We have $f:\mathbb{R}\rightarrow\mathbb{R}$ which is continunous and its derivative is also continuous. We define set $D=\{x:f'(x)=0\}$. Prove that set $f(D)$ is of 0 (Lebesgue) measure.  If we assume that $f$ is monotonic then every maximal interval on which $f$ is constant is landing on single point. Between such interval there is always a gap, so there are no intervals in $f(D)$. However, I believe that $f(D)$ can have "more" that countably infinite elements. If that's true, then I do not know how to proceed.
122576
Let $f:\mathbb{R}\to \mathbb{R}$ be $C^1$ function and $K = \{x : f'(x) = 0 \} $.   Show that $\mu \left(f\left(K\right)\right) = 0$, where $\mu$ is Lebesgue measure.   My attempt was following: $$\mu \left(f\left(K\right)\right)= \int_{f(k)} 1 dy \stackrel{(*)}{=} \int_{K}f'(x) dx = \mu\left(K\right) \cdot 0 = 0$$ but we cannot substitute $y = f(x)$ at $(*)$ like that. I was told that there exists quite elementary proof (not using Sard's theorem) so I'm looking for it.
11285
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.
80567
I came across the following challenging problem in my self-study:  Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be continuous. Then the set of points where $f$ is differentiable is a measurable set.  I am having trouble thinking of where to begin in proving this result, and wanted to see if anyone visiting had some suggestions on how to proceed.
44777
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
53589
If $f$ is continous in $[a,b] \subset \mathbb{R}$ and exist real numbers $\alpha \not= \beta$ such that $$\alpha\int_{a}^{c} f(x) dx + \beta\int_{c}^{b} f(x) dx  =0$$ for all $c \in [a,b]$ show that $f (x)=0$ for all $x \in [a,b].$
49805
As the title indicates, I'd like to prove the following:     If $f:\mathbb R\to\mathbb R$ is a continuous function on $[a,b]$, then $f$ attains its maximum.   Now, I do have a working proof: $[a,b]$ is a connected, compact space, which means that because $f$ is continuous, $f([a,b])$ is compact and connected as well.  Therefore, $f([a,b])$ is a closed interval, which means it has both a minimum and, as desired, a maximum.  What I would like, however, is a proof that doesn't require such general or sophisticated framework.  In particular, I'd like to know if there's a proof that is understandable to somebody beginning calculus, one that (at the very least) doesn't invoke compactness.  Any comments, hints, or solutions are welcome and apreciated.
103648
I have read that if we let $(X, \tau_x)$ and $(Y, \tau_y)$ be topological spaces, and  $f: X \rightarrow Y$ is constant, then f is continuous. Can anyone please explain why? I assume we must look at the preimage of the open sets in $(Y, \tau_y)$..?  Thanks in advance
39321
Let $f:[0,1]\to \mathbb{R}$ is continuous, such that $f(0)=f(1)$. For each $n\in\mathbb{N}$, prove that $\exists x\in [0,1]$ such that $x+\dfrac{1}{n}\in[0,1]$ and $f\left(x+\dfrac{1}{n}\right)=f(x)$.   My approach: The interval $[0,1]$ is connected, then if $f$ is continuous, we have $f([0,1])\subset\mathbb{R}$ is a interval. And I suppose that, I can define a closed path such that $f(0)=f(1)$. Any hint pls,
101469
Suppose that $f:\mathbb{R} \rightarrow \mathbb{R}$ is continuous on $\mathbb{R}$ and that $f(r)=0$ for all $r \in \mathbb{Q}$. Prove that $f(x)=0$ for all $x \in \mathbb{R}$.   My attempt: Define a sequence $(x_n)$ where $x_n \in \mathbb{Q}$ for all $n \in \mathbb{N}$ and assume that $(x_n) \rightarrow a \not\in \mathbb{Q}$. Since $f$ is continuous, we have $\lim_n{f(x_n)}=f(a)=0$. Since $a$ is arbitrary irrational number, we have $f(a)=0$ for all $a \not\in \mathbb{Q}$. Hence, we proved the statement.  Is my proof valid? or is there any flaw ?
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.
84314
Problem. Let $I,J$ be open intervals, $\,f:I\to \mathbb R$, continuous, $\,\varphi :J\to \mathbb R$,  continuously differentiable, with $\varphi[J]\subset I$, and $\varphi$ satisfying $$ \varphi'(t)=f\big(\varphi(t)\big),  \quad \text{for all $t\in J$}. $$ Show that $\varphi$ is monotonic.  This becomes rather straight-forward if we further assume the $f$ is continuously differentiable, or even locally Lipschitz continuous, as this implies that IVP for $x'=f(x)$ enjoy uniqueness, and hence if $f\big(\varphi(t_0)\big)=0$, for some $t_0\in J$, then $\varphi$ is constant. In fact something stronger holds in such case: $\varphi$ is either strictly monotonic or constant.  This problem is also straight-forward if $f(x)\ne 0$, for all $x\in I$, in which case uniqueness kicks in anew.  The hard part is to show monotonicity of $\varphi$ when $f$ is just continuous, and its values include zero.
43481
PROBLEM. Let $f:[a,b]\to\mathbb R$ be a continuous function, such that $f'(x)\ge 0$, for all $x\in [a,b]\setminus A$, where $A\subset [a,b]$ is a countable set. Show that $f$ is increasing.  Attention. In this problem, we DO NOT assume that $f$ is differentiable in the whole $[a,b]$.  Notes. (1) If we assume that $f$ is differentiable in the whole interval, then we can easily show that $f'(x)\ge 0$, everywhere. For otherwise, if $f'(x_0)=c&lt;0$, for some $x_0\in [a,b]$, then by virtue of , $(c,0)\subset f'([a,b])$, and hence, $f'(x)&lt;0$, for uncountably many $x$'s.   (2) The conclusion of the problem does not hold if we replace the assumption $A$ is countable with $A$ is a set of measure zero. Take for example the , with a negative sign in front.  (3) If the hypothesis $f'(x)\ge 0$, is replaced by $f'(x)=0$, then the conclusion becomes f is constant.
61316
The question is the following: Assume $f: \mathbb{R} \rightarrow \mathbb{R}$ is continuous, and for all $x \neq 0$, $f'(x)$ exists. If $\lim_{x \rightarrow 0} f'(x) = L$ exists, does it follow that $f'(0)$ exists?  My intuition is that this does not have to hold in general. However, I keep finding counter examples such that $f'(0)$ exists where $\lim_{x \rightarrow 0} f'(x) = L$ does not exist, which is obviously not the original question.
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..
53104
Let $X$ be a (Lebesgue-)measurable set of $\mathbb{R}^n$ and $f:X \to \mathbb{R}^n$ continuous function that preserves measurability ($A$ meausurable $\implies f(A)$ measurable).  Prove: for all $A \subset X$,$\space$$\lambda(A)=0 \implies \lambda(f(A)) = 0$ .  I'm totally stuck. Initially I made some progress but now I'm at a point where it feels like the statement shouldn't be true at all.  Additionally i'd like to know if the statement works for general topological measure spaces (or even just metric measure spaces). That is:   Let $X$ be a (Lebesgue-)measurable set of $Y$, Topological (or metric) measure space and $f:X \to Y$ continuous function that preserves measurability etc...
54565
If $f$ is an increasing function defined on $[a,b]$ . Then show that the set of discontinuities of $f$ is countable and that $f$ has points of continuity in every open subinterval of $[a,b]$  Attempt: I have trouble understanding why the number of discontinuities of $f$ should be countable. For example, if we define a function which is increasing but different at each point in the interval $[0,1]$, then, there can be uncountable number of discontinuities.  I think if we prove that the number of discontinuities is countable, then we can prove that the function has points of continuity in every open interval of $[a,b]$.  Please help me understand intuitively why $f$ must necessarily have countable number of discontinuities. Suppose, $f$ has an uncountable number of discontinuities, then what can happen?  Thank you for your help.
33427
Using the definition of continuity and the $\epsilon$-$\delta$ definition of a limit, how would you proceed to show that the following statement holds true?  Let $f : \Bbb{R} \rightarrow \Bbb{R}$ be a continuos function at $x=0$, with the property that if $x\neq 0$ then $f(x)\geq 0$. Then $f(0) \geq 0$.  My attempt so far has been to assume that $f(0) \lt 0$, which means that $$f(0)=\lim_{x\to 0} f(x)\lt 0,$$ because of the continuity at the given point. Furthermore, I think this means that when $x$ approaches $0$ from both sides, at some point the value $f(x)\lt 0$, thus forming a contradiction.   I can't seem to see what to do afterwards. Is this a way of approaching the problem? And how will the $\epsilon-\delta$ definition come into play if my approach is usable?