Let $f:\Bbb R \rightarrow \Bbb R$ be a Lebesgue measurable function in $L^2$. Show $F(x)=\int_0^x f(t)dt$ satisfies $|F(x)-F(y)|\leq C|x-y|^\frac 12$

Question

Let $f:\Bbb R \rightarrow \Bbb R$ be a Lebesgue measurable function that is in $L^2$. Show $F(x)=\int_0^x f(t)dt$ satisfies $|F(x)-F(y)|\leq C|x-y|^\frac 12$.

Here's what I have so far.

$f\in L^2 \Rightarrow (\int (|f|)^2d\lambda )^\frac12 \lt \infty \Rightarrow |f| \lt \infty$ a.e.

$|F(x)-F(y)|=|\int_0^xf(t)dt - \int_0^yf(t)dt|=|\int_y^xf(t)dt|=|\int f\mathbf 1_{[y,x]} d\lambda| \leq \int |f\mathbf 1_{[y,x]}| d\lambda \leq sup_{t \in [y,x]}f(t) |x-y|$

I'm not entirely sure where i've gone wrong, can someone please give me some pointers on how to get the $\frac 12$ in the exponent?

Answer 1

Alternatively, use $\text{H}\ddot{o}\text{lder}$'s Inequality: $f(x) = 1 f(x)$. Assume (WLOG) $y > x$.

$$\int_x^y |f(t)|\,\text{d}t \leq \left(\int_x^y |f(t)|^2\,\text{d}t \right)^{1/2} \left( \int_x^y |1|\,\text{d}t \right)^{1/2}$$

Answer 2

Use Cauchy Swartz inequality and that $f\in L^2$