**Theorem**

Let $h=\max_{1\leq i\leq n}(x_i-x_{i-1}).$ Then

$$

\parallel u-u_I\parallel_E\leq Ch\parallel u''\parallel

$$

for all $u\in V$, where $C$ is independent of $h$ and $u$.

**Proof**

Recalling the definition of two norms, it is clearly sufficient to prove the estimate piecewise, i.e., that

$$

\int_{x_j-1}^{x_j}(u-u_I)'(x)^2dx\leq c(x_j-x_{j-1})^2\int_{x_{j-1}}^{x_j}u''(x)^2dx

$$

as the stated result follows by summing over $j$, with $C=\sqrt{c}$. Let $e=u-u_I$ denote the error; since $u_I$ is a linear polynomial on the interval $[x_{j-1},x_j]$, the above is equivalent to

$$

\int_{x_{j-1}}^{x_j}e'(x)^2dx\leq c(x_j-x_{j-1})^2\int_{x_{j-1}}^{x_j}e''(x)^2dx.

$$

Changing variables by an affine mapping of interval $[x_{j-1},x_j]$ to the interval $[0,1]$, We see that this is equivalent to showing

$$

\int_0^1\widetilde{e}^2d\widetilde{x}\leq c\int_{x_{j-1}}^{x_j}\widetilde{e}''(x)^2d\widetilde{x},

$$

where $x=x_{j-1}+\widetilde{x}(x_j-x_{j-1})$ and

$$

\widetilde{e}(\widetilde{x})=e(x_{j-1}+\widetilde{x}(x_j-x_{j-1})).

$$

Note that we have arrived at an equivalent estimate that does not involve the mesh size at all. The technique of reducing a mesh-length dependent estimate to a mesh-independent one in this way is called a homogeneity argument (or scaling argument).

Let $w=\widetilde{e}$ to simplify the notation, and write $x$ for $\widetilde{x}$. Note that $w$ vanishes at both ends of the interval (the interpolation error is zero at all nodes). By Rolle's Theorem, $w'(\xi)=0$ for some $\xi$ satisfying $0 < \xi<1$. Thus,

$$

w'(y)=\int_{\xi}^yw''(x)dx.

$$

By Schwarz' inequality,

$$

\begin{aligned}|w'(y)|&=\left|\int_{\xi}^y w''(x)dx\right|\\&=\left|\int_{\xi}^y1\cdot w''(x)dx\right|\\&\leq\left|\int_{\xi}^y 1dx\right|^{\frac{1}{2}}\cdot\left|\int_{\xi}^y w''(x)^2dx\right|^{\frac{1}{2}}\\&=|y-\xi|^{\frac{1}{2}}\left|\int_{\xi}^y w''(x)^2dx\right|^{\frac{1}{2}}\\&\leq|y-\xi|^{\frac{1}{2}}\left(\int_0^1w''(x)^2dx\right)^{\frac{1}{2}}.\end{aligned}

$$

Squaring and integrating with respect to y completes the verification, with

$$

c=\sup_{0<\xi<1}\int_0^1|y-\xi|dy=\frac{1}{2}.

$$

as for

$$

\int_{x_j-1}^{x_j}(u-u_I)'(x)^2dx\leq c(x_j-x_{j-1})^2\int_{x_{j-1}}^{x_j}u''(x)^2dx

$$

we can directly deal with it as above

$$

\begin{aligned}(u-u_I)'(x)^2&=\int_{\xi}^{x_j}1\cdot (u-u_i)''(x)dx\\&\leq\int_{\xi}^{x_j}1^2dx\cdot\int_{\xi}^{x_j}u''(x)^2dx\\&\leq(x_j-x_{j-1})\int_{x_{j-1}}^{x_j}u''(x)^2dx\end{aligned}

$$

since the left term is a constant, the original inequality can be obtained by integrating.

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