Definition: Here a neighborhood of the point c → ∈ R n is the set N r ( c → ) = { x → ∈ R n : ‖ x → − c → ‖ 2 &lt; r } for some r &gt; 0, and a subset U ⊂ R n is said to be convex if x → , y → ∈ U , t ∈ [ 0 , 1 ] implies t y → + ( 1 − t ) x → ∈ U.My Attempt: Suppose x → , y → ∈ N r ( c → ) and t ∈ [ 0 , 1 ], we then have ‖ x → − c → ‖ 2 &lt; r and ‖ y → − c → ‖ 2 &lt; r. Now, ‖ t y → + ( 1 − t ) x → − c → ‖ 2 = ‖ t y → − t x → + x → − c → ‖ 2 ≤ ‖ t y → − t x → ‖ 2 + ‖ x → − c → ‖ 2 &lt; | t | ‖ y → − x → ‖ 2 + r ≤ | t | ( ‖ y → − c → ‖ 2 + ‖ c → − x → ‖ 2 ) + r &lt; | t | ( r + r ) + r = ( 1 + 2 | t | ) rScreeching halt.My Question: How should I make ‖ t y → + ( 1 − t ) x → − c → ‖ 2 less than r? My intuition is that I need the Cauchy-Schwarz inequality; but I am unsure where to apply it. Any hint would be greatly appreciated.

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Definition: Here a neighborhood of the point             c      →        ∈            R        n   is the set       N          r        (            c      →        )  =  {            x      →        ∈            R        n    :  ‖            x      →        −            c      →            ‖    2    &amp;lt;  r  } for some   r  &amp;gt;  0, and a subset   U  ⊂            R        n   is said to be convex if             x      →        ,            y      →        ∈  U  ,  t  ∈  [  0  ,  1  ] implies   t            y      →        +  (  1  −  t  )            x      →        ∈  U.My Attempt: Suppose             x      →        ,            y      →        ∈      N          r        (            c      →        ) and   t  ∈  [  0  ,  1  ], we then have   ‖            x      →        −            c      →            ‖    2    &amp;lt;  r and   ‖            y      →        −            c      →            ‖    2    &amp;lt;  r. Now,  ‖  t            y      →        +  (  1  −  t  )            x      →        −            c      →            ‖    2    =  ‖  t            y      →        −  t            x      →        +            x      →        −            c      →            ‖    2    ≤  ‖  t            y      →        −  t            x      →            ‖    2    +  ‖            x      →        −            c      →            ‖    2    &amp;lt;      |    t      |    ‖            y      →        −            x      →            ‖    2    +  r  ≤      |    t      |    (  ‖            y      →        −            c      →            ‖    2    +  ‖            c      →        −            x      →            ‖    2    )  +  r  &amp;lt;      |    t      |    (  r  +  r  )  +  r  =  (  1  +  2      |    t      |    )  rScreeching halt.My Question: How should I make   ‖  t            y      →        +  (  1  −  t  )            x      →        −            c      →            ‖    2   less than   r? My intuition is that I need the Cauchy-Schwarz inequality; but I am unsure where to apply it. Any hint would be greatly appreciated.

Miles Mueller · Accepted Answer

‖  t            y      →        +  (  1  −  t  )            x      →        −            c      →            ‖    2    =  ‖  t            y      →        +  (  1  −  t  )            x      →        −  t            c      →        −  (  1  −  t  )            c      →            ‖    2    =  ‖  t  (            y      →        −            c      →        )  +  (  1  −  t  )  (            x      →        −            c      →        )      ‖    2      ≤  ‖  t  (            y      →        −            c      →        )      ‖    2    +  ‖  (  1  −  t  )  (            x      →        −            c      →        )      ‖    2    =  t  ‖            y      →        −            c      →            ‖    2    +  (  1  −  t  )  ‖            x      →        −            c      →            ‖    2    &amp;lt;  t  r  +  (  1  −  t  )  r  =  r  .Concerning   &amp;lt; note that not both   t and   1  −  t can be zero.

Definition: Here a neighborhood of the point <mrow class="MJX-TeXAtom-ORD"> <mover>

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