Fractional integrals of continuous data are Hölder continuous

For each $t$, [F1] and $\beta >0$ give $\mathrm{\mid }{I}^{\beta }g(t)\mathrm{\mid }\le \mathrm{\parallel }g{\mathrm{\parallel }}_{\mathrm{\infty }}\mathrm{\Gamma }(\beta {)}^{-1}{\int }_0^t(t-r{)}^{\beta -1}dr=\mathrm{\parallel }g{\mathrm{\parallel }}_{\mathrm{\infty }}{t}^{\beta }\mathrm{/}(\beta \mathrm{\Gamma }(\beta ))$. By [F2] this is the first bound, and it also gives $I^{\beta }g(0)=0$.

Suppose $0\le s\le t\le T$ and put $\delta =t-s$. Since $\beta -1<0$, the kernel $(t-r{)}^{\beta -1}$ is no larger than $(s-r{)}^{\beta -1}$ for $0\le r<s$. Thus [F1] gives $\mathrm{\Gamma }(\beta )\mathrm{\mid }{I}^{\beta }g(t)-I^{\beta }g(s)\mathrm{\mid }\le \mathrm{\parallel }g{\mathrm{\parallel }}_{\mathrm{\infty }}\left({\int }_0^s\right((s-r{)}^{\beta -1}-(t-r{)}^{\beta -1})dr+{\int }_s^t(t-r{)}^{\beta -1}dr)\mathrm{.}$ The first integral is $(s^{\beta }-t^{\beta }+{\delta }^{\beta })\mathrm{/}\beta$. Its integrand is nonnegative, so the integral is nonnegative; and $t^{\beta }-s^{\beta }\ge 0$ by monotonicity of positive powers in [F3], so the same formula bounds it above by ${\delta }^{\beta }\mathrm{/}\beta$. The second integral is ${\delta }^{\beta }\mathrm{/}\beta$. Therefore the right side is at most $2\mathrm{\parallel }g{\mathrm{\parallel }}_{\mathrm{\infty }}{\delta }^{\beta }\mathrm{/}\beta$.

Divide step 1.2 by $\mathrm{\Gamma }(\beta )>0$ and use [F2]. This gives the stated Hölder bound when $s\le t$; interchanging $s,t$ gives it for every pair. The bound tends to zero with $\mathrm{\mid }t-s\mathrm{\mid }$, so $I^{\beta }g$ is continuous.