True Or False: If $f(x)$ is a positive continuous function in $[a,\infty)$ such that $\int_a^\infty f(x)\;dx$ converges then exists $ 0 < c < 1$ such that for every $c \le p \le 1$ $\int_a^\infty f^p(x)\;dx$ converges.
It feels true to me since the function will remain continuous after being raised to a power and all powers are less then 1 so it should not make the value too big.But I'm not sure I'm on the right track, and can't see how to prove it.
On
It's certainly good to become acquainted with $\log$-type examples, such as the one Mark Viola mentions. But I sometimes like to build these things up from the ground.
Sketch of a counterexample: For $x\ge 1,$ define
$$f(x) = \sum_{n=1}^{\infty}\frac{1}{2^nx^{1+1/n}}.$$
The series converges unifomly on $[1,\infty),$ hence $f$ is continuous, and $\int_1^\infty f < \infty.$
Claim: $\int_1^\infty f^p=\infty$ for every $p\in (0,1).$
Proof: Suppose $0<p<1.$ Choose $n_0\in \mathbb N$ such that $p(1+1/n_0) < 1.$ Then
$$\int_1^\infty f^p \ge \int_1^\infty \left (\frac{1}{2^{n_0}x^{1+1/n_0}}\right)^p\,dx = \int_1^\infty \frac{1}{2^{n_0p}x^{p(1+1/n_0)}}\,dx = \infty.$$
HINT:
Let $f(x)=\frac{1}{x\log^2(x)}$ for $x\in[a,\infty)$ for $a>1$.