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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 $x\in[0,1]$.


(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.


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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
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.

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