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I've tried to solve the following problem (1.1.30 from Berkeley Problems in Mathematics, no solution is provided therein), but I'm not sure if my solution is correct. I wanted to ask you for your opinion.

The problem: Let $S$ be the set of all real $C^1$ functions $f$ on $[0,1]$ such that $f(0)=0$ and $$ (*)\int_0^1f'(x)^2\mathrm{d}x\leq 1 $$ Define $$ J(f) = \int_0^1 f(x)\mathrm{d}x. $$ Prove that the function $J$ is bounded on $S$ and compute its supremum. Is there a function $f_0\in S$ at which $J$ attains its supremum? If so, give $f_0$.

My attempts: First note that for any $y\in[0,1]$, $\int_0^1 f'(x)^2\mathrm{d}x \geq \int_0^y f'(x)^2\mathrm{d}x$. Thus, using the Cauchy Schwartz inequality,

$$1\geq \int_0^y f'(x)^2\mathrm{d}x \cdot\underbrace{\int_0^1 1^2\mathrm{d}x}_{\geq\int_0^y 1^2\mathrm{d}x}\geq \left(\int_0^y f'(x)\cdot 1\mathrm{d}x\right)^2=(f(y)-f(0))^2=f(y)^2 $$ which implies that $|f(y)|\leq 1$. Therefore, $J(f)\leq|J(f)|\leq1$ so we have shown boundedness.

The rest is still a bit shaky:\ Now the supremum of $J$ on $S$ is attained when the equality in $(*)$ holds (why is that? I'm not sure I could show this). This is the case for $f_0(x)=x$. Now $J(f_0)=\frac{1}{2}$. My guess is that therefore $J(f)\leq \frac{1}{2}$ is the supremum and maximum.

Thank you for your suggestions.

Cyclone
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  • Clearly, it must be equality in $()$, otherwise you may increase $(1+\epsilon)f(x)$ to still satisfy $()$ which contradicts the maximum. 2) "This is the case for $f_0(x)=x$" is way too weak. This is the case for many functions, for example, for $f_1(x)=\sqrt{3}x(1-x/2)$ that has larger $J(f_1)=1/\sqrt{3}\approx 0.57>0.5$. Looks like the problem on calculus of variations (if not some smart trick).
    • – A.Γ. Aug 30 '15 at 13:06
  • @ A.G.: Thanks for your reply. Your argument why it needs to be equality in (*) seems perfect to me. I've also checked your example $f_1(x)$ and I agree that it has the properties you named. – Cyclone Aug 30 '15 at 13:56
  • I have posted the alternative solution via calculus of variations. You can find it if you follow the link to the duplicate question above. Just for you to see the standard technique which is not really hard to use. – A.Γ. Aug 30 '15 at 20:13