is a fairly common trick when the circumstances allow it.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.