Consider the set $E = \{ f \in C[0,1]\: f(1)=0, \int_0^1 tf(t)dt =1 \}$ What is the value of $\delta = \inf \{\|f\|_{\infty} : f \in E\}$? Consider the set $E = \{ f \in C[0,1]\: f(1)=0, \int_0^1 tf(t)dt =1 \}$
What is the value of $\delta =\inf \{\lvert\lvert f\rvert\rvert_{\infty} : f \in E\}$?
I was thinking it should be $\infty$ because when $t=0$ or $t=1$ then $tf(t)=0$  and so it seems that given that little bit of "wiggle room" that we could make $f(t)$ to be arbitrarily large and still have the integral equal to $1$... But the solution says that it's suppose to equal 2... I was wondering if somebody could help me wit this? I've been working on it for so long and really haven't made much progress... I fear I am getting tunnel vision...
Also, I need to prove that there is no $f \in E$ s.t. $||f||_{\infty}=\delta$
Thanks in advance I appreciate it so much don't know what I'd do without yall!!
 A: Without the $f(1)=0$ condition, we would try to minimize $\|f\|_\infty$ by making $f$ constant. Then the integral condition amounts to $f(t)=2$. Now we can enforce $f(1)=0$ by making $f$ fall down to $0$ very steeply a tiny bit before the right end of the interval. This makes the integral slightly smaller, so we multiply it by a factor just slightly above $1$ to fix this. Then we will have a valid $f$ with $\|f\|_\infty$ slightly above $2$.
The fact that no $f\in E$ with $\|f\|_\infty=2$ exists follows because for such $f$, in some neighbourhood $(1-\epsilon,1]$ of $1$, we must gave $f(x)<1$ (and $f(x)\le 2$ on the remaining $[0,1-\epsilon]$).
A: For any $f\in C[0,1]$ let $L(f)=\int_0^1 tf(t)dt.$ We have $$|L(f)|\le \int_0^1 t|f(t)|dt\le \int_0^1t\|f\|dt=\|f\|/2.$$ So if $L(f)=1$ then $\|f\|\ge 2.$
For $n\in \Bbb N$ let $g_n(x)=2$ for $x\in [0,1-1/(n+1)]$ and $g(1)=0$ and let $g$ be linear on $[1-1/(n+1),1].$
Let $f_n=g_n/L(g_n).$
We have $f\in E$ and $L(f)=1.$
And $\|f_n\|=\|g_n\|/L(g_n)=2/L(g_n)\to 2$ as $n\to \infty$ because $$1=\int_0^1 2tdt>L(g_n)>\int_0^{1-1/(n+1)}2tdt=(1-1/(n+1))^2.$$
