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Let $f$ is Lebesgue integrable function on $[0,1]$ with $\lim_{x\rightarrow 1}f(x) = c$, where $c$ is a constant. Then what would be the value of $n\int x^n f(x) \,dx?$

I think that this value should equal $0$ since $g_n = n x^n f(x) \rightarrow 0$, and $|g_n|\le f$. Hence, by dominated convergence theorem I must have $\int nx^n f(x)\,dx \rightarrow 0$ However, solution states that $\int nx^n f(x)\,dx = c$. I am not sure why dominated convergence theorem doe not hold.

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  • $\begingroup$ Why do you think $g_n \to 0$? I don't see where that came from. $\endgroup$ – David Bowman Sep 16 '17 at 22:42
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    $\begingroup$ It's also unclear what you want to compute. Do you want $n \int_0^1 x^n f(x) dx$ as $n \rightarrow \infty$? $\endgroup$ – parsiad Sep 16 '17 at 22:43
  • $\begingroup$ $nx^n \to n$ as $x \to 1$, so $nx^n f(x)$ is not dominated by $f$. $\endgroup$ – Chappers Sep 16 '17 at 22:43
  • $\begingroup$ @parsiad yes that is what I meant sorry.. $\endgroup$ – user1292919 Sep 17 '17 at 0:59
  • $\begingroup$ @David Bowman since $nx^n$ goes to 0 as $n$ goes to infinity. $\endgroup$ – user1292919 Sep 17 '17 at 1:00
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This is one type of behavior that shows us why we need a dominating function (or some condition, anyway) in order to be guaranteed that we can pass the limit inside: the contribution to the integral gets "pushed to the boundary" in the limit.

Here at every point in $x\in [0,1)$ we have $g_n\to 0$, and yet $g_n(1)$ diverges. The overall behavior is that the integral remains finite, just all the area gets squeezed onto a spike at the boundary as $n\to =\infty.$ (As Chappers noted in the comments, the fact that $g_n(1)$ diverges means you're mistaken about $f$ being a dominating function.)

A simple case to consider is when $f(x) = c$ so that $g_n(x) = cn x^n.$ In this case we have $\int_0^1cnx^ndx = c\frac{n}{n+1} \to c.$ And from the fact that the $nx^n$ factor essentially becomes a spike at the $x=1$ boundary, it shouldn't be too hard to see why the answer is $\lim_{x\to 1^{-}}f(x)$ more generally.

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This is not a case of the dominated convergene theorem failing to hold; rather it is a case of the conclusion of the dominated convergence theorem failing to hold, because the hypotheses are not satisfied.

The theorem is not applicable except when the dominating function is integrable, and "integrable" means the integral of its absolute value is finite. If $\displaystyle \int_{[0,1]} |f(x)|\,dx = \infty$ then $f$ is not integrable and the dominated convergence theorem cannot be applied with $f$ as the dominating function.

Knowing whether that applies to the particular function you've called $f$ depends on which function it is, and you haven't told us that.

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When $c\ne0$, the sequence $g_n$ is not dominated by any integrable function.

Define $h_n(x):=nx^n$. The value of $h_n(x)$ at $x=1-\frac1n$ is $ n\left(1-\frac 1n\right )^n. $ Since $(1-\frac1n)^n$ converges to $e^{-1}$, and each $h_n$ is increasing, there exists a universal constant $a>0$ such that for all large $n$, we have $h_n(x) > an$ when $x\ge1-\frac1n$.

Now if $c\ne0$, we know that $f$ is bounded away from zero near $x=1$, say $|f(x)|>c'>0$ for $x$ sufficiently close to $1$.

Combining the above, the net result is that for all large $n$, $|g_n(x)|=h_n(x)|f(x)|$ is at least as big as $c'an$ for all $x\ge 1-\frac1n$. Draw a picture of this situation, and convince yourself that any dominating function for the sequence $|g_n|$ cannot be integrable.

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As I said in my comment, the Dominated Convergence Theorem does not apply here because $nx^n f(x)$ is not bounded on $[0,1]$ by an integrable function. Indeed, by the property that $f$ has, it tends to $nc$ as $x \to 1$, which cannot be dominated by an integrable function for all $n$.

We can apply the DCT in a subset, however, to reduce the region on which the problem lies to a neighbourhood of $1$ (similar arguments are used when considering approximate identities, which is why this is an important case to understand). The key is to spot the property that $f$ has: it tends to $c$ as $x \to 1$, so for any $\varepsilon>0$ there is a $\delta > 0$ so that $$ c+\varepsilon > f(x) > c-\varepsilon $$ for $1>x>1-\delta$. So we can split the integral as $$ \int_0^{1-\delta} nx^n f(x) \, dx + \int_{1-\delta}^1 nx^n f(x) \, dx. $$ Now, $nx^n \to 0$ uniformly on $[0,1-\delta]$, so you can use the Dominated Convergence Theorem on this part (some multiple of $f$ will work as the dominating function: find the maximum of $nx^n$ on this interval).

For the other integral, monotonicity of the integral gives $$ \int_{1-\delta}^1 nx^n(c+\varepsilon) \, dx > \int_{1-\delta}^1 nx^nf(x) \, dx > \int_{1-\delta}^1 nx^n(c-\varepsilon) \, dx, $$ and the left and right integrals can be computed exactly: $\int_{1-\delta}^1 nx^n \, dx = \frac{n}{n+1}(1-(1-\delta)^n)$. Then $$ \frac{n}{n+1}(1-(1-\delta)^n)(c+\varepsilon) > \int_{1-\delta}^1 nx^nf(x) \, dx > \frac{n}{n+1}(1-(1-\delta)^n)(c-\varepsilon), $$ and taking the limit, $$ c+\varepsilon \geq \lim_{n\to\infty} \int_{1-\delta}^1 nx^nf(x) \, dx \geq c-\varepsilon. $$ This holds for any $\varepsilon>0$, and the result follows.

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