I would like to go back a bit and look at:
If \quad {\displaystyle G(x) = \int_a^{\psi(x)} f(t) dt } \quad then \quad G\,' (x) = \psi\,' (x) f(\psi(x)).
My attempt at the proof, which again mimic the proof of the basic case:
Proof
\begin{align*}
G(x+h) - G(x)
&=...
Thank you for the hint. This is my attempt at the proof which I mimic from that of the single variable case. Please check the rigour in the proof.
Let
\begin{equation*}
G(x) = \int_a^x F(x,t) dt.
\end{equation*}
Now
\begin{align*}
G(x+h) - G(x)
&= \int_a^{x+h} F(x+h,t) dt -...
Suppose a is a constant.
If G(x) = \int_a^x \Big [ f(t) \int_t^x g(u) du \Big ] dt, what is G\,'(x)?
My attempt,
G\,'(x) = f(x) \int_x^x g(u) du = 0,
and I am sure this is wrong.
Thank you for the answer, jgens. My feeling is that the concept can be found in topology. However the scarcity of examples in the literature obviously does not help to reinforce one's understanding.
Roughly speaking, a function is continuous if its graph can be drawn without lifting the...
Let f: X \rightarrow \mathbb{R}. The definition given for upper semicontinuity at a point a is for any positive \epsilon there is a positive \delta such that if |x - a| < \delta then f(x) < f(a) + \epsilon.
My understanding is, for ordinary continuity at the point a then the inequality...
Could you please check the statement of the theorem and the proof? If the proof is more or less correct, can it be improved?
Theorem
Let be a topological space and be the discrete space.
The space is connected if and only if for any continuous functions , the function is not onto...