Show $\int_{0}^{\infty} x^tf(x)dx$ is uniformly convergent for $t\in[a,b]$.

Question

Assume $\int_{0}^{\infty} x^tf(x)dx$ converges at $t=a$ and $t=b$. Show $$\int_{0}^{\infty} x^tf(x)dx$$ is uniformly convergent for $t\in[a,b]$.

It seems that the common tests cannot work. May I consider Cauchy's convergence test？

Answer 1

Absolute convegence case:

Here we consider the case where $\int^\infty_0 x^t\,|f(x)|\,dx<\infty$ for $t\in\{a,b\}$.

Since $\left|\int_A x^t\,f(x)\,dx\right|\leq \int_A x^t|f(x)|\,dx$ for all $A\subset [0,\infty)$, it suffices to assume $f\geq0$.

Given $\varepsilon>0$, there exists $C>0$ such that $$ \begin{align} \int^v_u(x^a + x^b)\,f(x)\,dx<\varepsilon\tag{0}\label{zero} \end{align} $$

Edit: One approach as suggested by "kimchi lover" above, is to use the obvious inequality

$$x^y f(x)\leq (x^a+x^b)f(x),\qquad x\geq0, \,a\leq y\leq b$$

This implies that

$$ \int^v_u x^y\, f(x)\,dx \leq \int^v_u (x^a+x^b)\,f(x)\,dx<\varepsilon $$ for all $u,v>C$ and $a\leq y\leq b$, which is a uniform bound in $y$.

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Another approach (my original solution in fact) relies on well known result in integration, namely Holder's inequality. measure space $(X,\mathscr{F},\mu)$, if $\frac1p+\frac1q=1$, and $f\in L_p(\mu)$ and $g\in L_q(\mu)$, then $$\int|f\,g |\,d\mu\leq \Big(\int_X|f|^p\,d\mu\Big)^{1/p}\Big(\int_X|g|^q\,d\mu\Big)^{1/q}$$

Remark: It is possible to accurate this inequality for the setting of proper Rieman integrals, but I will leave details out. [Here][1] is one posting that deals with this. With a little effort, that posting maybe adapted to the present setting: replace $[0,1]$ by $(0,\infty)$ and $dx$ by $f(x)\,dx$

For the OP, consider the measure space $([0,\infty),\mathscr{B}([0,\infty)),f(x)\,dx)$.

By assumption, the function $\phi(x)=x$ is in $L_a$ and $L_b$. Then, for any $a<t<b$, there is $0<\lambda<1$ such that $t=\lambda a+(1-\lambda)b$. The function $g(x)=x^{a\lambda}\in L_{1/\alpha}$ and $h(x)=x^{b(1-\lambda)}\in L_{1/(1-\lambda)}$.

Applying Holder's inequality and $\eqref{zero}$ $$\int^v_u x^t\,f(x)\,dx =\int^v_u x^{a\lambda}x^{b(1-\lambda)}f(x)\,dx\leq\Big(\int^u_v x^a f(x)\,dx\Big)^{\lambda}\Big(\int^u_v x^b f(x)\,dx\Big)^{1-\lambda}<\varepsilon$$ for all $v>v>C$.

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Conditional convergence case:

The case where $\int^\infty_0 x^t\,f(x)\,dx$, $t\in\{a,b\}$, converges conditionally ($\int^\infty_0x^t|f(x)|\,dx=\infty$), I have no definitive answer. I suspect that uniform convegence will depend on the function $f$.

The following classical example preserves uniform convergence: $ \phi(y):=\int^\infty_0 \frac{\sin}{t^y}\,dx$ where $0<y\leq 1$. Integration by parts gives $$\int^v_u \frac{\sin x}{x^y}\,dx= -\frac{\cos x}{x^y}\Big|^v_u - y\int^v_u\frac{\cos x}{x^{1+y}}\,dx$$ Hence, for $1\leq u<v$ $$ \Big|\int^v_u\frac{\sin x}{x^y}\,dx\Big|\leq \frac{1}{u^y}+\frac{1}{v^y} + y\int^v_u\frac{1}{x^{1+y}}\,dx = \frac{2}{u^y} $$

So, for $0<a\leq y\leq b\leq 1$, we have uniform convergence since $\frac{1}{u^y}\leq \frac{1}{u^a}$.

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At the moment, I have no a counterexample (an example of $f$ rather) where uniform convergence does not hold. [1]: Proof of Hölder's inequality for improper integrals