So, it is stated that if $C_1g(t)+C_2f(t)=0$ then $g(t)=\sin{t}$ and $f(t)=\cos{t}$ are linearly dependent, i.e., $C_1g(t)+C_2f(t)=0 \iff C_1=C_2=0$. It is proved simply by assuming two values of $t$, first $t=0$ and then $t=\frac{\pi}{2}$. In the first case $C_2=0$ and in the second case $C_1$ is $0$.  I don't get how is that a prove, if you can find other values of $t$, where you can express one function by the the other. Like if $t=\frac{\pi}{4}$, then if $C_1=1$ and $C_2=-1$ $\Rightarrow C_1g(t)+C_2f(t)=0$. What am I missing here?
46590
In the vector space of $f:\mathbb R \to \mathbb R$, how do I prove that functions $\sin(x)$ and $\cos(x)$ are linearly independent. By def., two elements of a vector space are linearly independent if $0 = a\cos(x) + b\sin(x)$ implies that $a=b=0$, but how can I formalize that? Giving $x$ different values? Thanks in advance.
85617
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.
83145
\begin{equation} g(f,t) = \begin{cases}          \bigl{b^Nt^N^-^1e^2^\pi^b^t cos(2\pi ft),  \text{t\geq 0;}\\          0 \text{ $ otherwise $}          \end{cases} \end{equation}
1567
On , one can see that   $$\|fg\|_r\leq \|f\|_p\|g\|_q,$$ when $$\frac{1}{r} = \frac{1}{p} + \frac{1}{q}, \ p,q\in(0,\infty].$$  I have two questions,   Is the statement true? If so, is it stated in some references? I need it for my thesis.
63378
$T(av_1 + bv_2) = aT(v_1) + bT(v_2)$  Why is this called linear? $f(x) =ax + b$, the simplest linear equation does not satisfy $f(x_1 + x_2) = f(x_1) + f(x_2)$.  Thank you.
107437
I'm trying to write a proof for this, but I don't know how to begin. Any help appreciated.  Suppose $g$ is a function  such that $g'(x) = \frac{1}{x}$ for all $x &gt; 0$ and $g(1) = 0$. Prove that $g(xy) = g(x) + g(y)$ for all $x, y &gt; 0$.    Let $h(x) = g(x,y)$
88733
Let $\psi: C([0,1]; \mathbb{R}) \mapsto C([0,1]; \mathbb{R})  $, be such that if $x \in [0,1]$ then $$\psi(f)(x) = \int\limits_{0}^{x} f(t) dt $$  I am asked to show that $\psi$ is not a contraction but $\psi^{2}$ is. And then conclude that $\psi$ has a unique fixed point.   I´ve managed to prove a statment that helps me conclude the last part but I´m having trouble proving that the function is not a contraction. This is what I've got.   If $f,g \in  C([0,1]; \mathbb{R})$  then $$ |\psi (f) - \psi(g)| \leq \int\limits_{0}^{x} |f(t)-g(t)| dt \leq \sup\limits_{x\in [0,1]} |f(x)-g(x)| \int\limits_{0}^{x} 1 dt = d(f,g) x  $$ Then $|\psi (f) - \psi(g)| \leq d(f,g) x $,  and taking supremum on the left hand side allows me to conclude that $d(\psi (f),\psi (g)) \leq d(f,g) x$  Is there any problem with this procedure? Does anybody has an insight on how to procede with the $\psi^{2}$ part?   Many Thanks!
90986
I've been given the following problem as homework:     Prove that there exists a linear transformation $T: \mathbb{R}^2 \to \mathbb{R}^3$ such that $T(1,1) = (1,0,2)$ and $T(2,3) = (1,-1,4)$.   Since it just says prove that one exists, I'm guessing I'm not supposed to actually identify the transformation.    One thing I tried is showing that it holds under addition/multiplication in the sense of:  1) $T(x + y) = T(x) + T(y)$  2) $T(cx) = cT(x)$  3) $T((1,1) + (2,3)) = T(1,1) + T(2,3)$? But that's not necessarily true.  I don't know how I'd accomplish this given my limited knowledge.  What's the approach to solving a problem like this?
15120
How do I show that if there are functions $f,g$ such that$$ f(g(x)+g(y))=bx+cy $$holds for all $b,c\in\mathbb{R}$, then we necessarily have $b=c$? Is this even true? It seems so, but I'm just not sure how to explain it in detail and using theory from functional equations. Any ideas?  Extra: In addition, I'd like to prove that there are no pair of functions $f,g$ such that$$ y^2+z^2=f(x,g(y-x)+g(z-x)). $$ Any ideas on how to prove this? Plus, does anyone know some literature regarding this type of problems?
93108
If $f(x) \in C(-\infty+\infty)$, $\;g(x)=f(x) \int_0^x f(t)\,dt\,$ and $\;g(x)$ is monotone-decreasing in $(-\infty,+\infty),$  Prove:$f(x)\equiv 0$.  It is easy to get $g(0)=0$,and I'm thinking about proving $g(x)\equiv 0$,then $f(x)\equiv 0$ follows.  But I havn't any ideas to prove $g(x)\equiv 0$.
55527
If $\alpha_1,\alpha_2,\alpha_3$ are linearly independent, and $(\beta_1,\beta_2,\beta_3) = (\alpha_1,\alpha_2,\alpha_3)C$. Prove that if $\mid C\mid \neq 0$, then $\beta_1,\beta_2,\beta_3$ are linearly independent.   This statement seems to be trivial, if $\alpha_1,\alpha_2,\alpha_3$ are linearly independent, then $\det{(\alpha_1,\alpha_2,\alpha_3)} \neq0$, and $\det{(\alpha_1,\alpha_2,\alpha_3) C} = \det{(\alpha_1,\alpha_2,\alpha_3)}*\det{}C\neq 0$, which implies $\det(\beta_1,\beta_2,\beta_3) \neq0$, and hence the columns are linearly independent.  I'm not sure how to prove it rigorously, or if I'm missing any steps here. Would really appreciate any help! Thank you!
70274
I know that $\frac{dy'(x)}{dy}=0$ (where $y'=\frac{dy(x)}{dx}$). The reason explained is that $y'$  does not depend explicitly on $y$. But intuitively, $y'$ depends on $y$, since if you vary $y$ you will modify $y'$. Why is my reasoning wrong (my reasoning sounds like it's related to functional calculus, instead of standard calculus)?  I tried to write  $\frac{dy'(x)}{dy}=\frac{d}{dy}\frac{dy}{dx}=\frac{d}{dx}\frac{dy}{dy}=\frac{d}{dx}1=0$, but this proof doesn't convince me.  I think that other way to see this is saying that if a function is of the form $f(y)$, it will not dependend on the variable $y'$.  But the same way you write $f(y)=y^2$, you could write $f(y)=\frac{d}{dx}y$, which clearly depends on $y'$. So I don't know if there are some types of operations which are restricted (for example, taking limits):  Note: The problem raises in the context of Classical Mechanics, where: $\frac{\partial L(\dot{x},t)}{\partial x}=0$.
60255
My brother brought me this same question, also on this .   $$\int_{0}^{x}f(t)dt = x+\int_{x}^{1}tf(t)dt \tag{1}$$  When I try to solve it, I also did same thing. Differentiation leads to  $$f(x) = 1-xf(x)\\  f(x) = \frac{1}{1+x} \\ \int_{0}^{1}f(x) dx = \ln 2 $$  But try putting $x = 1$ in original equation $(1)$,  $$\int_{0}^{1} f(t) dt = 1$$  so what has happened? We get two contradictory results.  Is it still true that $f(1) = 1/2$? Where is the mistake?
50379
Question :  Let $$f\ :\ [a,b]\rightarrow \mathbb R$$   be  continuous . And  there  exist  constants  $\alpha$  $\beta$  such  that  $\forall c\in [a,b]$  $$\alpha \int_a^c f(x)dx +\beta \int_c^b f(x)dx =0 $$   Prove  that  $f=0$  allover  $[a,b]$.  Attempt :(not much useful )   Since  the  equation  holds  for  all  $c\in [a,b]$   , if  take  $c=a$  then  we  have $$\beta \int_a^b f(x)dx =0 $$  And  taking  $c=b$  we  get $$\alpha \int_a^b f(x)dx=0 $$  See  if  both  $\alpha=0$  and  $\beta=0$  hold  then  there  is  nothing  to  do . If  only  one  of  them  is  $0$   and  the  other  is  non-zero  then  from  the  last  two  equations  we  get $$\int_a^b f(x)dx =0 $$ and  this  does   not  ensure   $f$'s  being  identically  $0$  on  $[a,b]$ (for  example : $f(x)=sin x$  and $a=0$,$b=2\pi $ ) .   So  our  only  possibility  is  that  both $\alpha \neq 0$  and $\beta \neq 0$  must  hold .  Now  I  cannot  proceed  any  further  from  this  point . Please  give  me  some   hints  as  to  what  to  do  next .  Thaanks.
121323
If we have:  $a&gt;0,a+b+c \ge 0,a-b+c \ge 0,a-c \ge 0,b^2-4ac \ge 0$  Then Prove that there exists two roots in $[-1,1]$.  We have $f(1),f(-1) \ge 0$ so if we could prove there is a number $c$ that $f(c)&lt;0$ and $-1&lt;c&lt;1$ we are done but I can't do that.  Edit:$f(x)=ax^2+bx+c$  Edit 2 I made a typo and corrected it now.Now using answer of @George Law It only remains to check the case $\Delta =0$ which is likely to cause problem with $a-c \ge o$ because we didn't use it.
40532
$f(x)$ is a differentiable function and $g(x)$ is a double differentiable function such that $|f(x)|\leqslant 1$ and $f'(x)=g(x)$. If $$f(0)^2+g(0)^2=9$$ then prove that there exists some $c\in(-3,3)$ such that$ \ \ g(c) \cdot g''(c)&lt;0$.  Attempt: Let us define a function $h(x) = g(x) g~'(x)$. Then $$h'(x) =  g(x)g''(x) + \left( g'(x) \right)^2 \tag 1$$ If we prove that for some $c \in (-3,3), h~'(c) &lt; 0,$ then $$g(c)g''(c) &lt;0 \tag 2$$ Also, $$\left|f(0)\right| &lt; 1 \implies f'(0) \in (-3,-2\sqrt 2 ) \cup (2\sqrt 2,3) $$ Could someone please advise me how do I move forward from here. Thank you very much for your help in this regard.
48398
Consider $f(x) = mx+b$. Let $b\ne 0$   If $f$ is linear, $f(0)$ should yield $0$ $f(0) = m(0)+b = b$ Therefore $f(x)=mx+b$ is nonlinear.    Question:  Why is $y=mx+b$ called a ""?     Remark:  Consider $f(x,y) = y + -mx = b$  \begin{align} \alpha f(x,y) &amp;= \alpha b \\ f(\alpha x,\alpha y) &amp;= \alpha y +-\alpha mx = \alpha (y +-mx) =\alpha b \end{align}  \begin{align} f(x+\beta,y+\beta) &amp;= y + \beta+ - m (x +\beta) \\&amp;= (y+-mx)+(\beta +-m\beta)\\ &amp;= f(x,y)+ f(\beta,\beta)  \end{align}  Therefore, $f(x,y) = y +-mx$ and $f(x,y) =b$ are linear.   Question:  Why is this not a contradiction?