Proving that $(Sf)(x)=1+\int_0^x t^2f(t)\,dt$ is $K$-Lipschitz

Question

I want to show that $S:C[0,1]\to C[0,1]$ where $$(Sf)(x) = 1+\int_{0}^{x} t^2\cdot f(t) \,dt$$

is a Lipschitz map for $x\in[0,1]$. There I use the Euclidean norm on $\mathbb{R}$ and the uniform norm om $C[0,1]$. This is what I have got:

For $f,g\in C[0,1]$, we have (by the triangle inequality)

\begin{eqnarray} \left|Sf(x)-Sg(x)\right| &=\left|\int t^2\cdot f(t)\, dt+\int t^2\cdot g(t)\, dt\right|\\ &=\left|\int_{0}^{x}t^2\cdot(f(t)-g(t))\,dt\right| \\ &\leq\int_{0}^{x}\left|t^2\right|\cdot\left|(f(t)-g(t))\right| \end{eqnarray}

How do I continue from here? I want to show that $\left|Sf(x)-Sg(x)\right|\leq K \cdot \lVert f-g\rVert_{\infty}$, where $0<K<1$.

Answer 1

Almost there...

You have $|Sf(x)-Sg(x)| \le \int_0^x t^2 dt \|f-h\|_\infty ={ x^3 \over 3} \|f-h\|_\infty$, and so $\|Sf-Sg\|_\infty \le { 1\over 3} \|f-h\|_\infty$.