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Suppose $g \colon[a,b] \rightarrow \mathbb{R}$ is continuous on $[a,b]$ and integrable with $g(a)=g(b)=0$ and that $g'' \colon (a,b)\rightarrow \mathbb{R}$ is continuous on $(a,b)$ and bounded by $K$. Prove that $|g(x)| \leq K|x-a||x-b|$ for all $x \in [a,b]$. I tried using the definition of derivatives and I'm leaning toward the Mean Value Theorem. Help please.

Siminore
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user115810
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    this is quite interesting.. Would you mind to write what you have tried in detail in the mean while some one will find a way to you.... I like this question +1 –  Dec 16 '13 at 10:39
  • $g(x)=1-x^2$, hence $g''(x)=-2$, $a=-1$, $b=1$, but the claim doesn't hold. I presume it should read $|g(x)|\le|K||x-a||x-b|$ – Michael Hoppe Dec 16 '13 at 10:55

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Using Taylor’s formula, we can get\begin{equation}g(a)=g(x)+g'(x)(a-x)+\frac{g''(x_1)(a-x)^2}{2}\end{equation}and\[g(b)=g(x)+g'(x)(b-x)+\frac{g''(x_2)(b-x)^2}{2}.\]Note that $g(a)=g(b)=0$, we have\[-(b-a)g(x)=\frac{g''(x_1)(a-x)^2(b-x)}{2}-\frac{g''(x_2)(b-x)^2(a-x)}{2},\]hence\begin{align}|(b-a)g(x)|=&\left|\frac{(a-x)(b-x)}{2}\right|(|g''(x_2)(a-x)|+|g''(x_1)(b-x)|)\\\leqslant&\left|\frac{(a-x)(b-x)}{2}\right|(K(x-a)+K(b-x))=K(b-a)\left|\frac{(a-x)(b-x)}{2}\right|\end{align}where $K=\sup|g''(x)|$.

Therefore\[|g(x)|\leqslant \left|\frac{K(a-x)(b-x)}{2}\right|.\]

Nirvanacs
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