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Let $[a,b]$ be a finite closed interval on $\mathbb{R}$, $f$ be a continuous differentible function on $[a,b]$. Prove that $$\max_{x\in [a,b]} |f(x)|\le \Bigg|\frac{1}{b-a} \int_a^b f(x)dx\Bigg|+\int_a^b |f'(x)|dx$$

I think this is similar to Sobolev embedding theorem but have no idea about how to use it. I have also tried to transform the inequality into $$(b-a)\int_a^b (|f(t)|-|f'(x)|)dx\le \Bigg|\int_a^b f(x)dx\Bigg|,$$ where $f(t)$ is the maximum, but still don't know how to proceed. Thanks for any help.

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There is a point $c\in (a,b)$ such that

$$f(c) = \frac{1}{b-a}\int_a^b f(x)\, dx$$

Let $x\in [a,b]$. Since

$$f(x) = f(c) + \int_c^x f'(t)\, dt$$

then

$$\lvert f(x)\rvert \le \lvert f(c)\rvert + \int_a^b \lvert f'(t)\rvert\, dt$$

Now finish the argument.

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although Kobe has explained well. I had already started writing the answer. So using triangle inequality we have $|f(c)| \leq |f(\xi)|+|f(c) - f(\xi)|$. By fundamental theorem of calculus $f(c) - f(\xi) = \int_{\xi}^cf'(x)dx$. Taking absolute values we have $$|f(c) - f(\xi)| = \int_{\xi}^cf'(x)dx \leq \int_{\xi}^c|f'(x)|dx \leq \int_a^b |f'(x)|dx \hspace{2mm} (A)$$. Using mean value theorem we also have $$ f(\xi) = \frac{\int_a^bf(x)dx}{b-a}$$ therefore $$ |f(\xi)| \leq \frac{\int_a^b|f(x)|dx}{b-a} \hspace{3mm}(B)$$ Hence by(A) and (B) we have the result.

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