Proof of Hardy's Classic Integral Inequality

Question

While searching for different proofs of Hardy integral inequality, I saw a proof that used Homogeneity of norm and a kernel function. The first line of the proof says:

Let $$F(x)=\frac{1}{x}\int_{0}^{x}{f(t)dt}= \int_{0}^{1}{f(tx)dt}.$$

The function $F(x)$ is the Hardy mean operator. I am unable to follow the equality of the two integrals above which is mentioned here in Proof 2 of the following paper https://kkms.org/index.php/kjm/article/download/289/204

Help: If anyone can please help me understand it, I will be very grateful.

Answer 1

Let $g(t) = tx$ for a fixed $x$ and note that $g’(t) = x$, which is constant w.r.t. $t$. Then read the equality from the right to the left and use a change of variables to deduce that $$\begin{align} \int_0^1 f(tx) dt &= \int_0^1 f(g(t)) dt \\ &= \int_0^1 f(g(t)) \frac{g’(t)}{g’(t)} dt \\ &= \frac 1x \int_0^1 f(g(t)) g’(t) dt \\&= \frac 1x \int_{g(0)}^{g(1)} f(t) dt \\ &= \frac 1x \int_0^x f(t) dt \end{align}$$

as desired.

Answer 2

This is just the substitution $s=x^{-1} t$. Then we get $dt = x ds$ and hence $$ \frac{1}{x} \int_0^x f(t) dt = \frac{1}{x} \int_{x^{-1}\cdot 0}^{x^{-1} \cdot x } f(sx) xdt = \int_0^1 f(sx) ds. $$