Given $f$ a real differentiable function, $2 \pi$ periodic such that $\int_{0}^{2 \pi} f(t) dt=0$ show that:  $\int_{0}^{2 \pi} (f(t))^2 dt \le \int_{0}^{2 \pi} (f'(t))^2 dt$. When does equality hold?
120740
Let $f$ be $C^1$ in $[-\pi, \pi]$ and satisfies $\int_{-\pi}^\pi f(x)dx=0$, periodic boundary condition.  Then, prove that $\int_{-\pi}^\pi (f(x))^2dx\le \int_{-\pi}^\pi (f'(x))^2dx$.  I try to prove it by Parseval's equality(with $X_n=\sin nx$) and Schwartz inequality,   but then some constants come out. Also why condition $\int_{-\pi}^\pi f(x)dx=0$ needs?  Give some advice.
92992
Let $S=\{f:\mathbb{R} \to  \mathbb{R}\}$ that satisfies: $\forall f\in S$, $f'$ exists and $f'$ is continuous and $f(0)=f(1)=0$.  Please prove that $\exists \delta :\forall f\in S$ s.t.  $\int_{0}^{1} (f(x))^2dx\leq \delta \int_{0}^{1} (f'(x))^2dx$. Thanks in advance.
50638
Suppose $f$ is positive and increasing on $[0, +\infty)$. Suppose that $F(x)=$$\int_0^x f(t)dt$. Prove that $$\int_0^{+\infty} \frac {1}{f(x)} dx$$ converges if and only if $$\int_0^{+\infty} \frac {x}{F(x)} dx$$ converges.   My thoughts, and attempt at solution -  Since $f$ is increasing, $F(x)=\int_0^xf(t)\,dt\le\int_0^xf(x)\,dt=xf(x).$ We can rearrange that to $\frac1{f(x)}\le\frac{x}{F(x)}.$ Hence if the integral of $\frac{x}{F(x)}$ converges, then so does the integral of $\frac1{f(x)}.$  However, the above working seems incomplete as it bounds the integral from the upper side only. In fact, it shows that if the second integral converges, then the first one may converge, but not vice versa. How do I place a lower bound on the integral, and go about the rest of the solution?   Haven't been able to proceed from here, please help.
26533
I'm trying to prove that $$\lim_{h \to 0}\frac{1}{h}\int_0^h{f(t)dt} = 0$$ where $$f(t) = \begin{cases}\cos{\frac{1}{t}} &amp;\text{ if } t \neq 0\\ 0&amp;\text{otherwise}\end{cases}.$$ Can someone give me a hint where to start? Darboux sums somehow seem to lead me nowhere.  NOTE: I cannot assume that $f$ has an antiderivative $F$.
20693
Let $f:[0,1] \longrightarrow \mathbb R$ be a continuous function, and let $g:\mathbb R \longrightarrow \mathbb R$ be a continuous and periodic function with period $1$. Prove that   $\displaystyle\lim_{n\to \infty}\int_0^1f(x)g(nx)dx=\left(\int_0^1f(x)dx\right)\left(\int_0^1g(x)dx\right)$.  any Ideas?
99631
I have tried the following: $$ f(x) = \sin x-\frac{2x}{\pi} \\ f'(x)= \cos x-\frac{2}{\pi} \\ f''(x) = -\sin x \leq 0 $$ But this doesn't seem to be heading in the right direction as it would appear $f'(x)$ is decreasing and has a zero within the domain, so I'm not sure how this might be used to prove the inequality.
113752
Let $f(x)$ be a strictly positive continuous function in $[0,1]$ such that $f^2(t) \leq 1 + 2\int_0^t f(s)\, \mathrm{d}s$. Prove that $f(t)\leq t+1$.    Obviously, if $f(x)$ can be differentiated, it is true. But I can't deal with the case that it is not differentiated.
63411
Let $f: \mathbb{R} \to \mathbb{R}$ be a differentiable function such that $f(0)=0$ and $\forall x \in \mathbb{R}$, we have $f'(x)=[f(x)]^2$. Show that $f(x)=0, \forall x \in \mathbb{R}$.
73737
What can be said about a complex valued, continuous function $f$, defined on $[0,1]$, such that: $$ \int_0^1{|f|^2}=\left|\int_0^1{f}\right|^2  $$ I encountered this form as part of an exercise. Obviously, the above holds for any constant $f$, and it seems intuitive that the converse also holds (i.e. that if the above equality is true, then $f$ is constant), but I could not prove it.  Any help will be appreciated.
57650
If $\int^x_0  f (t) dt =x+  \int^1_x  t f (t) dt$, find value of $f(1)$  solution:- $\int^x_0  f (t) dt =x+  \int^1_x  t f (t) dt$ $\int^x_0  f (t) dt =x+  \int^0_x  t f (t) dt$ +  $\int^1_0  t f (t) dt$ $\int^x_0  f (t) dt =x-  \int^x_0  t f (t) dt$ +  $\int^1_0  t f (t) dt$ $\int^x_0  f (t) dt + \int^x_0  t f (t) dt$ =$x  +  $$\int^1_0  t f (t) dt$  I think, I am not in the right track Help me to find the value of $f(1)$
114161
How to prove that $$ \int_{t_0}^{t_0+T} \sin(m\omega t)\sin(n\omega t)\,\mathrm{d}t$$  will equal to $0$ when $m\ne n$ and $\frac{T}{2}$ when $m=n\ne 0$? Besides $$ \int_{t_0}^{t_0+T} \cos(m\omega t)\cos(n\omega t)\,\mathrm{d}t$$ will equal to $0$ when $m\ne n$ and $\frac{T}{2}$ when $m=n\ne 0$ and $T$ when $m=n=0$?
89465
I am struggling to construct a simple proof of the following.  Suppose $f(x)$ is differentiable for any real value $x$ and that $f'(x) \le 4$ for all real values of x. Prove there is at most one real value $c&gt;2$ such that $f(c) =c^2$.  I am completely lost on where to start on this proof and any help would be greatly appreciated!
43439
Given $$f(x) = \int_{0}^{\phi (x)} g(t) dt$$ How could we find $f&#39;(x)$? Please explain your answer.
59868
Prove that for $ f(x+2\cdot\pi)=f(x), x \in \mathbb{R} $, there exists $ x_0 $ so $ f(x_0+\pi)=f(x_0) $.  This is supposed to be question about continuity, but I’m not sure exactly what they mean, perhaps I misunderstood some concept along the way.
98512
I want to show that if $f$ is non-increasing and $f\in L_{1}([a,\infty),m)$ where $m$ is Lebesgue measure then $\lim_{t\to\infty} t f(t)=0$. So far I've been able to show that $f\geq 0$ and that $\lim_{t\to\infty} f(t)=0$. Since monotone functions are differentiable a.e. I thought about using integration by parts but couldn't get anywhere with that. Any hints or suggestions would be greatly appreciated.
102094
let $$f(x)=\int_{x}^{x+1}\sin{(e^t)}dt$$  show that: $$e^x|f(x)|\le 2$$  My idea: let $$e^t=u$$ then $$|f(x)|=|\int_{e^x}^{e^{x+1}}\dfrac{1}{u}d\cos{u}|$$
9850
Let $f:\mathbb{R} \rightarrow \mathbb{R}$ be a differentiable function such that $f(0) = 0$ and $f'(x) = [f(x)]^{2}$, $\forall x \in \mathbb{R}$. Show that $f(x) = 0$,  $\forall x \in \mathbb{R} $.  I first (unsuccessfully) tried using the Mean Value Theorem, but this is in the Integrals chapter so the solution probably involves them. Can't really see where integrals come in here though.  What I've got so far:   (i) Since $f$ is differentiable, thus it is continuous and, hence,  integrable. Therefore $f^2$ is also integrable and as $f'=f^2$, $f'$ is too.   (ii) $f' \geq 0$, $\forall x \in \mathbb{R}$