I have a question and I'm really struggling to grasp what I need to do. I have Rolle's theorem and Mean Value Theorem given to me, and the question follows;  Let $f(x)$ be a continuous function on $[0,1]$ such that $$\int_{0}^{1} f(y) dy = 1$$  Prove that then $f(x)=1$ for some $x\in[0,1]$.  I don't know if it's the way this is worded but I've never been so confused in my entire life! Thanks in advance.
122847
We are asked the following:      let $f(x)$ be a continuous function on $[0, 1]$ such that   $$\int _ {0} ^ {1} \ f(y) dy = 1$$   prove that $f(x) = 1$ for some $x ∈ [0, 1]$   The question asks to take into account Rolle's Theorem and Mean Value Theorem.  I'm aware of the definitions, but I am really unsure how to answer this question and my tutors are not really responding till after easter.    Could someone please help? Why is there a "y" involved?  Thanks you
100435
Let $f\in C^2\{[0,1],\mathbb{R})$ and $m\{x\in [0,1]:f(x)=0\}&gt;0$. Prove that $$m\{x\in [0,1]:f'(x)=0\}&gt;0,$$ where $m$ denotes Lebesgue measure.  I don't have any clue to solve this exercise..
94975
Suppose $f$ is a continuous single-variable function, prove that:  $$2\int_a^b \int_a^x f(x)f(y) \, dy \, dx = \left[ \int_a^b f(x) \, dx \right]^2$$  This question was just on my Calculus III final exam and I couldn't figure it out, could someone provide some insight, please?
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$f:[0,1]\to \mathbb R$ , be continuous function then prove that    $$\int_0^1f^2(x)dx\geq \biggl(\int_0^1|f(x)| \biggr) ^2$$   I tried this for $x^2$  For that above is true   But I checked following proof Which is complete opposite to above.   Please help me to find that where is I am making wrong ?  Any help will be appreciated
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Prove that if $f:(0,\infty)→\mathbb{R}$ satisfying $f(xy)=f(x)+f(y)$, and if $f$ is continuous at $x=1$, then $f$ is continuous for $x&gt;0$.   I let $x=1$ and I find that $f(x)=f(x)+f(1)$ which implies that $f(1)=0$. So, $\lim_{x\to1}f(x)=0$, but how can I use this to prove continuity of $f$ for every $x \in \mathbb R$?   Any help would appreciated. Thanks
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I know that this can not be true due to counter-examples but I don't know where the error in my reasoning is.  Assumption: If $f(x)$ is differentiable in $\mathbb{R}$ then the derivative $f'(x)$ is continuous in  $\mathbb{R}$.  Faulty Proof: For every $c \in \mathbb{R}$, using the mean value theorem for $f(x),$ on the interval $x \in [c, c + h] $ where $h$ is positive.  $$ \frac{f(c + h) - f(c)}{h} = f'(\xi(h)) $$  Where $\xi(h) \in (c,c+h)$. Because this equation holds for every $h&gt;0$. It must hold in the limit as $h \rightarrow 0^+$.   $$ \lim_{h\to 0^+}\frac{f(c + h) - f(c)}{h} = \lim_{h\to 0^+}f'(\xi(h))  $$  But the left side of the equation is the right one sided derivative.  $$ f'_{+}(c) = \lim_{h\to 0^+}\frac{f(c + h) - f(c)}{h} = \lim_{h\to 0^+}f'(\xi(h))  $$  The same can be done for $h$ being negative, but because of differentiability at every point the left and right derivatives must be equal.    $$ f'(c) = f'_{+}(c) = f'_{-}(c) = \lim_{h\to 0^-}\frac{f(c + h) - f(c)}{h} = \lim_{h\to 0^-}f'(\xi(h))  $$  As $h \rightarrow 0^+$, $\xi(h) \rightarrow c$. So because the limit $\lim_{h\to 0^+}f'(\xi(h))$ exists and $\xi(h) \neq c$, it is equal to $\lim_{x\to c^+}f'(x)$  It follows that $\lim_{x\to c^+}f'(x) = \lim_{x\to c^-}f'(x) = f'(c)$ so the function $f'(x)$ is continuous.
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Theorem: Let $f:[0,1] \to [0,1]$. Assume $f$ is continuous. Then there exists   $c \in [0, 1]$ such that $f(c) = c$.         For any function $f$, let $g(x) = f(x) − x$. If the theorem is true, what does that say about $g(c)$?   Write a proof of the theorem. Do it by applying the Intermediate Value Theorem   in the appropriate way to the function $g(x) = f(x) − x$ with $a = 0$ and $b = 1$. Clearly state why the hypotheses of the IVT are satisfied here.        Well if the theorem is true, then there exists $c$ such that $f(c) = c$. Then $f(c) - c = 0$. Then $g(c) = 0$. Intermediate Value Theorem: Let $I$ be an interval and $f$ a function whose domain contains $I$. If $f$ is continuous, then for all $a, b \in I$ with $a &lt; b$ and all real numbers $k$, if $k$ is strictly between $f (a)$ and $f (b)$, then there exists $c$ such that $a &lt; c &lt; b$ and $f (c) = k$.   Proof Attempt: Take $a = 0$ and $b = 1$. Then $a,b \in[0,1]$ and $a &lt; b$. Since $f:[0,1] \to [0,1]$, it follows that for all real numbers $k$ such that $f(a) &lt; k &lt; f(b)$, then there exists $c$ such that $a &lt; c &lt; b$ so $0 &lt; c &lt; 1$ and $f(c) = k$. Then $f(c) - k = 0$. Thus, since $g(x)=f(x)-x$, it follows that $g(c) = 0$.   Looking to see if I did this correctly and/or if there is a more elegant way to prove problem 2. I am a little confused why the first theorem is true, but I went with it.
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Let $f:[0,1]\to\mathbb{R}$ be continuous with $f(0)=0$, $f(1)=1$ and $f$ is differentiable on $(0,1)$. Show that there are distinct $\xi,\eta\in(0,1)$ so that $f'(\xi)f'(\eta)=1$.  I think this requires mean value theorem. But this does not help since if apply the theorem to $f$ we can only get $f'(c)=1$ for some $c$.
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Let $g:[0,1] \rightarrow \mathbb R$ be a continuous function and assume that    $$ \int_{0}^{1} g(x) \phi'(x) dx = 0 $$   for all continuously differentiable functions $\phi: [0,1] \rightarrow \mathbb R$, where $\phi (0) = \phi(1) = 0$. Show that $g(x) = c$ for all $x \in [0,1]$ for some constant $c \in \mathbb R$.    My thoughts are applying integration by parts with $G(t) = \int_{0}^{t} g(x) dx $, since now we get  $$ 0 = \int_{0}^{1} G(t) \phi'(t) dt  + \int_{0}^{1} g(t) \phi (t) dt $$  which doesn't really go far. The second thought was applying MVT for definite integrals, which doesn't go far either as the conditions are hardly satisfied.   Any hints?
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I have been stuck for several days on this old Analysis problem (I am doing some study on my own).  I have tried several things (which I'll indicate below), but I cannot seem to figure it out.  Here is how the problem is presented: Problem: &quot;Let $f$ be a continuous real-valued function on $[a,b]$.  Suppose there exists a constant $M \geq 0 $ such that $$|f(x)| \leq M \int_a^x |f(t)| dt$$ for all $x \in [a,b]$.  Show that $f(x)=0$ for all $x \in [a,b]$.&quot; My Thoughts: I have tried using the mean value theorem iteratively, but that seems to always lead me down a dead end road.  I deduced that $f(a)=0$.  If only $f$ were assumed to be differentiable, then maybe I could play with trying to get the derivative to be $0$, but unfortunately it's only continuous.  My other thought was to (somehow) use the condition to show that $\int_a^b |f(x)| dx = 0$.  I also played around a bit with contradiction, but to no avail.  Even if one of these hair-brained thoughts is correct, I am not really sure what to do next. If you have any ideas, suggestions, or solutions, I would really appreciate it if you are willing to share them.  Thank you for your time.
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Determine all $a\in[0,1]$ such that for ${\it every}$ continuous function $f:[0,1]\to \mathbb{R}$ with $f(0)=f(1)$ there exists at least one $x$ where $f(x) = f(x+a)$.   First of all, $a=0,1/2,1$ are obviously good values. Secondly, no $1/2&lt;a&lt;1$ works (look at $f = \sin(2\pi x)$).  I remember trying to solve this several years ago. I think I started by dividing $[0,1]$ into regions where $f\geq 0$ and $f&lt;0$. Then studied each interval and all pairs of intervals with the same sign. However, this seems like a problem that should have a simple solution and I would love to see it.
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Suppose $f:[0,1]\to \mathbb{R}$ is continuous. Show that   $$\lim_{n\to\infty} n\int_0^1 f(x)x^n\,dx = f(1).$$  My answer so far: First I want to assume that $f\in C^1$. Then   $$n\int_0^1f(x)x^n\,dx = \left[\frac{n}{n+1}x^{n+1}f(x)\right]_0^1 - \frac{n}{n+1}\int_0^1 x^{n+1}f'(x)\, dx\\ \frac{n}{n+1}f(1) - \frac{n}{n+1}\int_0^1 x^{n+1}f'(x)\, dx,$$  which goes to $f(1)$ because the last integral goes to zero.  But approximating $f$ by $\phi\in C^1$ won't necessarily work, because $\phi(1)$ may not equal $f(1)$... how can we finish the argument?
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I only knew the standard mean value theorem for integrals. (i.e. $\int_a^b f(x)dx= f(c)(b-a)$ for some $c$ between $[a,b]$ where $f$ is continuous. This is directly derived by applying mean value theorem and Fundamental theorem of calculus) I'm taking numerical analysis this year and there is one theorem stated without a proof in my text. That is:  Let $f,g$ be continuous real-valued functions on $[a,b]$ such that $g$ is nonnegative. Then there exists some $c$ between $a,b$ such that $\int_a^b fg(x) dx = f(c) \int_a^b g(x) dx$  How do I prove this?
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Possible Duplicates:      I found a nice problem I would like to share. Problem: If $f$ is continuous on $[0,1]$, and if $$\int_0^1 f(x)x^n \ dx =0$$ for every positive integer $n$, prove that $f(x)=0$ on $[0,1]$. Source: W. Rudin, Principles of Mathematical Analysis, Chapter 7, Exercise 20. I have posted a proposed solution in the answers.
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Let $f:(a,b)\rightarrow\mathbb{R}$.  The statement to prove is that if $\forall x \in (a,b)$ Lebesgue integral $\int_{(a,x)}fd\lambda=0$, then $f(x)=0$ $\lambda$-almost everywhere.  So if it wouldn't be true then we would have $\forall x \in (a,b):\int_{(a,x)}f_+d\lambda=\int_{(a,x)}f_-d\lambda\neq0$, so every interval contains positive and negative values of the function. It is  such an $f$ so my guess would be that this function is not Lebesgue measurable and we can't take this integral.  Thanks in advance!
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This is ex. 14.2.7. from Terence Tao's Analysis II book.   Let $I:=[a,b]$ be an interval and $f_n:I \rightarrow \mathbb R$ differentiable functions with $f_n'$ converges uniform to a function $g:I \rightarrow \mathbb R$. Suppose $\exists x_0 \in I: \lim \limits_{n \rightarrow \infty} f_n(x_0) = L \in \mathbb R$. Then the $f_n$ converge uniformly to a differentiable function $f:I \rightarrow \mathbb R$ with $f' = g$.  We are not given that the $f_n'$ are continuous but he gives the hint that $$ d_{\infty}(f_n',f_m') \leq \epsilon \Rightarrow |(f_n(x)-f_m(x))-(f_n(x_0)-f_m(x_0))| \leq \epsilon |x-x_0| $$ This can be shown by the mean value theorem. My question is : How does this help me to prove the theorem ?
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Assume that $f$ be continuous on $\mathbb{R}$, $f'(x)$ exists for all $x\neq 0$, and $\lim_{x\rightarrow 0} f'(x)=1$. We need to show $f'(0)$ exist and is equal to $1$.  $f'(0)=\lim_{x\rightarrow 0}\frac{f(x)-f(0)}{x}$, $\lim_{x\rightarrow 0}f'(x)=1\Rightarrow\lim_{x\rightarrow 0}\lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}=1$...am I going in the right direction? Please help.