 Solution

For $\delta >0$ and $M=\Vert f^{\prime \prime }{\Vert }_{\infty }>0$, differentiate the bound $F(h)=\sqrt{6}\delta /h+Mh/2$. Its derivative vanishes at $h_{\ast }^2=2\sqrt{6}\delta /M$, and $F^{\prime \prime }(h)>0$. Consequently, when $h_{\ast }<1/2$,If $h_{\ast }\ge 1/2$, the infimum on the open allowed interval is approached as $h\uparrow 1/2$, rather than attained at an admissible endpoint. If $M=0$ and $\delta >0$, the bound decreases throughout the interval and has the same boundary behavior. At zero noise with $M>0$, its infimum is approached as $h\downarrow 0$.

A practical regularization parameter choice independent of $M$ isIt satisfies $h\to 0$ and $\delta /h\to 0$, hence gives a convergent regularization of an inverse problem on the domain of the Volterra operator inverse. This extends beyond $C^2$ exact data: if $f=Ku$ with $u\in L^2$, the differences are local averages of $u$. Extend $u$ by zero outside the interval; continuity of translations in $L^2$ makes these averages converge to $u$ on both halves. Together with $\Vert D_h(f^{\delta }-f){\Vert }_2\le \sqrt{6}\delta /h$ this proves convergence uniformly over the noise ball for every admissible exact datum.