How to prove lim $x\int \frac{f(t)}{(t)}dt$ from $x \to 1$ as $x\to 0^{+} = 0$ , as $f\in L[0,1]$?

Question

The hint just tells me to make use of eigenfunction

$X[x,1](t)=X[0,t](x)$ any $t$ and $x$ in $[0,1]$

How to prove this?

Any suggestion shall be appreciated!

Answer 1

If the actual question is as in the comments, for any positive $x<\frac{1}{\color{red}{2}}$

$$ \int_{x}^{1}f(t)\frac{x}{t}\,dt = \int_{x}^{\color{red}{2}x}f(t)\frac{x}{t}\,dt+\int_{\color{red}{2}x}^{1}f(t)\frac{x}{t}\,dt$$ is bounded, in absolute value, by $$ \int_{x}^{\color{red}{2}x}|f(t)|\,dt + \frac{1}{\color{red}{2}}\int_{\color{red}{2}x}^{1}|f(t)|\,dt $$ so $$ \limsup_{x\to 0^+}\left|x\int_{x}^{1}f(t)\frac{dt}{t}\right|\leq \frac{1}{\color{red}{2}}\int_{0}^{1}|f(t)|\,dt.$$ On the other hand the constant $\color{red}{2}$ can be replaced by any larger constant, so the wanted limit is zero.