Show that $\exists\alpha>0$ so that $|f(x_i)- \frac{1}{h}\int _0^1 f(x)\Phi_i (x)\,dx |\le\alpha h^2$.

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I am trying to prove the following statement. Let $f\in C^2([0,1])$ and let $\Phi _i\ for\ i=1,…N-1$ be the family of hat functions. Show that there exists a $\alpha >0$ such that \begin{equation} \left|f(x _i)- \frac{1}{h}\int _0^1 f(x)\Phi _i (x)\ dx \right|\leq \alpha h^2 \end{equation}

To prove the inequality I started by approximating the function f by a piecewise linear function that is continuous on the interval [0,1]. I divided this interval into N subintervals of equal length h, with N+1 nodes. $x_0=0, …,x_N=1$ The linear shape functions $\Phi _i(x)$ are defined such that $\Phi _i(x)$ is equal to 1 at the node $x _i$, and then decreases linearly to zero at the node $x _{i-1}$ and $x _{i+1}$. To prove the inequality I used the error estimate for the interpolation of a function by piecewise linear functions. This estimate tells us that the error in the interpolation of a function f at the nodes $x_i$ is bounded by a constant times $h^2$. We denote the interpolation of f an the node $x_i$ as $f _h(x _i)$. It holds that $f _h(x _i)=\frac{1}{h}\int _0^1 f(x)\Phi _i (x) dx$. The error can be expressed as $E=f(x _i)- f _h(x _i)$. From the error estimate it follows that $|E|\leq \alpha h^2$, where $\alpha$ is a constant. This inequality holds for each $x_i$. Therefore, we can rewrite the inequality as \begin{equation} |f(x _i)- \frac{1}{h}\int _0^1 f(x)\Phi _i (x) dx |\leq \alpha h^2 \end{equation} Is my solution correct or are there any mistakes? Thank you for your help!