Prove that $(T f)(x) = x^{-1/p}\int_0^x f(t)dt$ is continuous.

Question

Let $1 < p < \infty$ and define $$(T f)(x) = x^{-1/p}\int_0^x f(t)dt.$$ Let $q$ satisfy $\frac 1p + \frac 1q = 1$. Show that if $f \in L^q((0,\infty))$, then $Tf$ belongs to $C_0((0, \infty))$.

My work: Let $\varepsilon > 0$, $\delta = \delta(\varepsilon, x_0)$ to be determined. Let $x_0$ be fixed and $x$ be such that $|x - x_0| < \delta$. Without loss of generality, $x > x_0$. Then:

$|Tf(x) - Tf(x_0)| = |x^{-1/p} \int_0^x f(t) dt - x_0^{-1/p} \int_0^{x_0}f(t)dt|$

$ = |x^{-1/p}\int_0^{x_0}f(t)dt + x^{-1/p}\int_{x_0}^xf(t)dt - x_0^{-1/p} \int_0^{x_0}f(t)dt|$

$\leq |x^{-1/p}-x_0^{-1/p}||\int_0^{x_0}f(t)dt| + x^{-1/p}|\int_{x_0}^x f(t)dt|$

$\leq |x^{-1/p}-x_0^{-1/p}|||f||_qx_0^{1/p} + x_0^{-1/p} ||f||_q |x - x_0|^{1/p}$ (Hölder's inequality)

$\leq |x^{-1/p}-x_0^{-1/p}|||f||_qx_0^{1/p} + x_0^{-1/p} ||f||_q \delta^{1/p}$

This is were I get stuck. I need a bound on $|x^{-1/p}-x_0^{-1/p}|$, ideally in terms of $\delta$, but I can't seem to figure it out.

Answer 1

We to show that $$(T f)(x) = g(x)\int_0^x f(t)dt.$$ is continuous in $(0,+\infty)$ where $g(x)$ is any continuous function in $(0,+\infty)$ (in your case $g(x)=x^{-1/p}$).

Following your approach, for $x,x_0>0$, $$\begin{align} |Tf(x) - Tf(x_0)| &= \left|g(x)\int_0^{x_0}f(t)dt + g(x)\int_{x_0}^xf(t)dt - g(x_0) \int_0^{x_0}f(t)dt\right|\\ &\leq |g(x)-g(x_0)|\left|\int_0^{x_0}f(t)dt\right| + |g(x)|\left|\int_{x_0}^x f(t)dt\right|\\ &\leq |g(x)-g(x_0)|\cdot ||f||_qx_0^{1/p} + |g(x)|\cdot ||f||_q |x - x_0|^{1/p} \quad\text{(Hölder's inequality).} \end{align}$$ Now as $x\to x_0$, then $g(x)\to g(x_0)$ and the right-hand side goes to zero which implies that $Tf$ is continuous at $x_0$.

Answer 2

Since $x^{-1/p}$ is continuous on $(0,\infty)$ and the product of continuous functions is continuous, all you need for continuity is the fact that the indefinite integral of an integrable is (absolutely) continuous. Alternatively, use Hölder's inequality to get a quantitative continuity bound: $$ \left|\int_{x_1}^{x_2} f(t)\,dt \right| \le (x_2-x_1)^{1/p} \|f\|_q $$

Note that the relation of $p$ and $q$ is not relevant here.

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Next, let's show that $Tf(x)\to 0$ as $x\to 0$. Begin with the continuity of antiderivative: $$ \int_0^x |f(t)|^q\,dt \to 0 \quad \text{as } x\to 0 $$ and use Hölder's inequality: $$ \left|\int_{0}^{x} f(t)\,dt \right| \le x^{1/p} \left( \int_0^x |f(t)|^q\,dt\right)^{1/q} = o(x^{1/p}) \quad \text{as } x\to 0 $$

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Finally, let's show that $Tf(x)\to 0$ as $x\to \infty$. Given $\epsilon>0$, pick $M$ such that $\int_M^\infty |f(t)|^q\,dt < \epsilon$. For $x>M$, $$ Tf(x) = x^{-1/p}\int_0^M f(t)\,dt + x^{-1/p}\int_M^x f(t)\,dt $$ where the first term is $O(x^{-1/p})$ and the second is bounded, via Hölder's inequality, by $\epsilon^{1/q}$.