Continuity and Integration Let $f,g:[0,1]\rightarrow{\mathbb{R}}$ be continuous functions such that $f(0)=0$,  $\int_0^1 f(x) \,dx=0$, $g(0)=0$, $\int_0^1 g(x) \,dx=1$. Then I am having trouble showing that $max\{|f(x)-g(x)|:x\in {[0,1]}\} \neq{1}$. 
I would appreciate if you share your ideas if you have any!
I was trying to establish the result by contradiction, but I could not get it.
 A: Note $(g-f)$ is continuous and $(g-f)(0) = 0$.  Thus by continuity there is some $\delta>0$ such that $(g-f)(x) < \frac{1}{2}$ for all $0\leq x <\delta$. Then $$1 = \int_0^1 g(x)-f(x)\ dx = \int_0^\delta (g-f)(x)\ dx + \int_\delta^1 (g-f)(x)\ dx \leq \frac{\delta}{2} + \int_\delta^1 (g-f)(x)\ dx.$$  So $$\int_\delta^1 (g-f)(x)\ dx \geq 1-\frac{\delta}{2}.$$  By the mean value theorem, there is some $x\in (\delta, 1)$ such that $$(g-f)(x) = \frac{1}{1-\delta}\int_\delta^1 (g-f)(x)\ dx \geq \frac{1-\frac{\delta}{2}}{1-\delta} > 1.$$
A: Hint: argue by contradiction.  Use the fact $|\int_0^1 f(x)-g(x)\,dx| \leq \int_0^1 |f(x)-g(x)|\,dx$, the fact that $f$ and $g$ are continuous, and the other assumptions.
A: Follow these steps:


*

*Show that $\int_0^1 |f(x) - g(x)| \, dx \geq 1$

*Use continuity to show that $|f|, |g| < \frac{1}{2}$ on some neighborhood around $0$, say $[0, \delta]$. Hence $|f - g| < 1$ on $[0, \delta]$.

*Conclude that $\int_0^\delta |f(x) - g(x)| \, dx < \delta$. Combine this with step 1 to get $\int_\delta^1 |f(x) - g(x)| \, dx > 1 - \delta$.

*Suppose $\max\{|f(x) - g(x)|: x \in [0,1]\} = 1$ and show it contradicts $\int_\delta^1 |f(x) - g(x)| \, dx > 1 - \delta$.

