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Let $f: [0,1] \to \mathbb{R}$ be continuously differentiable with $f(0)=0$. Prove that $$\Vert f \Vert^{2} \leq \int_{0}^{1} (f'(x))^{2}dx$$

Here $\Vert f \Vert$ is given by $\sup\{|f(t)|: t \in [0,1]\}$.

I'm just a bit unclear how to proceed.

user8312
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2 Answers2

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$$ \begin{align} |f(x)|^2 &=\left|\int_0^xf'(t)\,\mathrm{d}t\right|^2\tag{1}\\ &\le\left(\int_0^x|f'(t)|\,\mathrm{d}t\right)^2\tag{2}\\ &\le\left(\int_0^1|f'(t)|\,\mathrm{d}t\right)^2\tag{3}\\ &\le\int_0^1|f'(t)|^2\,\mathrm{d}t\tag{4} \end{align} $$ Explanation:
$(1)$: integrating the derivative
$(2)$: convexity of $|x|$
$(3)$: integral over a larger domain
$(4)$: convexity of $x^2$

robjohn
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  • @Stefanos: I have added another line so that I can explain things better. $(2)$: the absolute value of an integral is no greater than the integral of the absolute value. That can be seen as an application of Jensen with the convex function $|x|$. It can also be seen as an application of the triangle inequality applied to the positive and negative parts of $f'$. – robjohn Mar 13 '14 at 22:45
  • @Stefanos: $(3)$: integrating a positive function over a larger domain gives a larger result. I think this is pretty self-explanatory. $(4)$: over a domain of unit measure, the square of the integral is no greater than the integral of the square. Again, this can be seen as Jensen's inequality, or simply as Hölder's Inequality applied to the functions $1$ and $f'$. – robjohn Mar 13 '14 at 22:46
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Let x = argsup f. Because f(0) = 0, f(x) is the integral up to x of f', which is less than or equal to the L2 norm of f' up to x by convexity of the squaring function, which is less than or equal to the L2 norm of f' on the entire interval. Now square both sides.

user135310
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  • Is there a more elementary approach I can extract from this? This question is from a chapter in my textbook which precedes any kind of measure theory (the chapter covers equicontinuous families and convergence in function $C^{0}$). I haven't yet been formally exposed to $L2$ norms or the concept of argsup. – user8312 Mar 13 '14 at 21:57
  • We can use the same concepts using simpler terminology. Let x be a value in the interval where f attains its maximum. The integral up to x of f' <= sqrt(integral up to x of of (f')^2) because g(x)=x^2 is convex. This quantity <= sqrt(integral over whole interval of (f')^2). But this is what the other answer says. – user135310 Mar 13 '14 at 22:11