I need help with proving / disproving something:     Look at the map      $$\Phi: (C([0,1], \mathbb R), ||\cdot||_{\infty}) \to (\mathbb R, |\cdot|); \,\,\,\,\,\,\Phi(u) := \int_0^1 u²(t) dt$$      a) Is it continuous?   b) Is uniformly continuus?   c) Can one find a non-empty subset of $C([0,1],\mathbb R)$ in which $\Phi$ is Lipschitz-continuous?   Can you help me with this?
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$\phi: (C([0,1];\Bbb R ),||\cdot||_\infty )\to (\Bbb R, |\cdot  | ); \: \: \: \: \: \:  \phi(u):=\int_0^1 u^2(t) dt $  Is this function continuous or even uniformly continuous? (I know that the function $g: M\to \Bbb R , g(x) := x^2$ is continuous but not uniformly)  Also, is there a non-empty subset of $ C([0,1];\Bbb R )$ so that $\phi$ is Lipschitz-continuous on that set?  Thanks in advance!
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Let $(X,d)$ be a metric space and $ a \in X $ Show that $f(x) = d(x,a)$ is a Lipschitz function from X to $\Re$ Use this fact to show $S$ is an open subset of $X$: Let $(X,d)$ be a metric space and $a \in X, r &gt; 0.$ Let $S$={$x \in X : d(a,x) &gt; r$}  &gt;Lipschitz Defn:  Let $f$ be a function from $(X,d)$ to $(Y,\rho)$ $f$ is said to be Lipschitz if: $\exists$ K $\geq$ 0 s.t. $\rho( f(x_1) , f(x_2) ) \leq$ Kd($x_1$,$x_2$) $\forall$ $x_1 x_2 \in X$  &gt;Continuity Defn:  Let $(X,d)$ and $(Y,\rho)$ be metric spaces, and $f: X \to Y$ Then $f$ is continuous $iff$ f$^{-1}$($B$) = { $x \in X : f(x) \in B $} is an open subset of $X$, whenever $B$ is an open subset of $Y$  &gt;Triangle Inequality for a Metric:  $d(x,z) \leq d(x,y) + d(y,z)$   My first question is, am I thinking about this correctly? If $f(x)$ is a Lipschitz function, it is uniformly continuous If $f(x)$ is uniformly continuous, it is continuous at every point If $f(x)$ is continuous at every point $\Rightarrow$ the inverse images of open sets are open Hence $S$ is an open subset of $X$.  Secondly, how would you formally construct this? We know $f(x) = d(x,a)$ s.t. $f:X \to \Re$ By definition of Lipschitz we have $\exists  K \geq 0$  s.t. $\rho(f(x_1),f(x_2)) \leq Kd(x_1,x_2) \forall x_1,x_2 \in X$ &amp; since our codomain is $\Re$ $\rho(f(x_1),f(x_2)) = | d(x_1,a) - d(x_2,a) | \leq Kd(x_1,x_2)$ Taking $K=1$ $\ldots$ Help in finishing this off would be appreciated. Thanks.
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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.
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Given a metric space $(X,d)$, how to prove that the function $d \colon X \times X \to \mathbf{R}$ is continuous?   If we take any two arbitrary real numbers $a$ and $b$ such that $a &lt; b$, then we need to show that the set $d^{-1} (a,b)$ given by   $$ d^{-1} (a,b) :=  \{ (x, y) \in X \times X | a &lt; d(x,y) &lt; b \} $$  is open in the product topology on $X \times X$.   A basis for this product topology is the collection of all cartesian products of open balls in $(X,d)$.
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Why does a Lipschitz function $f:\mathbb{R}^d\to\mathbb{R}^d$ map measure zero sets to measure zero sets?   It is easy to prove this statement if the domain is bounded. Is there any way to extend the argument to unbounded domains? Can anyone give me some hint, or show me a simple proof sketch?
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I'm stuck on a question which asks to prove that any continuous integer-valued function of a real variable is constant. In my lecture notes we are told that a map is continuous if the preimage of any open set is open.  How do I use this to answer the question? A full answer would be greatly appreciated as I can't seem to figure out how to solve it in my  notes, in analysis books or anywhere online.
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Prove that, there exists no continuous function $f:\mathbb R\rightarrow\mathbb R$ with $f=\chi_{[0,1]}$ almost everywhere.$\textbf(Make\ sure\ that\ your\ proof\ is\ completely\ rigorous)$.   I don't know, which property to use. (It is not allowed to show it with $\epsilon-\delta-criterion$, our last topics were: Lp-Spaces, Radon-Nikodym Theorem, Riesz Representation Theorem,Lipschitz-Functions, Product measures, Fubini Theorem)  but i can't find anything to do with continuity, which of them could be useful ?
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Let $\phi : E \to E'$ be a bi-Lipschitz map between metric spaces. Show that $E$ is complete if and only if $E'$ is complete.   I'm unsure of how to construct this proof. I know from a previous exercise that if $(x_n)$ is a Cauchy sequence in $E$, then $(\phi(x_n))$ is a Cauchy sequence in $E'$, if required.   Many thanks!
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The following is a problem from Berkeley's Grad Prelims:  Let  $f:\mathbb R^2 \to \mathbb R$ satisfy   (i) Given any $x_0,y_0 \in \mathbb R$, $y \mapsto f(x_0,y)$ and $x \mapsto f(x,y_0)$ are continuous.  (ii) For each compact $K$, $f(K)$ is compact.    Prove $f$ is continuous.  Attempt:  Let $(x_n,y_n) \to (x,y)$. We wish to show $f((x_n,y_n)) \to f((x,y))$.  Sequential compactness gives us a subsequence $$f((x_{n_k},y_{n_k})) \to f((\tilde x, \tilde y)) \in f(K).$$  I'm un sure how to proceed to show (i) $f((\tilde x, \tilde y)) = f((x,y))$, and (ii) how the existence of such a subsequence and the slice continuity give convergence of the entire sequence.   Originally I had hoped that the slice continuity would make $\{f((x_n,y_n))\}$ cauchy, but that doesn't seem to be the case.    I'm looking for a small hint as to how to proceed. Please, no answers.
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(UGC CSIR-2015, DECEMEMBER, MATHEMATICAL SCIENCES)        $f$ maps a bounded sequence to a bounded sequence.   $f$ maps a Cauchy sequence to a Cauchy sequence.   $f$ maps a convergent sequence to a convergent sequence.   $f$ is uniformly continuous.      I choose all of the options as possible answers because the condition $\sup_{x\in \Bbb R}|f'(x)| \lt \infty$ forces $f$ to be uniformly continuous.(Because $f$ becomes Lipschitz and Lipschitz condition implies uniform continuity)  i.e. $\frac {|f(x)-f(y)|}{|x-y|} \le \sup_{x\in \Bbb R}|f'(x)|$ $ \forall x,y$.  Hence all other options are bound to be true.  Am I correct?
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$f : S^1 \rightarrow  \mathbb{R}$ is a continuous map. Define $$A = \{(x, y) \in S^1 \times S^1: x \neq y, f(x) = f(y)\}.$$   We want to prove that $A$ has uncountably many points.    It seems very evident, but I want a rigorous argument. Thanks in advance.
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It is a Dugundji's exercise:  Show that if $X$ is $T_2$, $f \colon (X, \tau) \to (Y, \sigma)$ continuous in $X$ and $f|_{D}$ homeomorphism with $D$ dense in $X$ then $f(X - D) \cap f(D) = \varnothing$.  If I suppose that there is $y \in f(X-D) \cap f(D)$ then exists $x_1 \in D$ and $x_2 \in X-D$ s.t $f(x_1) = y$ and $f(x_2) = y$.  For every nbd $V \in \tau$ of $x_2$ $V \cap D \mapsto U \cap f(D)$ with $U \in \sigma$ because $f|_{D}$ maps open sets in $D$ to open sets in $f(D)$.   Because $f$ is continuos and $y \in U$ then I can find a nbd of $x_2$: $V_1$ s.t. $f(V_1) \subset U$.
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Show that if $f: \mathbb{R} \to \mathbb{R}$ is a continuous injective map, then it is strictly monotonic.    Could someone give me a proof for this? I have the intuition for why it's true - I'm just having trouble expressing that intuition in a rigorous manner. Basically consider two points $x_1, x_2 \in \mathbb{R}$. By the problem statement, $f$ is continuous on $[x_1, \, x_2]$. WLOG, assume that $f$ is strictly increasing. It there exists a point where it is not increasing, then $f$ hits a value twice, and it's not injective.
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Please help me to prove the following;  Let $ \ f:[0,1]^2\longrightarrow\mathbb{R}$ be such that: (i) $\ f(x,\cdot)$ is measurable for each fixed $x\in[0,1]$; (ii) $\ f(\cdot,y)$ is continuous for each fixed $y\in[0,1]$. Show that $\ f$ is measurable.
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Is $C^\infty(\mathbb{R})\subset C_b(\mathbb{R})$ dense? I.e. is any continuous bounded function $f:\mathbb{R}\to\mathbb{R}$ the uniform limit of smooth functions?   On any bounded interval this is true since by Stone-Weierstrass polynomials are smooth and dense in the continuous functions. I guess the result is still true for $\mathbb{R}$ but I don't know how to prove it. Since I don't have any integrability conditions on $f$ the technique of convoluting with mollifiers is probably not applicable.
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Suppose that $f:\Bbb R\to \Bbb R$ is continuous and that $f(x)\to 0$ as $x\to \pm\infty$. Prove that $f$ is uniformly continuous.   Since $f(x)\to 0$ as $x\to \pm\infty$ we have $|f(x)|&lt;\epsilon $ whenever $x&lt;-K$ and $x&gt;K$ where $K&gt;0$.  Also $f$ is uniformly continuous on $[-K,K]$ as the domain is compact.  Hence $f$ is uniformly continuous in $(-\infty,-K)\cup [-K,K]\cup (K,\infty)$.  But how to conclude that $f$ is uniformly continuous on $\Bbb R$ from above as I may chose $x\in (-\infty,-K);y\in [-K,K]$ ,how to conclude that $|f(x)-f(y)|&lt;\epsilon $ from above for a given $\epsilon$?  Please suggest the edits required.
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Given $f_{n} : [a,b]\rightarrow \mathbb R$ is uniformly Lipschitz and $f_{n}\rightarrow f$, how can I show that $f$ is Lipschitz?  $f_{n}$ is uniformly Lipschitz $\leftrightarrow \exists K &gt; 0$ such that for all $n$ and for every $x,y\in [a,b]$ we have $|f_{n}(x)-f_{n}(y)| &lt; K|x-y|$.  $f_{n}\rightarrow f \leftrightarrow \forall~\epsilon&gt;0~~\exists~N$ such that $\forall~n&gt;N$ we have $\mid f_{n}(x) - f(x)\mid&lt;\epsilon~~~(=\mid x-y\mid?)$  Then we have  \begin{align} |f(x)-f(y)| &amp;\leq |f(x)- f_{n}(x)| + |f_{n}(y)-f(y)| + |f_{n}(x)-f_{n}(y)| \\ &amp;\leq |x-y| + |x-y| + K|x-y| \\ &amp;\leq (K + 2) |x - y| \end{align}  which shows that $f$ is Lipschitz. Can anybody tell me if this makes sense? Sorry if layout is poor.