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Finite sum for infinite integral

Let $f\in L^1[0,1]$ such that $f(x)>0,\forall x\in[0,1]$. Then, for all $\alpha\ge0$, prove that $$\int_0^1 f^\alpha(x)\,dx \le \frac{1}{\alpha}\sum_{i=0}^\infty f^{\alpha-i}(0).$$

It’s easy for $1<\alpha<2$ for which using MCT is easy. But I'm not sure if above inequality is always true. I haven't found any counter example. Please help.

A:

Take $f=x$ if $x \in [0,1/3]$ and $f=0$ if $x \in [2/3,1]$.
Then $f$ is integrable, hence $f^{\alpha}$ is integrable for every $\alpha$ and
$$\int_0^1 f^{\alpha}(x) dx \leq \int_0^{1/3} x^{\alpha} dx + \int_{2/3}^1 0 dx.$$
Moreover, for each $i \geq 1$ you have $f^{\alpha-i}(0)=0$ when $\alpha-i <1$.
Thus,
$$\sum_{i=0}^{\infty} f^{\alpha-i}(0) = \sum_{i=1}^{\infty} f^{\alpha-i}(0).$$
If $f$ is not integ

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