# An inequality on $C^1$ periodic functions

Suppose $f \in C^1(\mathbb{R})$ and $f(x + 1) = f(x) \ \forall x \in \mathbb{R}$. Show that $$||f||_{\infty} \leq \int_0^1|f| + \int_0^1|f'|.$$

I have tried using techniques in Fourier Analysis such as Parseval's Identity and expanding $f$ since $f$ is periodic and $C^1$; I have also tried Mean Value Theorem, but I have yet to produce anything useful. Both answers and hints are appreciated and thanks in advance!

• What does the mean value theorem tell you about the first integral? How does the second integral relate to the distance between $f(x)$ and $f(y)$ for $x,\,y \in [0,\,1]$? – Daniel Fischer Jul 8 '13 at 14:15

1. $\|f\|_{\infty}=\max_{0\le x\le 1}|f(x)|=|f(x_0)|$ for some $x_0\in[0,1]$
2. $|f(y_0)|=\min_{0\le y\le 1}|f(y)|$ for some $y_0\in[0,1]$(Note $|f(y_0)|\le \int_0^1|f(x)|\;\mathbb{d}x$)
3. $|f(x_0)|=|f(x_0)-f(y_0)+f(y_0)|=|\int_{y_0}^{x_0}f^{\prime}(x)\;\mathbb{d}x+f(y_0)|\le \int_0^1|f^{\prime}(x)|\;\mathbb{d}x+|f(y_0)| \\ \le \int_0^1|f^{\prime}(x)|\;\mathbb{d}x+\int_0^1|f(x)|\;\mathbb{d}x$
By the mean-value theorem there is a $0 \leq \xi \leq 1$ such that $$\int_{0}^{1} f(x) dx = f(\xi)$$ Therefore $$f(x) = f(x) - f(\xi) + \int_{0}^{1} f(y) dy = \int_{\xi}^{x} f'(y) dy + \int_{0}^{1} f(y) dy$$ Taking absolute values we get $$|f(x)| \leq \int_{0}^{1} |f'(y)| dy + \int_{0}^{1} |f(y)| dy$$ This inequality can be refined. I think you can put an $\tfrac 12$ in front of the derivative term.