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I'm looking for differentiable functions $f:\Bbb R\to\Bbb R$ such that

$$\left(\int_0^1 |f(t)|dt\right)^2> \frac{f(0)^2+f(1)^2}2$$

I found $f(x)=k$, for some constant $k$, $f(x)=x$, $f(x)=x^2$ and $f(x)=e^x$ that hold the opposite, but I couldn't find any function with this property. What does a function need to have in order for the statement to hold?

Garmekain
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    Any function with $f(0) = f(1) = 0$ and $f(x) \not= 0$ for some $x \in (0,1)$ will also do. – MSDG Dec 01 '18 at 20:42
  • @MisterRiemann Are they the only ones? – Garmekain Dec 01 '18 at 20:46
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    Certainly not. Take e.g. $f(x) = \pi\sin(\pi x) + 1$. Then the LHS is $(2+1)^2=9$, and the RHS is $1$, but $f(0)=f(1)=1.$ There are a bunch of functions satisfying your property. – MSDG Dec 01 '18 at 20:51

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Roughly speaking, any function with a large enough hump in the middle will satisfy this. The right side only considers the function values at the endpoints. If $f(\frac 12)$ is very large and the function is continuous the left side can be made as large as you want. To make a specific one, just use a triangle $$f(x)=\begin {cases} 1+100x&0 \le x \lt \frac 12\\51-100(x-\frac 12)&\frac 12 \le x \le 1\end {cases}$$ We have $f(0)=f(1)=1$ so the right side is $1$. The integral is $26$ so the left side is $676$

Ross Millikan
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